Ошибка синтаксиса примерное положение integer - TopOshibok.ru - решение и исправление самых разных ошибок

I am very new to postgres. I got this error when try to run the following script:

CREATE OR REPLACE FUNCTION xyz(text) RETURNS INTEGER AS
'DECLARE result int;
BEGIN
    SELECT count(*) into result from tbldealercommissions
    WHERE 
    txtdealercode = $1;

    if result < 1 then returns 1; 
    else returns 2 ;
    end if;
END;
    '
LANGUAGE sql VOLATILE;

The error is

ERROR:  syntax error at or near "int"
LINE 3: 'DECLARE result int;

not sure what cause this error. Any help is appreciated.

asked Oct 6, 2011 at 12:07

This is unsuitable:

LANGUAGE sql

use this instead:

LANGUAGE plpgsql

The syntax you are trying to use is not pure SQL language but the procedural PL/pgSQL language. In PostgreSQL you can install different languages and PL/pgSQL is only primus inter pares in that regard. This also means that you might get the error message, that this language is not installed. In that case use

CREATE LANGUAGE plpgsql;

which actives it. Depending on the version of PostgreSQL you might need superuser rights to do this step.

Have fun.

answered Oct 6, 2011 at 12:19

A.H.A.H.

64k15 gold badges93 silver badges126 bronze badges

Not only are you using the wrong language (as noted by A.H.) but there is returns keyword, you want return. You might want to use a different delimiter to avoid running into problems with string literals in your functions, $$ is pretty common. I think your function should look more like this:

CREATE OR REPLACE FUNCTION xyz(text) RETURNS INTEGER AS $$
DECLARE result int;
BEGIN
    select count(*) into result
    from tbldealercommissions
    where txtdealercode = $1;

    if result < 1 then return 1; 
    else return 2;
    end if;
END;
$$ LANGUAGE plpgsql VOLATILE;

answered Oct 6, 2011 at 17:37

mu is too shortmu is too short

427k70 gold badges834 silver badges801 bronze badges

Источник

I am very new to postgres. I got this error when try to run the following script:

CREATE OR REPLACE FUNCTION xyz(text) RETURNS INTEGER AS
'DECLARE result int;
BEGIN
    SELECT count(*) into result from tbldealercommissions
    WHERE 
    txtdealercode = $1;

    if result < 1 then returns 1; 
    else returns 2 ;
    end if;
END;
    '
LANGUAGE sql VOLATILE;

The error is

ERROR:  syntax error at or near "int"
LINE 3: 'DECLARE result int;

not sure what cause this error. Any help is appreciated.

asked Oct 6, 2011 at 12:07

This is unsuitable:

LANGUAGE sql

use this instead:

LANGUAGE plpgsql

CREATE LANGUAGE plpgsql;

which actives it. Depending on the version of PostgreSQL you might need superuser rights to do this step.

Have fun.

answered Oct 6, 2011 at 12:19

A.H.A.H.

62.8k14 gold badges91 silver badges122 bronze badges

CREATE OR REPLACE FUNCTION xyz(text) RETURNS INTEGER AS $$
DECLARE result int;
BEGIN
    select count(*) into result
    from tbldealercommissions
    where txtdealercode = $1;

    if result < 1 then return 1; 
    else return 2;
    end if;
END;
$$ LANGUAGE plpgsql VOLATILE;

answered Oct 6, 2011 at 17:37

mu is too shortmu is too short

421k69 gold badges826 silver badges789 bronze badges

Syntax errors are quite common while coding.

But, things go for a toss when it results in website errors.

PostgreSQL error 42601 also occurs due to syntax errors in the database queries.

At Bobcares, we often get requests from PostgreSQL users to fix errors as part of our Server Management Services.

Today, let’s check PostgreSQL error in detail and see how our Support Engineers fix it for the customers.

What causes error 42601 in PostgreSQL?

PostgreSQL is an advanced database engine. It is popular for its extensive features and ability to handle complex database situations.

Applications like Instagram, Facebook, Apple, etc rely on the PostgreSQL database.

But what causes error 42601?

PostgreSQL error codes consist of five characters. The first two characters denote the class of errors. And the remaining three characters indicate a specific condition within that class.

Here, 42 in 42601 represent the class “Syntax Error or Access Rule Violation“.

