Delete part of src after a certain character

Question

I am using IFTTT with Wordpress, so every time that I create a post on instagram, the recipe creates a post on Wordpress.

The code that IFTTT creates:

<div>
   <img src='https://scontent.cdninstagram.com/t51.2885-15/sh0.08/e35/12724750_1719815604930604_2078818546_n.jpg?ig_cache_key=MTE5MTQxNTkwMjE5ODM2NzYwOQ%3D%3D.2' style='max-width:600px;' /><br/>
   <div>
      <a href="http://ift.tt/1XKFC24Q7" target="_blank">See</a>
   </div>
</div>

For some reason a plugin that I am using to add this image like featured image, doesn't like this end of the src of the image, so I need to delete the code after ".jpg":

"?ig_cache_key=MTE5MTQxNTkwMjE5ODM2NzYwOQ%3D%3D.2").

I've found a Wordpress filter that modify the post before save. So I tried this code but this delete all the other code after the ?

add_filter( 'wp_insert_post_data' ,  __NAMESPACE__ . '\\filter_post_data' , '99', 2 );

function filter_post_data( $data , $postarr ) {

  $data['post_content'] = substr( $data['post_content'], 0, strpos( $data['post_content'], "?")); delete all after ?

  return $data;
}

How can say delete everything after the question mark (?) but ONLY inside the src of the img tag?

This is the reference for $data['post_content']:

https://codex.wordpress.org/Plugin_API/Filter_Reference/wp_insert_post_data

Could you please add some further detail to the question. What exactly is in post_content? You may require a regex if you're looking to extract/replace a certain part of the code — puppyFlo, Feb 24 '16 at 15:40

puppyFlo · Answer 1 · 2016-02-24T16:11:58.487

0

If you are looking to replace the content in that specific HTML then maybe try this:

$a="<div><img src='https://scontent.cdninstagram.com/t51.2885-15/sh0.08/e35/12724750_1719815604930604_2078818546_n.jpg?ig_cache_key=MTE5MTQxNTkwMjE5ODM2NzYwOQ%3D%3D.2' style='max-width:600px;' /><br/><div><a href='http://ift.tt/1XKFC24Q7'target='_blank'>See</a></div></div>";
echo preg_replace('/(\?.*)(?=\')/U','',$a);

<div><img src='https://scontent.cdninstagram.com/t51.2885-15/sh0.08/e35/12724750_1719815604930604_2078818546_n.jpg' style='max-width:600px;' /><br/><div><a href='http://ift.tt/1XKFC24Q7'target='_blank'>See</a></div></div>

Good luck

edited Feb 24 '16 at 16:11

answered Feb 24 '16 at 15:43

puppyFlo

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4
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I added more details – mattia Feb 24 '16 at 15:48
Does the regex resolve your problem? It may need some enhancement depending on the specific contents of your post_content variable. – puppyFlo Feb 24 '16 at 15:54

Cârnăciov · Answer 2 · 2016-02-24T16:24:36.403

Here's a script I wrote that does that.

$string = "img src='https://scontent.cdninstagram.com/t51.2885-15/sh0.08/e35/12724750_1719815604930604_2078818546_n.jpg?ig_cache_key=MTE5MTQxNTkwMjE5ODM2NzYwOQ%3D%3D.2' style='max-width:600px;' /><br/>";
$string2 = "asd";
$count = 0;
for($i=0;$i<strlen($string);$i++)
{
    if($string[$i]=='?')
    {
        while($string[$i]!='\'')
            $i++;
    }
    $string2[$count] = $string[$i];
    $count++;
}
echo $string2;

It will basically copy the contents of one string into another, but will skip anything following a ? until it finds a '

It's up to you to implement it, should be pretty simple. The code I posted outputs

img src='https://scontent.cdninstagram.com/t51.2885-15/sh0.08/e35/12724750_1719815604930604_2078818546_n.jpg' style='max-width:600px;' />

Bear in mind this code is designed for this task only, it will only work here, be careful what you use it for as it may produce unexpected results, if for example you have other question marks in the content.

However it would be much easier to edit the link before it becomes HTML, if it's a plugin that's fetching the link it shouldn't be very hard.

score 0 · Accepted Answer · answered Feb 24 '16 at 16:44

Thanks you all, I solved with this code:

add_filter( 'wp_insert_post_data' ,  __NAMESPACE__ . '\\filter_post_data' , '99', 2 );

function filter_post_data( $data , $postarr ) {
    $data['post_content'] = preg_replace('/(\?.*)(?=\' style)/','', $data['post_content']);
    return $data;
}

but now I have another issue, how can I sayto apply this filter ONLY if the category is "Instagram"?

this condition seems doesn't work:

add_filter( 'wp_insert_post_data' ,  __NAMESPACE__ . '\\filter_post_data' , '99', 2 );

function filter_post_data( $data , $postarr ) {

  if ( in_category('instagram') ) {
    $data['post_content'] = preg_replace('/(\?.*)(?=\' style)/','', $data['post_content']);
    return $data;
  }
}

thanks

If this solved your problem, please accept the answer to close the post. Don't ask new questions in answers. — isherwood, Jun 14 '19 at 20:24

Delete part of src after a certain character

3 Answers3