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Home/ Questions/Q 6645069
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T00:14:28+00:00 2026-05-26T00:14:28+00:00

I have some jQuery that looks like this: $.ajax({ type: POST, url: /problems/vote.php, dataType:

  • 0

I have some jQuery that looks like this:

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



                }
                error : function() 
                {
                    alert("ajax error");
                }
            });

when I am writing my AJAX code, in case of some validation or database errors, I want to stop execution and return to the jQuery error block. What do I have to do in the PHP AJAX code in order to return things to the error block instead of the success block?

Thanks!!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T00:14:29+00:00Added an answer on May 26, 2026 at 12:14 am

    you’d have to use the success function and capture what is returned for example:

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

    you’ll only use error: if there is a problem with the AJAX request

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