In short, this error mainly occurs due to the syntax errors in the queries executed. A typical error shows up as:

Here, the syntax error has occurred in position 119 near the value “parents” in the query.

How we fix the error?

Now let’s see how our PostgreSQL engineers resolve this error efficiently.

Recently, one of our customers contacted us with this error. He tried to execute the following code,

CREATE OR REPLACE FUNCTION prc_tst_bulk(sql text)
RETURNS TABLE (name text, rowcount integer) AS
$$
BEGIN
WITH m_ty_person AS (return query execute sql)
select name, count(*) from m_ty_person where name like '%a%' group by name
union
select name, count(*) from m_ty_person where gender = 1 group by name;
END
$$ LANGUAGE plpgsql;

But, this ended up in PostgreSQL error 42601. And he got the following error message,

ERROR: syntax error at or near "return"
LINE 5: WITH m_ty_person AS (return query execute sql)

Our PostgreSQL Engineers checked the issue and found out the syntax error. The statement in Line 5 was a mix of plain and dynamic SQL. In general, the PostgreSQL query should be either fully dynamic or plain. Therefore, we changed the code as,

RETURN QUERY EXECUTE '
WITH m_ty_person AS (' || sql || $x$)
SELECT name, count(*)::int FROM m_ty_person WHERE name LIKE '%a%' GROUP BY name
UNION
SELECT name, count(*)::int FROM m_ty_person WHERE gender = 1 GROUP BY name$x$;

This resolved the error 42601, and the code worked fine.

[Need more assistance to solve PostgreSQL error 42601?- We’ll help you.]

Conclusion

In short, PostgreSQL error 42601 occurs due to the syntax errors in the code. Today, in this write-up, we have discussed how our Support Engineers fixed this error for our customers.

PREVENT YOUR SERVER FROM CRASHING!

Never again lose customers to poor server speed! Let us help you.

Our server experts will monitor & maintain your server 24/7 so that it remains lightning fast and secure.

GET STARTED

var google_conversion_label = «owonCMyG5nEQ0aD71QM»;

Здравствуйте! Довольно распространенный вопрос, находил решения, но все же компилятор выводит ошибку. Как-то неправильно задаю первичный ключ.

Использую Postgresql

Java

@Entity
@Table(name="users")
public class Users{
    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(columnDefinition="serial")
    private Long id;
    @Column(name="login")
    private String login;
    @Column(name="password")
    private String password;
}

SQL

CREATE TABLE IF NOT EXISTS users (
  id  SERIAL PRIMARY KEY NOT NULL,
  login VARCHAR(20) UNIQUE NOT NULL,
  password VARCHAR(20) NOT NULL
);

Ошибка:
org.hibernate.tool.schema.spi.CommandAcceptanceExc eption: Error executing DDL via JDBC Statement
Caused by: org.postgresql.util.PSQLException: ОШИБКА: ошибка синтаксиса (примерное положение: «auto_increment»)

Добавлено через 27 минут
SQL-запрос пишу в консоли postgresql. Таблица без записей.
Также в проекте имеется обычный конфигурационный файл для JPA. Думаю, что роли он здесь не играет

__________________
Помощь в написании контрольных, курсовых и дипломных работ, диссертаций здесь

Пытаюсь создать табличку, вот такую

CREATE TABLE screens_items (
	screenitemid             bigint                                    NOT NULL,
	screenid                 bigint                                    NOT NULL,
	resourcetype             integer         DEFAULT '0'               NOT NULL,
	resourceid               bigint          DEFAULT '0'               NOT NULL,
	width                    integer         DEFAULT '320'             NOT NULL,
	height                   integer         DEFAULT '200'             NOT NULL,
	x                        integer         DEFAULT '0'               NOT NULL,
	y                        integer         DEFAULT '0'               NOT NULL,
	colspan                  integer         DEFAULT '0'               NOT NULL,
	rowspan                  integer         DEFAULT '0'               NOT NULL,
	elements                 integer         DEFAULT '25'              NOT NULL,
	valign                   integer         DEFAULT '0'               NOT NULL,
	halign                   integer         DEFAULT '0'               NOT NULL,
	style                    integer         DEFAULT '0'               NOT NULL,
	url                      varchar(255)    DEFAULT ''                NOT NULL,
	dynamic                  integer         DEFAULT '0'               NOT NULL,
	sort_triggers            integer         DEFAULT '0'               NOT NULL,
	application              varchar(255)    DEFAULT ''                NOT NULL,
	PRIMARY KEY (screenitemid)
);

Получаю
Error: ОШИБКА: ошибка синтаксиса (примерное положение: "application")

psql --version
psql (PostgreSQL) 9.4.9

Вроде слово «application» не зарезервировано?

