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Home/ Questions/Q 893627
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Editorial Team
  • 0
Editorial Team
Asked: May 15, 20262026-05-15T14:15:09+00:00 2026-05-15T14:15:09+00:00

look at this function please $(.menu_tree img.edit).click(function() { id = this.id; lang = ‘<?=$lang_id?>’;

  • 0

look at this function please

$(".menu_tree img.edit").click(function()
    {
        id = this.id;
        lang = '<?=$lang_id?>';
        var body_width = $("body").width();
        var body_height = $("body").height();
        $("#shadow").width(body_width);
        $("#shadow").height(body_height);
        $("#shadow").show();

        var width = $("#edit_title").width();
        var height = $("#edit_title").height();
        $("#edit_title").height(0);
        $("#edit_title").width(0);
        $("#edit_title").animate(
        {
            width: width,
            height: height
        },600);
        $.post
        (
            "get_title.php",
            {id: id, lang: lang},
            function(data)
            {
                alert("qqq");
            },
            "json"
        );
    });

in get_title.php i generate json object, something like {name:"name",val:"value"}
it works fine if i don’t wrote "json", but with "json" it even doesn’t alert my qqq🙁

Any ideas?

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T14:15:10+00:00Added an answer on May 15, 2026 at 2:15 pm

    The 1.4.2 parser is more strict than earlier versions. As noted by michal, that json is not valid because the property names are not double quoted. I was bit by this issue recently when upgrading a site to jQuery 1.4.2.

    I strongly suggest allowing PHP to take care of json encoding for you. My problem, which I suspect is yours as well, was that I was putting together json strings manually in PHP, and jQuery was rejecting it because some were single quoted.

    So, for the PHP rather than something like

    echo "{name:'$val',val:'$val'}";
    exit;
    

    let PHP do the encoding:

    header('Content-type: application/json');
    echo json_encode(array('name'=>$val,'val'=>$val));
    exit;
    

    also, adding a Content-Type header for JSON can’t hurt if you aren’t already.

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