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Home/ Questions/Q 7576815
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Editorial Team
  • 0
Editorial Team
Asked: May 30, 20262026-05-30T17:00:51+00:00 2026-05-30T17:00:51+00:00

I have an AJAX function as so: function admin_check_fn(type) { //$(#notice_div).show(); //var allform =

  • 0

I have an AJAX function as so:

function admin_check_fn(type)
{

    //$("#notice_div").show();

    //var allform = $('form#all').serialize();

    $.ajax({
        type: "POST",
        //async: false,
        url: "<?php bloginfo('template_url'); ?>/profile/adminquery_check.php",
        data: { type: type },
        //data: 'code='+code+'&userid='+userid,
        dataType: "json",
        //dataType: "html",

        success: function(result){
            var allresult = result.res
            $('#result').html(  allresult  );

            alert(allresult);
            //$("#notice_div").hide();  
        }
    })
}

And server-side:

$queryy="SELECT * FROM wp_users";
$name = array();    
$resultt=mysql_query($queryy) or die(mysql_error()); ?>

<?php while($rowss=mysql_fetch_array($resultt)){  

    $name = $rowss['display_name']; 

}

echo json_encode( array( 

    "res" =>  array($rowss['display_name']),
    "fvdfvv" => "sdfsd"

    ) 
);

Basically for some reason it is not displaying all of the returned values from the query to the users table in the database. It works when I query another table with just one entry in it, so im thinking it could be something to do with the fact there is an array it is not parsing correctly?

Just wondered if anyone else has came accross this problem?

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T17:00:53+00:00Added an answer on May 30, 2026 at 5:00 pm

    Your ajax success is treating the returned data as html but it is json.

    var allresult = result.res
     /* assumes allResult is html and can be inserted in DOM*/
     $('#result').html(  allresult  );
    

    You need to parse the json to create the html, or return html from server

    Also php loop:

    $name = $rowss['display_name']; 
    

    Should be more like:

    $name[] = $rowss['display_name']; 
    
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