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Having trouble using ajax to retrieve info from a database. I was getting the code from here:

  1. http://openenergymonitor.org/emon/node/107 the code from here worked but would only output 1 item
  2. Simple Ajax Jquery script- How can I get information for each of the rows in the table? I tried to add this to my code but I could not get it to work.

I am getting everything from a table with the following php:

$result = mysql_query("SELECT * FROM voterecords");  

$data = array();
while ( $row = mysql_fetch_row($result) )
{
  $data[] = $row;
}
echo json_encode($data);

which outputs the following if I navigate to that php page:

[["68","1234","0","1234",""],["69","added with ajax","0","this item was added using ajax",""]]

The format of the above is as follows: id, title, votes, description, owner

I think that bit all works but I cant be sure because i dont know what JSON is supposed to look like.

Ok now here is the jquery which is supposed to retrieve the info from the JSON and put it into the html element #output

$(function () 
  {

    $.ajax({                                      
      url: 'retrieve.php', data: "", dataType: 'json',  success: function(rows)
      {
        for (var i in rows)
        {
         var row = rows[i];          

          var id = row[0];              
          var name = row[1];
          var votes = row[2];  
          var info = row[3]; 

          $('#output').append("<b>id: </b>"+id+"<b> name: </b>"+name+"<b> votes: </b>"+votes+"<b> info: </b>"+info)
                      .append("<hr />");
    } 
  } 
});

I was expecting this to output all the info but nothing happens.

Community
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cohen
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1 Answers1

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Your code is fine except you have a missing closing ) from the callback function.

Also, in JavaScript, it's better to place opening braces on the same line, not on the next, as is common in some other languages.

Corrected/cleaned-up code:

$(function () {
    $.ajax({url: 'retrieve.php', dataType: 'json'}).done(function(rows) {
        for (var i in rows) {
            var row = rows[i];
            var id = row[0];
            var name = row[1];
            var votes = row[2];
            var info = row[3];
            $('#output')
                .append("<b>id: </b>"+id+"<b> name: </b>"+name+"<b> votes: </b>"+votes+"<b> info: </b>"+info)
                .append("<hr />");
        }
    });
});
Mitya
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  • Always check the console if something isn't working as expected in JS - it will tell you have syntax errors like this and where the error (may) have occurred. I've edited my post to include a cleaned-up, corrected version of your code. Also note you're not really making any saving by committing the various JSON properties to local variables. If you do, though, consider a multi-var pattern (Google it) rather than specifying `var` each time. – Mitya Jul 15 '12 at 09:24
  • I have not done any JS or jquery before so ill look up how to debug it to stop this from happening in the future. I get what you mean with saving JSON properties to variables, I could just use the JSON variable itself. Its not my code so Ill probably look at it more closely once its working and take out those redundant bits. Thanks for your help – cohen Jul 15 '12 at 10:08