Question 40120 in PHP Question posted in
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im stuck with onerror / php | Get a space in output

Have this as a part in a ajax search page

index.php

    $(document).ready(function(){  
        $('#user').keyup(function(){  
            var query = $(this).val();  
            if(query != '')  
            {  
                $.ajax({  
                    url:"search.php",  
                    method:"POST",  
                    data:{query:query},  
                    success:function(data)  
                    {  
                        $('#postList').fadeIn();  
                        $('#postList').html(data);  
                    }  
                });  
            }  
        });  
  
    });

search.php

$output .= '<div class="bild"><img src="'.$row["URL"].'" width="200" height="300" alt="'.$row["Titel"].'" onError="this.onerror=null;this.src="loading.jpg";"/></div>';
$output .= '</div>';

The problem is that i get a space in index.php when i activate it

this.src=" loading.jpg"

Anyone have any good explanation for why this is and also im happy if someone can get me in right path 🙂

Thanks

enter image description here

Need to get away the space ..

this.src="loading.jpg"
Tags:

2

Answers


  1. 0

    Here you have double quotes inside double quotes, change one of the to single quotes or escape the ones inside:

    onError="this.onerror=null;this.src="loading.jpg";"

    Try to do this:

    onError="this.onerror=null;this.src="loading.jpg";"

    or this:

    onError="this.onerror=null;this.src='loading.jpg';"

    or this:

    onError="this.onerror=null;this.src='loading.jpg';"

    One of them should fix it, just try with your code which one is best.

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  2. 0

    You quotations marks for loading.jpg are not escaped, thereby it is treating the loading.jpg as an attribute of the div tag and not as a src parameter of this. Replace the quotes around loading.jpg with ' and it should work.

    I quickly typed the example in an interactive PHP shell to show you the effect.

    First what your result was (you see that loading.jpg is in a different color highlighted):

    php > $output = '<div class="bild"><img src="'.$row["URL"].'" width="200" height="300" alt="'.$row["Titel"].'" onError="this.onerror=null;this.src="loading.jpg";"/></div>';
php > echo($output);
<div class="bild"><img src="http:demo.com" width="200" height="300" alt="Demo" onError="this.onerror=null;this.src="loading.jpg";"/></div>

    And now my fix:

    php > $output = '<div class="bild"><img src="'.$row["URL"].'" width="200" height="300" alt="'.$row["Titel"].'" onError="this.onerror=null;this.src='loading.jpg';"/></
div>';
php > echo($output);
<div class="bild"><img src="http:demo.com" width="200" height="300" alt="Demo" onError="this.onerror=null;this.src='loading.jpg';"/></div>
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