Источник

Содержание

PostgreSQL error 42601- How we fix it
What causes error 42601 in PostgreSQL?
How we fix the error?
Conclusion
PREVENT YOUR SERVER FROM CRASHING!
10 Comments
[Error] 42601: syntax error at or near «SELECT» on pgsql 9.2.5 #813
Comments
PostgreSQL – SQL state: 42601 syntax error
Solution
Related Articles

PostgreSQL error 42601- How we fix it

by Sijin George | Sep 12, 2019

Syntax errors are quite common while coding.

But, things go for a toss when it results in website errors.

PostgreSQL error 42601 also occurs due to syntax errors in the database queries.

At Bobcares, we often get requests from PostgreSQL users to fix errors as part of our Server Management Services.

Today, let’s check PostgreSQL error in detail and see how our Support Engineers fix it for the customers.

What causes error 42601 in PostgreSQL?

PostgreSQL is an advanced database engine. It is popular for its extensive features and ability to handle complex database situations.

Applications like Instagram, Facebook, Apple, etc rely on the PostgreSQL database.

But what causes error 42601?

PostgreSQL error codes consist of five characters. The first two characters denote the class of errors. And the remaining three characters indicate a specific condition within that class.

Here, 42 in 42601 represent the class “Syntax Error or Access Rule Violation“.

In short, this error mainly occurs due to the syntax errors in the queries executed. A typical error shows up as:

Here, the syntax error has occurred in position 119 near the value “parents” in the query.

How we fix the error?

Now let’s see how our PostgreSQL engineers resolve this error efficiently.

Recently, one of our customers contacted us with this error. He tried to execute the following code,

But, this ended up in PostgreSQL error 42601. And he got the following error message,

This resolved the error 42601, and the code worked fine.

[Need more assistance to solve PostgreSQL error 42601?- We’ll help you.]

Conclusion

In short, PostgreSQL error 42601 occurs due to the syntax errors in the code. Today, in this write-up, we have discussed how our Support Engineers fixed this error for our customers.

PREVENT YOUR SERVER FROM CRASHING!

Never again lose customers to poor server speed! Let us help you.

Our server experts will monitor & maintain your server 24/7 so that it remains lightning fast and secure.

SELECT * FROM long_term_prediction_anomaly WHERE + “‘Timestamp’” + ‘”BETWEEN ‘” +
2019-12-05 09:10:00+ ‘”AND’” + 2019-12-06 09:10:00 + “‘;”)

Hello Joe,
Do you still get PostgreSQL errors? If you need help, we’ll be happy to talk to you on chat (click on the icon at right-bottom).

У меня ошибка drop table exists “companiya”;

CREATE TABLE “companiya” (
“compania_id” int4 NOT NULL,
“fio vladelca” text NOT NULL,
“name” text NOT NULL,
“id_operator” int4 NOT NULL,
“id_uslugi” int4 NOT NULL,
“id_reklama” int4 NOT NULL,
“id_tex-specialist” int4 NOT NULL,
“id_filial” int4 NOT NULL,
CONSTRAINT “_copy_8” PRIMARY KEY (“compania_id”)
);

CREATE TABLE “filial” (
“id_filial” int4 NOT NULL,
“street” text NOT NULL,
“house” int4 NOT NULL,
“city” text NOT NULL,
CONSTRAINT “_copy_5” PRIMARY KEY (“id_filial”)
);

CREATE TABLE “login” (
“id_name” int4 NOT NULL,
“name” char(20) NOT NULL,
“pass” char(20) NOT NULL,
PRIMARY KEY (“id_name”)
);

CREATE TABLE “operator” (
“id_operator” int4 NOT NULL,
“obrabotka obrasheniya” int4 NOT NULL,
“konsultirovanie” text NOT NULL,
“grafick work” date NOT NULL,
CONSTRAINT “_copy_2” PRIMARY KEY (“id_operator”)
);

