PHP preg_replace remove specific parts from string

Question

im having problems with understanding regex in PHP. I have img src:

src="http://example.com/javascript:gallery('/info/2005/image.jpg',383,550)"

and need to build from it this:

src="http://example.com/info/2005/image.jpg"

How it it possible to cut first and last part from string to obtain clear link without javascript part?

Right now im using this regex:

$cont = 'src="http://example.com/javascript:gallery('/info/2005/image.jpg',383,550)"'

    $cont = preg_replace("/(src=\")(.*)(\/info)/","$1http://example.com$3", $cont);

and output is:

src="http://example.com/info/2005/image.jpg',383,550)"

(src=\")(.*)(\/info[^']*)[^"]* will work with the string you have. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 13, 2020 at 20:18

The fourth bird · Accepted Answer · 2020-11-14 10:54:28Z

3

As an alternative solution, you might also capture the src="http://example.com part by matching the protocol in group 1, so you can use it in the replacement.

(src="https?://[^/]+)/[^']*'(/info[^']*)'[^"]*

Explanation

(src="https?://[^/]+)/ Capture group 1, match src="http, optional s, :// and till the first /
[^']*' Match any char except ', then match '
(/info[^']*) Capture group 2, match /info followed by any char except '
'[^"]* Match the ' followed by matching any char except "

Regex demo | Php demo

$cont = 'src="http://example.com/javascript:gallery(\'/info/2005/image.jpg\',383,550)"';
$cont = preg_replace("~(src=\"https?://[^/]+)/[^']*'(/info[^']*)'[^\"]*~", '$1$2', $cont);
echo $cont;

Output

src="http://example.com/info/2005/image.jpg"

answered Nov 14, 2020 at 10:54

The fourth bird

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1 Comment

PiciuU Over a year ago

Thanks for giving me an alternative solution :) Finally, I understand a little more about how regex works

Ryszard Czech · Accepted Answer · 2020-11-13 22:17:40Z

2

Use

preg_replace("/src=\"\K.*(\/info[^']*)'[^\"]*/", 'http://example.com$1', $cont)

See regex proof.

Explanation

--------------------------------------------------------------------------------
  src=                     'src='
--------------------------------------------------------------------------------
  \"                       '"'
--------------------------------------------------------------------------------
  \K                       match reset operator
--------------------------------------------------------------------------------
  .*                       any character except \n (0 or more times
                           (matching the most amount possible))
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    \/                       '/'
--------------------------------------------------------------------------------
    info                     'info'
--------------------------------------------------------------------------------
    [^']*                    any character except: ''' (0 or more
                             times (matching the most amount possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  '                        '\''
--------------------------------------------------------------------------------
  [^\"]*                   any character except: '\"' (0 or more
                           times (matching the most amount possible))

answered Nov 13, 2020 at 22:17

Ryszard Czech

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2 Comments

PiciuU Over a year ago

It works also, thanks for this precise explanation :)

Ryszard Czech Over a year ago

@PiciuU Glad to hear, please kindly accept the answer by clicking ✓ on the left.

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