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Access для Microsoft 365 Access 2021 Access 2019 Access 2016 Access 2013 Access 2010 Access 2007 Еще…Меньше

Возвращает значение типа Boolean, которое указывает, является ли выражение значением ошибки.

Синтаксис

IsError
(

выражение

)

Обязательный аргумент выражениеаргумент может быть любым допустимым выражением.

Замечания

Значения ошибок создаются путем преобразования реальных чисел в значения ошибок с помощью функции CVErr. Функция IsError используется для определения того, числовое выражение представляет ошибку. IsError возвращает true, если аргумент выражения указывает на ошибку; в противном случае возвращается false.

Пример

Примечание: В примерах ниже показано, как использовать эту функцию в модуле Visual Basic для приложений (VBA). Чтобы получить дополнительные сведения о работе с VBA, выберите Справочник разработчика в раскрывающемся списке рядом с полем Поиск и введите одно или несколько слов в поле поиска.

В следующем примере с помощью функции IsError проверяется, является ли числовое выражение значением ошибки. Функция CVErr возвращает значение типа Error Variant из определяемой пользователем функции. Предполагается, что UserFunction — это определяемая пользователем функция, возвращающая значение ошибки, заданное, например, строкой UserFunction = CVErr(32767), где 32767 — определяемое пользователем число.





  


                       
  




Dim ReturnVal, MyCheck
ReturnVal = UserFunction()
MyCheck = IsError(ReturnVal)    ' Returns True.

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Access for Microsoft 365 Access 2021 Access 2019 Access 2016 Access 2013 Access 2010 Access 2007 More…Less

Returns a Boolean value indicating whether an expression> is an error value.

Syntax

IsError
(

expression

)

The required expressionargument can be any valid expression.

Remarks

Error values are created by converting real numbers to error values using the CVErr function. The IsError function is used to determine if a numeric expression represents an error. IsError returns True if the expression argument indicates an error; otherwise, it returns False.

Example

Note: Examples that follow demonstrate the use of this function in a Visual Basic for Applications (VBA) module. For more information about working with VBA, select Developer Reference in the drop-down list next to Search and enter one or more terms in the search box.

This example uses the IsError function to check if a numeric expression is an error value. The CVErr function is used to return an Error Variant from a user-defined function. Assume UserFunction is a user-defined function procedure that returns an error value; for example, a return value assigned with the statement UserFunction = CVErr(32767), where 32767 is a user-defined number.


Dim ReturnVal, MyCheck
ReturnVal = UserFunction()
MyCheck = IsError(ReturnVal)    ' Returns True.

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Access for Microsoft 365 Access 2021 Access 2019 Access 2016 Access 2013 Access 2010 Access 2007 More…Less

Returns a Boolean value indicating whether an expression> is an error value.

Syntax

IsError
(

expression

)

The required expressionargument can be any valid expression.

Remarks

Example


Dim ReturnVal, MyCheck
ReturnVal = UserFunction()
MyCheck = IsError(ReturnVal)    ' Returns True.

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Access для Microsoft 365 Access 2021 Access 2019 Access 2016 Access 2013 Access 2010 Access 2007 Еще…Меньше

Возвращает значение типа Boolean, которое указывает, является ли выражение значением ошибки.

Синтаксис

IsError
(

выражение

)

Обязательный аргумент выражениеаргумент может быть любым допустимым выражением.

Замечания

Пример


Dim ReturnVal, MyCheck
ReturnVal = UserFunction()
MyCheck = IsError(ReturnVal)    ' Returns True.

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Нужны дополнительные параметры?

Изучите преимущества подписки, просмотрите учебные курсы, узнайте, как защитить свое устройство и т. д.

Всем привет, сегодня разбираем функцию IIF в запросах Access. Если вы не знаете, как создать условие «ЕСЛИ ТО» в базе данных Microsoft Access, то вы пришли по адресу.
На простых запросах мы разберем принцип работы встроенной функции IIF. С помощью IIf можно определить, является ли выражение истиной или ложью. Если выражение истинно, IIf вернет одно значение; если ложно, IIf вернет другое значение.

Синтаксис функции IIF: IIf (условие; если истина; если ложь).
1 аргумент — наше условие. Если это условие выполняется, то функция IIF в запросах Access вернет значение второго аргумента, если же условие не выполняется, то функция IIF в запросах Access вернет значение третьего аргумента.

