VB6 Split Function

[Googlrr]

k, so I have this right here.

"digna_kratowicz" : "dhdhd" : "digna_kratowicz@mailin8r.com"


On one line, on a list. I call this line of the list to a string "strUsr".

I want only the first part, "digna_kratowicz" (without the quotations). Then, in a separate string, the password "dhdhd" (again without the quotations). I want the last email part to gtfo (its not important.).

How would I make use of this split function to get it. I've looked around but most of the things I try don't seem to work.


+rep to helperz?

[|G3|]

Just use getbetween.

[CheeSie]

Your better off removing the quotations then spliting it. Example

Dim strUser() as String
strUser = Replace(strUser, chr$(38), "") 'I think " is chr38 check on that
strUser = Split(strUser, " : ")

strUser(0) would = digna_kratowicz
strUser(1) would = dhdhd
struser(2) would = digna_kratowicz@mailin8r.com

should work.

[|G3|]

Originally Posted by CheeSie

Your better off removing the quotations then spliting it. Example

Dim strUser() as String
strUser = Replace(strUser, chr$(38), "") 'I think " is chr38 check on that
strUser = Split(strUser, " : ")

strUser(0) would = digna_kratowicz
strUser(1) would = dhdhd
struser(2) would = digna_kratowicz@mailin8r.com

should work.

Or that D:
I forgot about the split function DD:

[rcadble]

Quotations are chr(34), not chr(38). Otherwise, it should work.