Question

我想将文本框中的字符串拆分为多个文本框，每个框中包含一个单词，在Access中使用vba

所以我们说文本框名称text1，它包含一个像hello I'm asking a question

这样的字符串

我想将该字符串拆分为文本框，这就像

text2 = hello
text3 = I'm
text4 = asking
 ...etc

有一种简单的方法吗？

Answer 1

Access有一个内置的split（）命令可以执行此操作。

因此，此代码将起作用：

Dim v        As Variant
Dim sOne     As Variant
Dim i        As Integer

v = Split(Me.TextBox0, " ")
i = 0
For Each sOne In v
   i = i + 1
   Me("text" & i) = sOne
Next

因此，上面将从表单上的textbox0获取字符串，并将3个值放入文本框1到3中。

因此不需要大量的代码。

Answer 2

这是一个简洁的解决方案：ParseWord() Function

下面的代码有点冗长，但一旦实现就很容易使用。这段代码的作用是在数据库项目中创建一个函数。此函数称为：ParseWord（）。它不是内置功能。因此，您需要将代码添加到数据库项目中的模块。

Function ParseWord(varPhrase As Variant, ByVal iWordNum As Integer, Optional strDelimiter As String = " ", _
    Optional bRemoveLeadingDelimiters As Boolean, Optional bIgnoreDoubleDelimiters As Boolean) As Variant
On Error GoTo Err_Handler 'I COMMENTED THIS OUT AND THE REFERENCE AT THE BOTTOM
    'Purpose:   Return the iWordNum-th word from a phrase.
    'Return:    The word, or Null if not found.
    'Arguments: varPhrase = the phrase to search.
    '           iWordNum = 1 for first word, 2 for second, ...
    '               Negative values for words form the right: -1 = last word; -2 = second last word, ...
    '               (Entire phrase returned if iWordNum is zero.)
    '           strDelimiter = the separator between words. Defaults to a space.
    '           bRemoveLeadingDelimiters: If True, leading delimiters are stripped.
    '               Otherwise the first word is returned as null.
    '           bIgnoreDoubleDelimiters: If true, double-spaces are treated as one space.
    '               Otherwise the word between spaces is returned as null.
    'Author:    Allen Browne. http://allenbrowne.com. June 2006.
    Dim varArray As Variant     'The phrase is parsed into a variant array.
    Dim strPhrase As String     'varPhrase converted to a string.
    Dim strResult As String     'The result to be returned.
    Dim lngLen As Long          'Length of the string.
    Dim lngLenDelimiter As Long 'Length of the delimiter.
    Dim bCancel As Boolean      'Flag to cancel this operation.

    '*************************************
    'Validate the arguments
    '*************************************
    'Cancel if the phrase (a variant) is error, null, or a zero-length string.
    If IsError(varPhrase) Then
        bCancel = True
    Else
        strPhrase = Nz(varPhrase, vbNullString)
        If strPhrase = vbNullString Then
            bCancel = True
        End If
    End If
    'If word number is zero, return the whole thing and quit processing.
    If iWordNum = 0 And Not bCancel Then
        strResult = strPhrase
        bCancel = True
    End If
    'Delimiter cannot be zero-length.
    If Not bCancel Then
        lngLenDelimiter = Len(strDelimiter)
        If lngLenDelimiter = 0& Then
            bCancel = True
        End If
    End If

    '*************************************
    'Process the string
    '*************************************
    If Not bCancel Then
        strPhrase = varPhrase
        'Remove leading delimiters?
        If bRemoveLeadingDelimiters Then
            strPhrase = Nz(varPhrase, vbNullString)
            Do While Left$(strPhrase, lngLenDelimiter) = strDelimiter
                strPhrase = Mid(strPhrase, lngLenDelimiter + 1&)
            Loop
        End If
        'Ignore doubled-up delimiters?
        If bIgnoreDoubleDelimiters Then
            Do
                lngLen = Len(strPhrase)
                strPhrase = Replace(strPhrase, strDelimiter & strDelimiter, strDelimiter)
            Loop Until Len(strPhrase) = lngLen
        End If
        'Cancel if there's no phrase left to work with
        If Len(strPhrase) = 0& Then
            bCancel = True
        End If
    End If

    '*************************************
    'Parse the word from the string.
    '*************************************
    If Not bCancel Then
        varArray = Split(strPhrase, strDelimiter)
        If UBound(varArray) >= 0 Then
            If iWordNum > 0 Then        'Positive: count words from the left.
                iWordNum = iWordNum - 1         'Adjust for zero-based array.
                If iWordNum <= UBound(varArray) Then
                    strResult = varArray(iWordNum)
                End If
            Else                        'Negative: count words from the right.
                iWordNum = UBound(varArray) + iWordNum + 1
                If iWordNum >= 0 Then
                    strResult = varArray(iWordNum)
                End If
            End If
        End If
    End If

    '*************************************
    'Return the result, or a null if it is a zero-length string.
    '*************************************
    If strResult <> vbNullString Then
        ParseWord = strResult
    Else
        ParseWord = Null
    End If

Exit_Handler:
    Exit Function

Err_Handler: 'I COMMENTED OUT THESE 4 LINES
    Call LogError(Err.Number, Err.Description, "ParseWord()")
    Resume Exit_Handler
End Function

将模块中的函数添加到数据库后，您就可以像在内置函数中一样在VBA代码中调用它。

示例（点击事件）：

我还必须注释掉“On Error GoTo Err_Handler”行，因为我没有那些设置。（我在代码中引用了这些）

MS Access：使用VBA将字符串从文本框拆分为其他文本框

2 个答案: