Does any one know how to search for the Asterisk?

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According to the online help (search for help on Like Operator), you need to put square brackets around the asterisk, i.e. Like “[*]”.

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My solution may not be the most elegant, but it works

1. In the table that contains the data create a new ‘yes/no’ field (called it something like ‘FLAG’)

2. Create a new module

3. Add the following Code to the module (replace the , and with the names of the table you wish to search and the field you wish to search and use as the flag.

4. Run this routine. VIOLA, all the fields that have an * at the end of them are CHECKED TRUE in the flag field!

Sub SearchAst()
Dim rstSearchTable As Recordset
Dim strFieldSearch As String
Dim strFieldFlag As String

Set rstSearchTable = CurrentDb.OpenRecordset(“”, dbopentable)
strFieldSearch = “”
strFieldFlag = “”

With rstSearchTable
Do While Not .EOF
If Right(.Fields(strFieldSearch), 1) = “*” Then
.edit
.Fields(strFieldFlag) = True
.Update
End If
.MoveNext
Loop
End With
End Sub

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Here’s a routine you could use to clean up those offending values:

Public Function ExtractNumFromStr(ByVal strIn As String, _
                            Optional intStartPos As Integer) As Variant
  'Created 2/16/2000 by Charlotte Foust
  On Error GoTo ExtractNum_err
  Dim varNum As Variant
  Dim strNum As String
  Dim strTemp As String
  Dim intLoop As Integer

  ' Get the number portion of the string
  
  'if they didn't pass in a start position
  'start at position 1
  If intStartPos = 0 Then
    intStartPos = 1
  End If ' intStartPos = 0 
  
  'grab the portion of the string to use
  strTemp = Right(strIn, 1 + Len(strIn) - intStartPos)

  'if the string is all numeric
  'return the whole string
  If IsNumeric(strTemp) Then
    strNum = strTemp

  'otherwise, extract the numbers
  'from the string
  Else 'IsNumeric(strTemp)

    'step through the string and test
    'each character, only returning the numbers
    For intLoop = intStartPos To Len(strIn)

      If IsNumeric(Mid(strIn, intLoop, 1)) Then

        strNum = strNum & Mid(strIn, intLoop, 1)
      End If 'IsNumeric(Mid(strIn, intLoop, 1))
    Next intLoop '  = intStartPos To Len(strIn)

    'if no numbers were found, return zero
    If Len(strNum) = 0 Then

      strNum = "0"
    End If 'Len(strNum) = 0 
  End If 'IsNumeric(strTemp)

  varNum = clng(strNum)
ExtractNum_exit:
  On Error Resume Next
  ExtractNumFromStr = varNum
  Exit Function
ExtractNum_err:
  Resume Next
End Function 'ExtractNumFromStr(ByVal strIn As String, _
                            Optional intStartPos As Integer) As Variant

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