QRegExp Not Repeated More Than Four Times.

[anda_skoa]

The Above Experisson it Wont Take "2"'s more than four times.

As Chris67 said: your expression accecpts up to 9 digits after the first.
If you only want to accept 4, then don't add the remaining 5 options.

Cheers,
_

[d_stranz]

QRegExp rx("^(?:[7-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9])$");

As I read your regular expression, it says:

Match:

• from the start of the string ("^")
• the sub-expression: ("(?:")
• zero or one of 7, 8, or 9 ("[7-9]?")
• followed by zero or one of 0 through 9, up to eight times ("[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?[0-9]?")
• followed by a single digit 0 through 9 ("[0-9]")
• and this must be the entire string (")$")

So your regex will match the following:

• Any string containing only a single digit 0 - 9 (including strings containing only "7", "8", or "9")
• Any string starting with 7, 8, or 9 and ending with any single digit 0 - 9
• Any string starting with 7, 8, or 9 followed by up to eight more digits 0 - 9, ending with one more 0 - 9
• Any string containing one through nine digits between 0 and 9

For example:

• "0" (or "1", "2", ... "9")
• "70" (or "71", "72", ... "99")
• "70" (or "701", "7012", "7013", ... "7012345678")
• "0" (or "01", "012", "0123", ... "012345678"

Note that it will not match "7 22222 222" since this contains embedded whitespace. It will match "722222222".

Edit: So, your regex will match any string of one through eight digits between "0" and "99999999" and also any string of one through nine digits that starts with a "7", "8", or "9".

[ChrisW67]

Nice explanation d_stranz. Way more effort than I put in (no way i was going type type that on my tablet ). I hope it is not lost on the OP.