MID function in Visual Basic 6.0

[akaaryan]

hello friends....

please help me out on this.........i write ma hole question with explaination . .... . . please tell me the solution . . . .thank u/......


i hv one textbox with value = "alkim25plt r"

now i have to write this word in second textbox(text2) . . . . . .with only
first 5 letters in this second textbox(text2) . . . . . . . .

then next 2 wrods in textbox3 . . . . . .

then last 4 words in textbox4 . . . . . . . . . . . .



please give me the full code of this question . . .. .on my E-Mail ID : E-mail address removed


please help me out . . . . . .thanks

[QVeen72]

Hi,

Did you Try to use Left() and Mid() Functions...?

Regards
Veena

[jamesd0142]

[code=text]
dim abc as string
abc = mid(textbox1.te xt,1,5)
[/code]
Clearly this will give you the first five characters in textbox1.

1 stands for the start possition
5 stands for the length

[akaaryan]

[code=text]
dim abc as string
abc = mid(textbox1.te xt,1,5)
[/code]
Clearly this will give you the first five characters in textbox1.

1 stands for the start possition
5 stands for the length

ya i know that....bt u didn't understand ma query ......
i scaned one string in one textbox like "KFA20LJKJK ". .. . then i want fitst 5 char in second textbox2....... then next......then next.......plz help me out its very important. .. . if possible write full code...so i can understand. . .. .

[mafaisal]

Hello,
Try To Use Left, Right, mid fns

eg,

Text1.Text = "alkim25pltr"
Text2.Text = left(Text1.Text,5)
Text3.Text = mid(Text1.Text,5,7)
text4.text = right(Text1.Text,7)

Try like this

Faisal

hello friends....

please help me out on this.........i write ma hole question with explaination . .... . . please tell me the solution . . . .thank u/......


i hv one textbox with value = "alkim25plt r"

now i have to write this word in second textbox(text2) . . . . . .with only
first 5 letters in this second textbox(text2) . . . . . . . .

then next 2 wrods in textbox3 . . . . . .

then last 4 words in textbox4 . . . . . . . . . . . .



please give me the full code of this question . . .. .on my E-Mail ID : E-mail address removed


please help me out . . . . . .thanks