CREATE TABLE “polsovateli” (
“id_user” int4 NOT NULL,
“id_companiya” int4 NOT NULL,
“id_obrasheniya” int4 NOT NULL,
“id_oshibka” int4 NOT NULL,
CONSTRAINT “_copy_6” PRIMARY KEY (“id_user”)
);

CREATE TABLE “reklama” (
“id_reklama” int4 NOT NULL,
“tele-marketing” text NOT NULL,
“soc-seti” text NOT NULL,
“mobile” int4 NOT NULL,
CONSTRAINT “_copy_3” PRIMARY KEY (“id_reklama”)
);

CREATE TABLE “tex-specialist” (
“id_tex-specialist” int4 NOT NULL,
“grafik” date NOT NULL,
“zarplata” int4 NOT NULL,
“ispravlenie oshibok” int4 NOT NULL,
CONSTRAINT “_copy_7” PRIMARY KEY (“id_tex-specialist”)
);

CREATE TABLE “uslugi” (
“id_uslugi” int4 NOT NULL,
“vostanavlenia parola” int4 NOT NULL,
“poterya acaunta” int4 NOT NULL,
CONSTRAINT “_copy_4” PRIMARY KEY (“id_uslugi”)
);

ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_operator_1” FOREIGN KEY (“id_operator”) REFERENCES “operator” (“id_operator”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_uslugi_1” FOREIGN KEY (“id_uslugi”) REFERENCES “uslugi” (“id_uslugi”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_filial_1” FOREIGN KEY (“id_filial”) REFERENCES “filial” (“id_filial”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_reklama_1” FOREIGN KEY (“id_reklama”) REFERENCES “reklama” (“id_reklama”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_tex-specialist_1” FOREIGN KEY (“id_tex-specialist”) REFERENCES “tex-specialist” (“id_tex-specialist”);
ALTER TABLE “polsovateli” ADD CONSTRAINT “fk_polsovateli_companiya_1” FOREIGN KEY (“id_companiya”) REFERENCES “companiya” (“compania_id”);

ERROR: ОШИБКА: ошибка синтаксиса (примерное положение: “”companiya””)
LINE 1: drop table exists “companiya”;
^

Источник

[Error] 42601: syntax error at or near «SELECT» on pgsql 9.2.5 #813

When trying to remove or update an object from DB i get this exception on pgsql 9.2.5. On 9.4 it works fine.

Npgsql: npgsql-3.1-alpha0058
EF: 6.1.3
posgresql: postgresql92-9.2.5-1PGDG.rhel6.x86_64

[Error] 42601: syntax error at or near «SELECT» — at Npgsql.NpgsqlConnector.DoReadSingleMessage (DataRowLoadingMode dataRowLoadingMode, Boolean returnNullForAsyncMessage, Boolean isPrependedMessage) [0x00000] in :0
at Npgsql.NpgsqlConnector.ReadSingleMessageWithPrepended (DataRowLoadingMode dataRowLoadingMode, Boolean returnNullForAsyncMessage) [0x00000] in :0

