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Home/ Questions/Q 5953801
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T17:52:41+00:00 2026-05-22T17:52:41+00:00

The code below is returned to an html page in an object using AJAX.

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The code below is returned to an html page in an object using AJAX.

It doesn’t seem to do what It does when I am not echoing it as PHP, which is to sort the DIVS as they are created based on the attribute ‘data-sort’

I know that the function is correct, am I perhaps escaping it incorrectly?

echo "

        newdiv = $(\'<div id=\'eventdiv\' class=\'shadow curved_sml gradient\' data-sort=\'".$row_display['TimeStamp']."\'>".$row_new['Title']." TIME: ".$row_display['TimeStamp']."</div>\');

        div_id_after = Math.floor(parseFloat(newdiv.attr('data-sort')));

        $('div[data-sort='+div_id_after+']').after(newdiv);

        ";

Regards,
Taylor

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T17:52:41+00:00Added an answer on May 22, 2026 at 5:52 pm

    Yes, you are escaping it incorrectly. What you have is:

    • first level of quotes: double ones, in PHP,
    • second level of quotes: single ones, in JavaScript,

    but you have incorrect quotes in the third level – HTML passed as string in JS, which in turn is passed as string in PHP. Also you should not escape '‘s, because you enclosed them in "‘s.

    Try the following:

    echo "
        newdiv = $('<div id=\"eventdiv\" class=\"shadow curved_sml gradient\" data-sort=\"" . $row_display['TimeStamp'] . "\">" . $row_new['Title'] . " TIME: " . $row_display['TimeStamp'] . "</div>');
    
        div_id_after = Math.floor(parseFloat(newdiv.attr('data-sort')));
    
        $('div[data-sort=\"'+div_id_after+'\"]').after(newdiv);
    ";
    

    I think this should solve your problem. But I did not test it.

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