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Home/ Questions/Q 6652857
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T01:11:36+00:00 2026-05-26T01:11:36+00:00

I have some PHP AJAX code that is supposed to validate some parameters sent

  • 0

I have some PHP AJAX code that is supposed to validate some parameters sent by jQuery and return some values. Currently, it consistently returns invokes the jQuery error case, and I am not sure why.

Here is my jQuery code:

$('.vote_up').click(function() 
{        
    alert ( "test: " + $(this).attr("data-problem_id") );
    problem_id = $(this).attr("data-problem_id");

    var dataString = 'problem_id='+ problem_id + '&vote=+';

    $.ajax({
                type: "POST",
                url: "/problems/vote.php",
                dataType: "json",
                data: dataString,
                success: function(json)
                {           
                    // ? :)
                    alert (json);


                },
                error : function(json) 
                {
                alert("ajax error, json: " + json);

                //for (var i = 0, l = json.length; i < l; ++i) 
                    //{
                    //  alert (json[i]);
                    //}
                }
            });


    //Return false to prevent page navigation
    return false;
});

and here is the PHP code. The validation errors in PHP do occur, but I see no sign that the error that is happening on the php side, is the one that is invoking the jQuery error case.

This is the snippet that gets invoked:

if ( empty ( $member_id ) || !isset ( $member_id ) )
{
    error_log ( ".......error validating the problem - no member id");
    $error = "not_logged_in";
    echo json_encode ($error);
}

But how do I get the “not_logged_in” to show up in my JavaScript of the jQuery so that I know it is the bit that got returned? And if it isn’t, how do I make it that that error is what comes back to the jQuery?

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T01:11:38+00:00Added an answer on May 26, 2026 at 1:11 am

    Don’t echo $error in the json_encode() method just simply echo $error like so. Also, don’t use the variable json, use the variable data. Edited code below:

    PHP

    if ( empty ( $member_id ) || !isset ( $member_id ) )
    {
        error_log ( ".......error validating the problem - no member id");
        $error = "not_logged_in";
        echo $error;
    }
    

    jQuery

    $('.vote_up').click(function() 
    {        
        alert ( "test: " + $(this).attr("data-problem_id") );
        problem_id = $(this).attr("data-problem_id");
    
        var dataString = 'problem_id='+ problem_id + '&vote=+';
    
        $.ajax({
                    type: "POST",
                    url: "/problems/vote.php",
                    dataType: "json",
                    data: dataString,
                    success: function(data)
                    {           
                        // ? :)
                        alert (data);
    
    
                    },
                    error : function(data) 
                    {
                    alert("ajax error, json: " + data);
    
                    //for (var i = 0, l = json.length; i < l; ++i) 
                        //{
                        //  alert (json[i]);
                        //}
                    }
                });
    
    
        //Return false to prevent page navigation
        return false;
    });
    
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