StringRegExpReplace

[needhelpplease12]

Hi, I want to replace a fixed substring and return the rest of the string.

I have the string

$string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"

and want to remove the first part of the string which is

http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=

I have a few of these links and the first part always is the same. The part after it changes (the image name) so I want to extract that url to the image.

I tried with the following code but it returns the entire original string unchanged:

$new = StringRegExpReplace($string, "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=", '')

MsgBox(0, "", $new)

How to get the last part

Please?

Edited May 24, 2014 by needhelpplease12

[Iczer]

$string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"

$new = StringRegExpReplace($string, ".+url=", '')

MsgBox(0, "", $new)

[somdcomputerguy]

Or this, which will return exactly the same thing. AutoIt can skin a cat in many ways!

$string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"
$newstring = StringSplit($string, "url=", 1)
MsgBox(0, "", $newstring[2])

[needhelpplease12]

Thanks.

Do you know why my approach didn't work? Do I have to escape the dot or something?

Edited May 24, 2014 by needhelpplease12

[somdcomputerguy]

Do I have to escape the dot or something?

Well escaping the dot in the original string returns the whole string. So it's probably something else.. My RegEx knowledge is very limited, although I have been trying to improve it a bit. But I'm glad cat's have nine lives..

$string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"
$replaced = StringReplace($string, "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=", '')
MsgBox(0, "", $replaced)

[Malkey]

Hi, I want to replace a fixed substring and return the rest of the string.

I have the string

$string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"

and want to remove the first part of the string which is

http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=

I have a few of these links and the first part always is the same. The part after it changes (the image name) so I want to extract that url to the image.

I tried with the following code but it returns the entire original string unchanged:

$new = StringRegExpReplace($string, "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=", '')

MsgBox(0, "", $new)

How to get the last part

Please?

For correctness the dots should be escaped. But because the dot stands for any one character, including the dot "." character, not escaping the dot character still works.

However the "?" character in the regular expression pattern makes the previous character greedy. So if the question mark "?" is not escaped "?" the RE pattern will not match because of the missing "?" character in the RE pattern.

Local $string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"

Local $new = StringRegExpReplace($string, "(http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php\?url=)", "")

MsgBox(0, "", $new)

Edited May 24, 2014 by Malkey

[UEZ]

You don't need necessarily regex to cut the right part. It will work also with the string functions.

E.g.

$string = "http://domain.com/asd/cbcc507d86c16fec3947370ac5849ac0/http://www.domainxyz.org/out.php?url=http://www.otherdomain.org/a/s/d/123.jpg"

ConsoleWrite(StringTrimLeft($string, StringInStr($string, "url=", 0, -1) + 3) & @CRLF) 

Br,

UEZ

[needhelpplease12]

Thanks guys

[somdcomputerguy]

My RegEx knowledge is very limited, although I have been trying to improve it a bit.

For correctness the dots should be escaped. But because the dot stands for any one character, including the dot "." character, not escaping the dot character still works.

However the "?" character in the regular expression pattern makes the previous character greedy. So if the question mark "?" is not escaped "?" the RE pattern will not match because of the missing "?" character in the RE pattern.

My knowledge improves a bit more.. Thanks Malkey.

Edited May 24, 2014 by somdcomputerguy

[mikell]

My 2 cents

Using the lazy '?' to get the last link in the string

$new = StringRegExpReplace($string, '(?s).*(http.+?$)', '$1')