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    <title>Re: 字串分解的小问题.....</title>
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      <h1>Re: 字串分解的小问题.....</h1>
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Posted by <a href="mailto:honey0@tcts1.seed.net.tw">Honey</a> on November 14, 1998 at 20:48:26:<p>
In Reply to: <a href="7346.html">Re: 字串分解的小问题.....</a> posted by Honey on November 14, 1998 at 16:46:43:<p>
提供以下程式给你参考<br>在Form1上 放个Command1 Text1<br>在Text1中输入字串<p>Private Sub Command1_Click()<br>Dim Lngi As Long, k As Long, u As Long<br>Dim yy(), o As Long, PString() As String<br>ReDim yy(1 To 1)<p>Me.Cls<br>yy(1) = 0<br>o = 1<br>Text1.Text = Trim(Text1.Text)<br>'/*去除左右有空白的部分*/<br>k = Len(Text1.Text) '/*字串总长度*/<br>For Lngi = 1 To k<br>    u = InStr(Lngi, Text1.Text, " ") '/*寻找空白字串位置*/<br>    '/*u若不为零表示有空白的字串*/<br>    If u Then<br>        If Not Mid(Text1.Text, u + 1, 1) = " " Then<br>            '/*若此空白字串右边不为空白字串*/<br>            o = o + 1<br>            ReDim Preserve yy(1 To o)<br>            yy(o) = u<br>            Lngi = u<br>        End If<br>    Else<br>        o = o + 1<br>        ReDim Preserve yy(1 To o)<br>        yy(o) = k<br>        Exit For<br>    End If<br>Next<br>'/*经以上程序 yy阵列中存放了空白字串位置*/<br>For Lngi = 1 To o<br>Form1.Print yy(Lngi)<br>Next<p>k = UBound(yy) - 1<br>ReDim PString(k)<br>For Lngi = 1 To k<br>    PString(Lngi) = Mid(Text1.Text, yy(Lngi) + 1, yy(Lngi + 1) - yy(Lngi))<br>Next<p>For Lngi = 1 To k<br>    Form1.Print PString(Lngi)<br>Next<br>End Sub<p>其实这个方法 和先前的比起来 速度也差不多<br>不过程式比原来长了一些<br>
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