Рассмотрим базу данных «Студенты». Если вы хотите более подробно ознакомится с тем, как создавалась база данных Access Студенты, то переходите по ССЫЛКЕ.

Рассмотрим 1 пример.
У нас есть таблица СТУДЕНТЫ, известно, что некоторые студенты получают стипендию. Предположим, что сумма базовой стипендии 5000 руб.

Создадим такой запрос, который начислит 5000 руб тем студентам, кто получает стипендию, а всем остальным — 0!
Перейдем в конструктор запросов, нам понадобится дополнительное вычисляемое поле, а также функция IIF в запросах Access.

Результат запроса с функцией IIF представлен ниже:

Рассмотрим 2 пример.
Для 2 примера необходимо в таблицу СТУДЕНТЫ добавить новое логическое поле МЕДАЛЬ.

Студентам, которые имеют стипендию и медаль начислим 7000 рублей, у кого просто стипендия — 5000 руб, остальным -0!
Снова нам поможем функция IIF в запросах Access.

Обратите внимание, что функцию IIF можно использовать в качестве аргумента другой функции IIF.

Рассмотрим 3 пример.
Работаем с таблицей ПРЕПОДАВАТЕЛИ. Мы знаем, сколько получает каждый преподаватель и знаем, сколько у него детей.

Сделаем так, чтобы преподавателям, у которых больше 2 детей, начислялась надбавка 10000 рублей, тем у кого от 1 до 2 детей — 5000 рублей, у кого нет детей — 0!

Хотите больше примеров, где разбирается функция IIF в запросах Access? Переходите по ССЫЛКЕ.

Если вам нужна готовая база данных Access, то ознакомьтесь со списком представленным ЗДЕСЬ.

This MSAccess tutorial explains how to use the Access iif function with syntax and examples.

Description

The Microsoft Access iif function returns one value if a specified condition evaluates to TRUE, or another value if it evaluates to FALSE.

Syntax

The syntax for the iif function in MS Access is:

iif ( condition, value_if_true, value_if_false )

Parameters or Arguments

condition: The value that you want to test.
value_if_true: The value that is returned if condition evaluates to TRUE.
value_if_false: The value that is return if condition evaluates to FALSE.

Returns

The iif function returns value_if_true when the condition is TRUE.
The iif function returns value_if_false when the condition is FALSE.

Applies To

The iif function can be used in the following versions of Microsoft Access:

Access 2019, Access 2016, Access 2013, Access 2010, Access 2007, Access 2003, Access XP, Access 2000

Example

Let’s look at how to use the iif function in MS Access:

iif ([Qty] > 10, "large", "small")

This example would return «large» if the value in the Qty field is greater than 10. Otherwise, it would return «small».

This is equivalent to the following IF statement in VBA code.

If [Qty] > 10 Then
   result = "large"
Else
   result = "small"
End If

Example in SQL/Queries

You can use the iif function in a query in Microsoft Access.

For example:

In this query, we have used the iif function as follows:

Hours: IIf([Time Out]<#12:00:00 PM#,([Time Out)-[Time In])*24,(([Time Out]-[Time In])*25)-0.5)

In this example, if the [Time Out] field is less than or equal to 12 o’clock noon, then the iif function will return the number of hours that have elapsed between [Time Out] and [Time In].

If [Time Out] is greater than 12 o’clock noon, then the iif function will subtract 30 minutes from the time worked.

The results of the iif function will be displayed it in a column called Hours.

Frequently Asked Questions

Question: How would I use the iif function if I wanted to have more than one condition?

Answer: You could use the AND keyword to include multiple conditions.

In the example above, the iif function will return «Yes» if both the ContactTitle = «Owner» and City = «Madrid». If one or both of these conditions is not met, it will return «No».

I am trying to accomplish the following in an Access 2010 query

Select  
    UNIT, DATE, Sum(IIF(ISERROR(A),NULL,A)) AS DLP_PERCENTAGE  
From  
    tableA;

where

A = (ACT-BASE)/BASE

I get a generic OVERFLOW error. I am missing something obvious. I am trying to catch an error in the calculation and return NULL if an error exists or the result if no error. I have to do it in a query. Any ideas what I have overlooked?

asked Apr 29, 2011 at 15:58

SUM((ACT-BASE)/IIF(BASE=0, Null, BASE)) AS DLP_PERCENTAGE

answered Apr 29, 2011 at 20:33

Saic SiquotSaic Siquot

6,5134 gold badges34 silver badges56 bronze badges

Источник

I have a database which, among other things, tracks specific actions over time. Within this database, I have pages that run percentile statistics comparing specific actions (noted here for privacy as Action1 and Action2), within specified years. When there are records where both types of actions have taken place in the specified year, the queries and reports work flawlessly. But I am designing the database to be ready to accommodate years where there have been no actions as of yet. This is resulting in the usual error message one gets when one attempts to divide by zero. I’d like to be able to use the IFERROR function.

