Question

请求帮助！在下面的代码部分（工作得很好btw），我一直在实际的文件名和工作表名称中获得.txt。我一直在寻找解决这个问题需要解决的问题，但我无法解决这个问题。我对VBA很新，所以我尽我所能请帮忙？！

 'Opens the folder to location to select txt file
  fileToOpen = Application.GetOpenFilename("Text Files (*.txt), *.txt")
    If fileToOpen <> False Then
    Workbooks.OpenText fileName:=fileToOpen, _
    DataType:=xlDelimited, Tab:=True
    End If
 'Creates the file name based on txt file name
  fileName = Mid(fileToOpen, InStrRev(fileToOpen, "\") + 1)
 'Creates the sheet name based on the active txt file
  sheetName = Left(ActiveWorkbook.Name, Len(ActiveWorkbook.Name) - 4)

 'Rename the original text file
   ActiveWorkbook.SaveAs ("S:\MERIT OUTPUTS FOLDER\MSI Recruitment 
  Limited\" & "DNU_" & fileName & ".txt")

Answer 1

也许你也可以使用你的工作表名作为文件名，因为你已经从工作表名中删除了“.txt”。

'Opens the folder to location to select txt file
  fileToOpen = Application.GetOpenFilename("Text Files (*.txt), *.txt")
    If fileToOpen <> False Then
    Workbooks.OpenText fileName:=fileToOpen, _
    DataType:=xlDelimited, Tab:=True
    End If

 'Creates the sheet name based on the active txt file
  sheetName = Left(ActiveWorkbook.Name, Len(ActiveWorkbook.Name) - 4)

 'Rename the original text file
   ActiveWorkbook.SaveAs ("S:\MERIT OUTPUTS FOLDER\MSI Recruitment 
  Limited\" & "DNU_" & sheetName & ".txt")

Answer 2

您可以使用正则表达式并提取不带扩展名的文件名并使用它

public String extractFileName(String fullPathFile){
        try {
            Pattern regex = Pattern.compile(EXTRACT_FILE_NAME_REGEX);
            Matcher regexMatcher = regex.matcher(fullPathFile);
            if (regexMatcher.find()){
                return regexMatcher.group(1);
            }
        } catch (PatternSyntaxException ex) {
            LOG.info("extractFileName::pattern problem <"+fullPathFile+">",ex);
        }
        return fullPathFile;
    }

从我的文件名和工作表名称中删除.txt

2 个答案: