vb里一个字符串分别怎么取它的前几位,中间,和后面呢?假设123456789,要12345,34567,56789,教教我吧 怎样取单元格中间几位
VB \u600e\u4e48\u8f93\u51fa 5 456 34567 2345678 123456789Private Sub Command1_Click()
X = "123456789"
Y = Space(4)
For I = 1 To 5
Z = Mid(X, 6 - I, I * 2 - 1)
Print Right(Y, 5 - I) + Z
Next
End Sub
\u7528MID\u51fd\u6570\uff0c\u6bd4\u5982A1\u4e2d\u8f93\u5165123456789
A2\u5355\u5143\u683c\u4e2d\u8f93\u5165=MID\uff08A1,3,3)
A2\u5c31\u4f1a\u663e\u793a345
\u7b2c\u4e00\u4e2a3\u8868\u793a\u4ece\u7b2c\u4e09\u4f4d\u5f00\u59cb\u53d6\u6570\uff0c\u7b2c\u4e8c\u4e2a3\u8868\u793a\u53d6\u4e09\u4f4d
\u5982\u679c\u4f60\u8f93\u5165=MID\uff08A1,3,5\uff09
A2\u5c31\u4f1a\u663e\u793a34567
\u5e0c\u671b\u91c7\u7528\uff0c\u90fd\u662f\u4e00\u4e2a\u4e00\u4e2a\u5b57\u6253\u51fa\u6765\u7684\u3002
比如left("123456789",5),就是取前五位，得12345
right("123456789",5)，就是以后五位，得56789
mid也类似，具体可以看帮助。
12345
left("123456789",5)
34567
mid("123456789",3,5)
56789
right("123456789",5)
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