2015-10-07 19:13:17.082 CEST [9318]: LOG: statement: DISCARD ALL
2015-10-07 19:13:17.083 CEST [9318]: ERROR: syntax error at or near «SELECT» at character 4857
2015-10-07 19:13:17.083 CEST [9318]: STATEMENT: SELECT «Project4».»Id1″ AS «Id», «Project4″.»Id» AS «Id1», «Project4″.»Name», «Project4″.»DatastoreRef», «Project4″.»DatastoreName», «Project4″.»TestNetworkName», «Project4″.»TestNetworkRef», «Project4″.»TestNetworkVLanId», «Project4″.»RecoveryNetworkName», «Project4″.»RecoveryNetworkRef», «Project4″.»RecoveryNetworkVLanId», «Project4″.»ResourcePoolName», «Project4″.»ResourcePoolRef», «Project4″.»ComputeResourceName», «Project4″.»ComputeResourceRef», «Project4″.»FolderName», «Project4″.»FolderRef», «Project4″.»PowerOnTimeoutMin», «Project4″.»DestinationPath», «Project4″.»Status», «Project4″.»MaxRecoveryPoint», «Project4″.»ServerId», «Project4″.»Enabled», «Project4″.»StartDateTime», «Project4″.»DaysOfWeek», «Project4″.»Interval», «Project4».»C1″, «Project4″.»IntervalUnit», «Project4″.»EndDateTime», «Project4″.»DateTimeFormat_DateFormat», «Project4″.»DateTimeFormat_TimeFormat», «Project4″.»DateTimeFormat_TimeZoneId», «Project4».»C35″ AS «C2», «Project4».»C3″, «Project4».»C4″, «Project4».»C5″, «Project4».»C6″, «Project4».»C7″, «Project4».»C8″, «Project4».»C9″, «Project4».»C10″, «Project4».»C11″, «Project4».»C12″, «Project4».»C13″, «Project4».»C14″, «Project4».»C2″ AS «C15», «Project4».»C15″ AS «C16», «Project4».»C16″ AS «C17», «Project4».»C17″ AS «C18», «Project4».»C18″ AS «C19», «Project4».»C19″ AS «C20», «Project4».»C20″ AS «C21», «Project4».»C21″ AS «C22», «Project4».»C22″ AS «C23», «Project4».»C23″ AS «C24», «Project4».»C24″ AS «C25», «Project4».»C25″ AS «C26», «Project4».»C26″ AS «C27», «Project4».»C27″ AS «C28», «Project4».»C28″ AS «C29», «Project4».»C29″ AS «C30», «Project4».»C30″ AS «C31», «Project4».»C31″ AS «C32», «Project4».»C32″ AS «C33», «Project4».»C33″ AS «C34», «Project4».»C34″ AS «C35» FROM (SELECT «Alias1″.»Id», «Alias1″.»Name», «Alias1″.»DatastoreRef», «Alias1″.»DatastoreName», «Alias1″.»TestNetworkName», «Alias1″.»TestNetworkRef», «Alias1″.»TestNetworkVLanId», «Alias1″.»RecoveryNetworkName», «Alias1″.»RecoveryNetworkRef», «Alias1″.»RecoveryNetworkVLanId», «Alias1″.»ResourcePoolName», «Alias1″.»ResourcePoolRef», «Alias1″.»ComputeResourceName», «Alias1″.»ComputeResourceRef», «Alias1″.»FolderName», «Alias1″.»FolderRef», «Alias1″.»PowerOnTimeoutMin», «Alias1″.»DestinationPath», «Alias1″.»Status», «Alias1″.»MaxRecoveryPoint», «Alias1″.»ServerId», «Alias1».»Id1″, «Alias1″.»Enabled», «Alias1″.»StartDateTime», «Alias1″.»DaysOfWeek», «Alias1″.»Interval», «Alias1″.»IntervalUnit», «Alias1″.»EndDateTime», «Alias1″.»DateTimeFormat_DateFormat», «Alias1″.»DateTimeFormat_TimeFormat», «Alias1″.»DateTimeFormat_TimeZoneId», «Alias1».»C1″, «UnionAll1».»C1″ AS «C2», «UnionAll1″.»Id» AS «C3», «UnionAll1».»Id1″ AS «C4», «UnionAll1″.»InstanceId» AS «C5», «UnionAll1″.»Name» AS «C6», «UnionAll1″.»Ref» AS «C7», «UnionAll1″.»IsPhysical» AS «C8», «UnionAll1″.»BackupId» AS «C9», «UnionAll1″.»JobId» AS «C10», «UnionAll1″.»OSUsername» AS «C11», «UnionAll1″.»OSUserPwd» AS «C12», «UnionAll1″.»OSType» AS «C13», «UnionAll1″.»SnapshotRef» AS «C14», «UnionAll1».»Id2″ AS «C15», «UnionAll1».»Id3″ AS «C16», «UnionAll1″.»MacAdresss» AS «C17», «UnionAll1″.»NicLabel» AS «C18», «UnionAll1″.»NetworkConfigId» AS «C19», «UnionAll1″.»VirtualMachine_Id» AS «C20», «UnionAll1».»C2″ AS «C21», «UnionAll1».»C3″ AS «C22», «UnionAll1».»C4″ AS «C23», «UnionAll1».»C5″ AS «C24», «UnionAll1».»C6″ AS «C25», «UnionAll1».»C7″ AS «C26», «UnionAll1».»C8″ AS «C27», «UnionAll1».»C9″ AS «C28», «UnionAll1».»C10″ AS «C29», «UnionAll1».»C11″ AS «C30», «UnionAll1».»C12″ AS «C31», «UnionAll1».»C13″ AS «C32», «UnionAll1».»C14″ AS «C33», «UnionAll1».»C15″ AS «C34», CASE WHEN («UnionAll1″.»Id» IS NULL) THEN (CAST (NULL AS int4)) ELSE (1) END AS «C35» FROM (SELECT «Extent1″.»Id», «Extent1″.»Name», «Extent1″.»DatastoreRef», «Extent1″.»DatastoreName», «Extent1″.»TestNetworkName», «Extent1″.»TestNetworkRef», «Extent1″.»TestNetworkVLanId», «Extent1″.»RecoveryNetworkName», «Extent1″.»RecoveryNetworkRef», «Extent1″.»RecoveryNetworkVLanId», «Extent1″.»ResourcePoolName», «Extent1″.»ResourcePoolRef», «Extent1″.»ComputeResourceName», «Extent1″.»ComputeResourceRef», «Extent1″.»FolderName», «Extent1″.»FolderRef», «Extent1″.»PowerOnTimeoutMin», «Extent1″.»DestinationPath», «Extent1″.»Status», «Extent1″.»MaxRecoveryPoint», «Extent1″.»ServerId», «Extent2″.»Id» AS «Id1», «Extent2″.»Enabled», «Extent2″.»StartDateTime», «Extent2″.»DaysOfWeek», «Extent2″.»Interval», «Extent2″.»IntervalUnit», «Extent2″.»EndDateTime», «Extent2″.»DateTimeFormat_DateFormat», «Extent2″.»DateTimeFormat_TimeFormat», «Extent2″.»DateTimeFormat_TimeZoneId», CASE WHEN («Extent2″.»Id» IS NULL) THEN (CAST (NULL AS int2)) ELSE (CAST («Extent2″.»IntervalValue» AS int2)) END AS «C1» FROM «public».»jobs» AS «Extent1» LEFT OUTER JOIN «public».»schedules» AS «Extent2» ON «Extent1″.»Id» = «Extent2″.»Id» WHERE «Extent1″.»Id» = $1 LIMIT 1) AS «Alias1» LEFT OUTER JOIN LATERAL (SELECT «UnionAll1».»C1″, «UnionAll1″.»Id», «UnionAll1».»Id1″, «UnionAll1″.»InstanceId», «UnionAll1″.»Name», «UnionAll1″.»Ref», «UnionAll1″.»IsPhysical», «UnionAll1″.»BackupId», «UnionAll1″.»JobId», «UnionAll1″.»OSUsername», «UnionAll1″.»OSUserPwd», «UnionAll1″.»OSType», «UnionAll1″.»SnapshotRef», «UnionAll1».»Id2″, «UnionAll1».»Id3″, «UnionAll1″.»MacAdresss», «UnionAll1″.»NicLabel», «UnionAll1″.»NetworkConfigId», «UnionAll1″.»VirtualMachine_Id», «UnionAll1».»C2″, «UnionAll1».»C3″, «UnionAll1».»C4″, «UnionAll1».»C5″, «UnionAll1».»C6″, «UnionAll1».»C7″, «UnionAll1».»C8″, «UnionAll1».»C9″, «UnionAll1».»C10″, «UnionAll1».»C11″, «UnionAll1».»C12″, «UnionAll1».»C13″, «UnionAll1».»C14″, «UnionAll1».»C15″ FROM ((SELECT CASE WHEN («Extent4″.»Id» IS NULL) THEN (CAST (NULL AS int4)) ELSE (1) END AS «C1», «Extent3″.»Id», «Extent3″.»Id» AS «Id1», «Extent3″.»InstanceId», «Extent3″.»Name», «Extent3″.»Ref», «Extent3″.»IsPhysical», «Extent3″.»BackupId», «Extent3″.»JobId», «Extent3″.»OSUsername», «Extent3″.»OSUserPwd», «Extent3″.»OSType», «Extent3″.»SnapshotRef», «Extent4″.»Id» AS «Id2», «Extent4″.»Id» AS «Id3», «Extent4″.»MacAdresss», «Extent4″.»NicLabel», «Extent4″.»NetworkConfigId», «Extent4″.»VirtualMachine_Id», CAST (NULL AS int4) AS «C2», CAST (NULL AS int4) AS «C3», CAST (NULL AS text) AS «C4», CAST (NULL AS text) AS «C5», CAST (NULL AS bool) AS «C6», CAST (NULL AS int4) AS «C7», CAST (NULL AS int4) AS «C8», CAST (NULL AS int4) AS «C9», CAST (NULL AS int4) AS «C10», CAST (NULL AS int4) AS «C11», CAST (NULL AS int4) AS «C12», CAST (NULL AS text) AS «C13», CAST (NULL AS text) AS «C14», CAST (NULL AS int4) AS «C15» FROM «public».»vms» AS «Extent3» LEFT OUTER JOIN «public».»vm_net» AS «Extent4» ON «Extent3″.»Id» = «Extent4″.»VirtualMachine_Id» WHERE «Alias1″.»Id» = «Extent3″.»JobId») UNION ALL (SELECT 2 AS «C1», «Extent5″.»Id», «Extent5″.»Id» AS «Id1», «Extent5″.»InstanceId», «Extent5″.»Name», «Extent5″.»Ref», «Extent5″.»IsPhysical», «Extent5″.»BackupId», «Extent5″.»JobId», «Extent5″.»OSUsername», «Extent5″.»OSUserPwd», «Extent5″.»OSType», «Extent5″.»SnapshotRef», CAST (NULL AS int4) AS «C2», CAST (NULL AS int4) AS «C3», CAST (NULL AS text) AS «C4», CAST (NULL AS text) AS «C5», CAST (NULL AS int4) AS «C6», CAST (NULL AS int4) AS «C7», «Extent6″.»Id» AS «Id2», «Extent6″.»Id» AS «Id3», «Extent6″.»Name» AS «Name1», «Extent6″.»Command», «Extent6″.»IsCustom», «Extent6″.»TimeoutMin», «Extent6″.»Priority», «Extent6″.»VirtualMachine_Id», CASE WHEN («Extent7″.»Id» IS NULL) THEN (CAST (NULL AS int4)) ELSE (1) END AS «C8», «Extent7″.»Id» AS «Id4», «Extent7″.»Id» AS «Id5», «Extent7″.»Name» AS «Name2», «Extent7″.»Value», «Extent7″.»VirtualMachineApplicationTest_Id» FROM «public».»vms» AS «Extent5» INNER JOIN «public».»vm_test» AS «Extent6» LEFT OUTER JOIN «public».»vm_testparam» AS «Extent7» ON «Extent6″.»Id» = «Extent7″.»VirtualMachineApplicationTest_Id» ON «Extent5″.»Id» = «Extent6″.»VirtualMachine_Id» WHERE «Alias1″.»Id» = «Extent5″.»JobId»)) AS «UnionAll1») AS «UnionAll1» ON TRUE) AS «Project4» ORDER BY «Project4″.»Id1″ ASC ,»Project4″.»Id» ASC ,»Project4″.»C35″ ASC ,»Project4″.»C4″ ASC ,»Project4″.»C2″ ASC ,»Project4″.»C22″ ASC ,»Project4″.»C29″ ASC
2015-10-07 19:13:18.221 CEST [9318]: LOG: statement: DISCARD ALL
2015-10-07 19:13:18.221 CEST [9318]: ERROR: syntax error at or near «SELECT» at character 4857

