1

I'm trying to get my code to:

  • Get a post's thumbnail url
  • or (if no thumbnail there) to
  • Echo a default url

I'm not sure if I am going the easiest way or right way about this.

This is the code so far. $postid being the attribute for the id of the post I want the url from.

<?php
// Add Shortcode
function friend_pic( $atts , $content = null ) {

    // Attributes
    extract( shortcode_atts(
        array(
            'postid' => '',
        ), $atts,
            'f-p'
        )
    );

    $postid = $atts['postid'];


ob_start();

// shortcode contents from here

 echo "<img src=\"";
     if ( $postid ) {
      $thumb_url = wp_get_attachment_url( get_post_thumbnail_id( $postid, 'thumbnail', false) );
      $thumb_url = $thumb_url[0];
      }else{
      echo "http://www.website.com/default.jpeg";
     }                          
    echo "\">"; 

// Shortcode edit ends here

$output = ob_get_contents();
ob_end_clean();   

    /// FINAL OUTPUT ////
    return $output;
}
add_shortcode( 'f-p', 'friend_pic' );

**Aditional Details* The point of this is to display the pictures of my friends by using a shortcode in my posts. The pages of my website are about my family and friends, and when I want to reference one I just wanna show the thumb of their page, its their picture, and if there isnt a thumb then show a default picture (a N/A picture).

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3
  • can you post the full code of the shortcode function? btw, shortcode functions should not echo the result. Commented Jul 10, 2014 at 21:16
  • Hi Michael. I didn't add it all cuz the rest of the shortcode don't affect this. And the end of the code is not echoed. I can add it if you want. However all this current code does it trying to find the thumb of a wp post and echo the URL if there isn't to echo a default one. Commented Jul 11, 2014 at 1:09
  • Have to disagree with the reason this question was put on hold. "specific to wordpress" is clearly met if he's trying to use a wp function. I could be wrong. Commented Jul 11, 2014 at 2:00

2 Answers 2

Reset to default
1

I checked it out further and there were a couple of problems with your code.

  • your if statement checks for a post ID, but you don't have a check for whether or not there is actually a featured image set, so I added a condition to the if statement
  • your closing bracket for get_post_thumbnail_id() was misplaced
  • you were not actually echoing $thumb_url
  • wp_get_attachment_url() returns the image src (not an array) and does not take any additional parameters beyond the attachment ID, which meant that $thumb_url[0] was returning just the first letter of the image src. Looks like you meant to use wp_get_attachment_image_src() which takes an image size parameter. (See codex: Function Reference/wp get attachment url, Function Reference/wp get attachment image src)

Here my modified version of your code:

echo "<img src=\"";
if ( $postid && get_post_thumbnail_id( $postid)) {
  $thumb_url = wp_get_attachment_image_src( get_post_thumbnail_id( $postid), 'thumbnail', false );
  $thumb_url = $thumb_url[0];
  echo $thumb_url;
}
else{
  echo "http://www.website.com/default.jpeg";
}                          
echo "\">";

I would add that in the future, a good rule of thumb when coding is to take things a step at a time, rather than writing a whole bunch of code at once. Do it one line at a time and make sure that your variables contain the expected output before moving on to the next step. Otherwise, you can end up with a chunk of code like this that actually has problems in several spots and it can be difficult and/or time consuming to try and figure out why it's not working.

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1
  • Its working perfect. I thank you for your time and the explanations. You're right about my coding "behaviour", its a bad habit I have, definitely learning that doing it right works wonders. Thanks again Kisabelle Commented Jul 11, 2014 at 23:58
1

You need to escape the quotes in your echo statements:

echo "<img src="";

should be changed to:

echo "<img src=\"";

or you could use single quotes so that you don't have to escape the double-quotes:

echo '<img src="';

Similarly, your last line of code should be changed to:

echo "\">";
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1
  • Thx Kisabelle. Already are there. I just didn't add them to my code on the site😊 sry! Commented Jul 11, 2014 at 1:05

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