php5 and jQuery ajax return array

Question

I'm trying to figure out how to use jQuery's .ajax() to query a MySQL table and return the results as an array. This is what I have so far:

jQuery:

$.ajax({url: 'procedures?',
    data: {Action: 'GetSuggestList' },
    dataType: "json",
    success: function(oData){
        if(oData.errormsg != null)
              alert('failed: ' + oData.errormsg);
        alert(oData.results);
}});

PHP code (with the assumption that my mysql_connect() and mysql_select_db() are valid):

 $pcAction = isset( $_REQUEST['Action'] ) ? $_REQUEST['Action'] : "" ;
 if($pcAction=='GetSuggestList'){
     $tb_name = 'suggestions';
     echo json_encode( DoGetSuggestList($tb_name) ) ;
 }

 function DoGetSuggestList($ptb_name){
     $qry_suggest = mysql_query('SELECT * FROM ' . $ptb_name . ' WHERE Active ORDER BY updated DESC');
     $rsl_suggest = mysql_fetch_array($qry_suggest);
     $jSONreturn = array("errormsg" => $jSONerror, "results" => $rsl_suggest);

     return $jSONreturn;
 };

I'm not sure what else needs to be in my AJAX success function in order to iterate through each resulting row in the array...

If you want all of the results, shouldn't you have that mysql_fetch_array call in a while loop? What does the json you're getting back look like now? (you can use console.log(oData) if you have firebug installed) — machinemessiah
– machinemessiah, Commented Apr 20, 2011 at 16:15
I would echo $pcAction... the table name $tb_name might not be set — colinmarc
– colinmarc, Commented Apr 20, 2011 at 16:20
@machinemessiah - I think you are right about having mysql_fetch_array in a loop because console.log(oData) is returning the results of just 1 record — sadmicrowave
– sadmicrowave, Commented Apr 20, 2011 at 16:23

Alfred · Accepted Answer · 2013-03-09 13:31:21Z

3

I think you need to have the results from the query dumped into an associate array before you json_encode() them.

while( ... ) {
    $a_results[] = array( 's_column' => $row['column'] );
}

return json_encode( array('errormsg' => $jSONerror, 'results' => $a_results) );

edited Mar 9, 2013 at 13:31

Alfred

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answered Apr 20, 2011 at 16:32

Donnie

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