The text was updated successfully, but these errors were encountered:

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PostgreSQL – SQL state: 42601 syntax error

Posted By: Anonymous

I would like to know how to use a dynamic query inside a function. I’ve tried lots of ways, however, when I try to compile my function a message SQL 42601 is displayed.

The code that I use:

Error message I receive:

What is wrong? How can I solve this problem?

Solution

Your function would work like this:

You cannot mix plain and dynamic SQL the way you tried to do it. The whole statement is either all dynamic or all plain SQL. So I am building one dynamic statement to make this work. You may be interested in the chapter about executing dynamic commands in the manual.

The aggregate function count() returns bigint , but you had rowcount defined as integer , so you need an explicit cast ::int to make this work

I use dollar quoting to avoid quoting hell.

However, is this supposed to be a honeypot for SQL injection attacks or are you seriously going to use it? For your very private and secure use, it might be ok-ish – though I wouldn’t even trust myself with a function like that. If there is any possible access for untrusted users, such a function is a loaded footgun. It’s impossible to
make this secure.

Craig (a sworn enemy of SQL injection!) might get a light stroke, when he sees what you forged from his piece of code in the answer to your preceding question. 🙂

The query itself seems rather odd, btw. But that’s beside the point here.

Answered By: Anonymous

Disclaimer: This content is shared under creative common license cc-by-sa 3.0. It is generated from StackExchange Website Network.

Источник

Syntax errors are quite common while coding.

But, things go for a toss when it results in website errors.

PostgreSQL error 42601 also occurs due to syntax errors in the database queries.

At Bobcares, we often get requests from PostgreSQL users to fix errors as part of our Server Management Services.

Today, let’s check PostgreSQL error in detail and see how our Support Engineers fix it for the customers.

What causes error 42601 in PostgreSQL?

PostgreSQL is an advanced database engine. It is popular for its extensive features and ability to handle complex database situations.

Applications like Instagram, Facebook, Apple, etc rely on the PostgreSQL database.

But what causes error 42601?

PostgreSQL error codes consist of five characters. The first two characters denote the class of errors. And the remaining three characters indicate a specific condition within that class.

Here, 42 in 42601 represent the class “Syntax Error or Access Rule Violation“.

In short, this error mainly occurs due to the syntax errors in the queries executed. A typical error shows up as:

Here, the syntax error has occurred in position 119 near the value “parents” in the query.

How we fix the error?

Now let’s see how our PostgreSQL engineers resolve this error efficiently.

Recently, one of our customers contacted us with this error. He tried to execute the following code,

CREATE OR REPLACE FUNCTION prc_tst_bulk(sql text)
RETURNS TABLE (name text, rowcount integer) AS
$$
BEGIN
WITH m_ty_person AS (return query execute sql)
select name, count(*) from m_ty_person where name like '%a%' group by name
union
select name, count(*) from m_ty_person where gender = 1 group by name;
END
$$ LANGUAGE plpgsql;

But, this ended up in PostgreSQL error 42601. And he got the following error message,

ERROR: syntax error at or near "return"
LINE 5: WITH m_ty_person AS (return query execute sql)

RETURN QUERY EXECUTE '
WITH m_ty_person AS (' || sql || $x$)
SELECT name, count(*)::int FROM m_ty_person WHERE name LIKE '%a%' GROUP BY name
UNION
SELECT name, count(*)::int FROM m_ty_person WHERE gender = 1 GROUP BY name$x$;

This resolved the error 42601, and the code worked fine.

[Need more assistance to solve PostgreSQL error 42601?- We’ll help you.]

Conclusion

In short, PostgreSQL error 42601 occurs due to the syntax errors in the code. Today, in this write-up, we have discussed how our Support Engineers fixed this error for our customers.

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var google_conversion_label = «owonCMyG5nEQ0aD71QM»;

Источник

I am just moving to some PostgreSQL from MS-SQL and have checked numerous pages on how to use variables in a script but am getting nowhere

Within pg-admin I have my database and have a new script. I have stripped the SQL right back to a single line as follows:

DECLARE v_syncId integer;

Running this script gives me:

ERROR:  syntax error at or near "integer"
LINE 1: DECLARE v_syncId integer;
                         ^
SQL state: 42601
Character: 18

I am obviously doing something wrong here but can’t see what

What I have tried:

I have tried setting default values but no luck (same error)

DECLARE v_syncId integer DEFAULT 50;

DECLARE v_syncId integer = 50;

Both give the same error

I have tried other data types:

DECLARE v_syncId VARCHAR;

and also same result.

Any references here would be appreciated.

Источник

What causes error 42601 in PostgreSQL?

How we fix the error?

Conclusion

PREVENT YOUR SERVER FROM CRASHING!

PostgreSQL error 42601- How we fix it

What causes error 42601 in PostgreSQL?

How we fix the error?

Conclusion

PREVENT YOUR SERVER FROM CRASHING!

[Error] 42601: syntax error at or near «SELECT» on pgsql 9.2.5 #813

PostgreSQL – SQL state: 42601 syntax error

Posted By: Anonymous

Solution

Answered By: Anonymous

Related Articles

What causes error 42601 in PostgreSQL?

How we fix the error?

Conclusion

PREVENT YOUR SERVER FROM CRASHING!

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