I did a Google search regarding the IFERROR function in Access and, not surprisingly, was directed to a question in this forum. Unfortunately, the question is from 2011, so I’m not just replying to it here. I tried applying what I found and am not getting what I want. The responder to the 2011 question indicated that the best way to do this was to use a combination of the IIf function and Iserror function. I attempted to follow their guidance and came up with the following:

=IIf(IsError([txt#ActualAction1s]/[txt#ActualAction2s]*100),»»,txt#ActualAction1s]/[txt#ActualAction2s*100))

I’m getting a notice that there’s an error in my expression syntax. Can someone help?

pbaldy

If that’s a copy/paste, you having missing brackets in the second part.

Two missing brackets and too many closing parentheses.

Code:

=IIf(IsError([txt#ActualAction1s]/[txt#ActualAction2s]*100),"",[txt#ActualAction1s]/[txt#ActualAction2s]*100)

If you avoid special characters in the object names you wouldn’t need the brackets.

Wow…once again I’m ecstatic about how quick you folks are to reply. Thanks a bunch!

I corrected the issues with brackets and closing parentheses, and still end up getting my error message like I was getting before (#Num!). Here’s the updated formula:

=IIf(IsError([txt#Action1s]/[txt#Action2s]*100),»»,[txt#Action1s]/[txt#Action2s]*100)

The special characters were added by the system…the actual text box names are txtAction1s and txtAction2s (modified for privacy/confidentiality).

And while I’m in the mood for thanks…thanks for being gentle about the brackets and parentheses errors in the first place. That’ll teach me to write «code» when I’m sleepy enough to drop off any second.

jdraw

Are you telling us that your field/control names are
txtAction1s and txtAction2s
and Access added the #?

First thing I’d do is remove the «#», or track down why it’s there. This isn’t normal behavior.

Wow…that was interesting. I went back in and looked at my textbox names and discovered that at some point I may have decided to use those symbols to further describe what’s in the textboxes themselves. I have changed all the names to eliminate those symbols. I must’ve been asleep when I did it in the first place. I agree…it’s not a good idea to use them, even if they don’t have a direct effect, since they are used in programming. My formula, which still produces the error #Num!, is:

=IIf(IsError([txtAction1s]/[txtAction2s]*100),»»,[txtAction1s]/[txtAction2s]*100)

Thanks for the best-practices heads up!

jdraw

I’m wanting to hide the error code that results from creating the report this comes from for those years where actions and records haven’t been recorded/entered yet. As it is now, since the Action2 will be empty, I get the #NUM! error message. I’d like not to see that.

apart from anything else, if there is an error, you are returning a text value («») and if not you are returning a number — bound to cause problems further down the line. Suggest return either 0 or null if there is an error if you want a numeric type to the column.

If your concern is that [txtAction2s] is either null or 0 then try

=IIf(nz([txtAction2s],0),null,[txtAction1s]/[txtAction2s]*100)

I tried the formula above using nz, but I’m still getting the same result.

looks like I missed a bit off

=IIf(nz([txtAction2s],0)=0,null,[txtAction1s]/[txtAction2s]*100)

looks like I missed a bit off

=IIf(nz([txtAction2s],0)=0,null,[txtAction1s]/[txtAction2s]*100)

This was EXACTLY what I was looking for. Thank you so much. It’s all working exactly how I wanted it to. I’m looking for the place to mark this as «solved» but haven’t found it yet. Again, thank you!!!!

looks like I missed a bit off

=IIf(nz([txtAction2s],0)=0,null,[txtAction1s]/[txtAction2s]*100)

Is there a way to add an OR type statement to this? Something along the lines of =IIf(nz([txtAction2s],0)=0 OR nz([txtAction1s],0)=0 ,null,[txtAction1s]/[txtAction2s].

I am working with a similar statement but I have the possibility of either number being 0

pbaldy

What you posted looks valid. Did you try it?

