这是我的代码:
try{
Font font1=new Font("SansSerif",Font.ITALIC,20);
JTextField levi=new JTextField(20);
levi.setBounds(107,10,246,100);
levi.setForeground(Color.black);
levi.setHorizontalAlignment(SwingConstants.BOTTOM);
levi.setFont(font1);
levi.setBorder(nue);
lin.add(levi);
} catch(Exception ex) {
JOptionPane.showMessageDialog(null, ex);
}
答案 0 :(得分:2)
如果您检查Sub firstCoord()
Dim fpath As String, fname As String
Dim dateCount As Integer, strDate As Date
Dim i As Integer, j As Integer, k As Integer, lastRow As Integer, lastRow2 As Integer
Dim ws As Worksheet, allws As Worksheet
Dim seg As String
Dim strNum As String
Dim strRow As Integer
lastRow = Sheet1.Range("A" & Sheet1.Rows.Count).End(xlUp).Row
seg = Mid(ThisWorkbook.Name, 34, 1)
With Application.WorksheetFunction
For i = 2 To lastRow
fpath = "_______\"
strDate = Sheet1.Range("B" & i)
strNum = seg & Format(Mid(Sheet1.Range("A" & i), 4, 3), "000") & "000"
dateCount = 0
Do While Len(Dir(fpath & "_____-" & Format(strDate - dateCount, "ddmmmyyyy") & ".xlsx")) = 0 And dateCount < 35
dateCount = dateCount + 1
Loop
fname = "____-" & Format(strDate - dateCount, "ddmmmyyyy") & ".xlsx"
Workbooks.Open (fpath & fname)
For Each ws In Workbooks(fname).Worksheets
If ws.Name Like "*all*" Then
Set allws = Workbooks(fname).Worksheets(ws.Name)
ws.Activate
End If
Next ws
lastRow2 = ActiveSheet.Range("A" & ActiveSheet.Rows.Count).End(xlUp).Row
ThisWorkbook.Activate
k = 1
Do While (.CountIf(Sheet1.Range("C" & i & ":" & "E" & i), "") <> 0 Or Sheet1.Range("F" & i) = "") And k <= lastRow2
If Left(allws.Range("A" & k), 7) = strNum Then
Sheet1.Range("C" & i) = allws.Range("D" & k)
Sheet1.Range("D" & i) = allws.Range("C" & k)
Sheet1.Range("E" & i) = allws.Range("E" & k)
ElseIf k = lastRow2 And Sheet1.Range("C" & i) = "" Then
Sheet1.Range("F" & i) = "Not Found"
End If
k = k + 1
Loop
Workbooks(fname).Close
Next i
End With
End Sub
JTextField
,则setHorizontalAlignment()
方法的唯一有效密钥为:
JTextField.LEFT JTextField.CENTER JTextField.RIGHT JTextField.LEADING JTextField.TRAILING
传入SwingConstants.BOTTOM
会抛出异常。