I agree — it looks valid. But do add a closing parenthesis

pbaldy

But do add a closing parenthesis

Oops, didn’t notice that.

Oops, didn’t notice that.

Slim pickins’ on new questions, so I was digging LOL

with IIF, both arguments must be valid … so even though the expression wouldn’t be used if nz(txtAction2s,0) results in zero, perhaps do this in the expression:
… / nz( [txtAction2s],1)
even though that answer wouldn’t be returned … then it won’t throw an error

pbaldy

Slim pickins’ on new questions, so I was digging LOL

I know what you mean.

But do add a closing parenthesis

What am I missing? I see 2 lefts and 2 rights, and they appear to be in the correct places? Perhaps it was edited when I wasn’t looking.

=IIf(nz([txtAction2s],0)=0,null,[txtAction1s]/[txtAction2s]*100)

Источник

I’m executing a query like this

select field from table;

In that query, there is a loop running on many tables. So, if the field is not present in a table I get a

Runtime Error 3061

How can I by pass this error such as that on this error flow should go to another point?

This is the code I have recently after going through this forum.

Option Explicit

Private Sub UpdateNulls()
 Dim rs2 As DAO.Recordset
  Dim tdf As DAO.TableDef
  Dim db As Database
  Dim varii As Variant, strField As String
  Dim strsql As String, strsql2 As String, strsql3 As String
  Dim astrFields As Variant
  Dim intIx As Integer
  Dim field As Variant
  Dim astrvalidcodes As Variant
  Dim found As Boolean
  Dim v As Variant


  Open "C:\Documents and Settings\Desktop\testfile.txt" For Input As #1
  varii = ""
  Do While Not EOF(1)
    Line Input #1, strField
    varii = varii & "," & strField
  Loop
  Close #1
  astrFields = Split(varii, ",")  'Element 0 empty




        For intIx = 1 To UBound(astrFields)


        'Function ListFieldDescriptions()
                            Dim cn As New ADODB.Connection, cn2 As New ADODB.Connection
                            Dim rs As ADODB.Recordset, rs3 As ADODB.Recordset
                            Dim connString As String
                            Dim SelectFieldName

                            Set cn = CurrentProject.Connection

                            SelectFieldName = astrFields(intIx)

                            Set rs = cn.OpenSchema(adSchemaColumns, Array(Empty, Empty, Empty, SelectFieldName))

                           'Show the tables that have been selected '
                            While Not rs.EOF

                           'Exclude MS system tables '
                            If Left(rs!Table_Name, 4) <> "MSys" Then
                            strsql = "Select t.* From [" & rs!Table_Name & "] t Inner Join 01UMWELT On t.fall = [01UMWELT].fall Where [01UMWELT].Status = 4"
                            End If

                            Set rs3 = CurrentDb.OpenRecordset(strsql)

            'End Function

            strsql2 = "SELECT label.validcode FROM variablen s INNER JOIN label ON s.id=label.variablenid WHERE varname='" & astrFields(intIx) & "'"

            Set db = OpenDatabase("C:\Documents and Settings\Desktop\Codebook.mdb")
            Set rs2 = db.OpenRecordset(strsql2)

                With rs2
                .MoveLast
                .MoveFirst
                 astrvalidcodes = rs2.GetRows(.RecordCount)
                .Close '
                End With


                    With rs3
                    .MoveFirst
                    While Not rs3.EOF
                        found = False
                        For Each v In astrvalidcodes
                        If v = .Fields(0) Then
                        found = True
                        Debug.Print .Fields(0)
                        Debug.Print .Fields(1)


              Exit For
                    End If
                    Next
                If Not found Then
                msgbox "xxxxxxxxxxxxxxxx"

                End If
                End If
                .MoveNext


                Wend
                End With

           On Error GoTo 0        'End of special handling

    Wend



Next intIx


  End Sub

I’m getting a

Type Mismatch Runtime Error

in Set rs3 = CurrentDb.OpenRecordset(strsql)

I guess I’m mixing up ado and dao but I’m not certainly sure where it is.

Источник

Пример

Нужна дополнительная помощь?

Нужны дополнительные параметры?

Example

Need more help?

Want more options?

Example

Need more help?

Want more options?

Пример

Нужна дополнительная помощь?

Нужны дополнительные параметры?

Description

Syntax

Parameters or Arguments

Returns

Applies To

Example

Example in SQL/Queries

Frequently Asked Questions

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