Passing array values from php to javascript data returns only one value

Question

This function works fine, but it only returns one value. What I'm missing here?

Here's my code:

  <script type='text/javascript'>
$(function () {

<?php
//Get the names and id's
$get_info=mysql_query("select * from table where id = '1'");

if($get_info){
while($row_info=mysql_fetch_array($get_info))
{
$username=$row_info['name'];
$user_id=$row_info['profile_id'];
?>


onDataRequest:function (mode, query, callback) {
 var data = [

{ id:<?php echo $user_id;?>, name:'<?php echo $username;?>', 'avatar':'http://cdn0.4dots.com/i/customavatars/avatar7112_1.gif', 'type':'contact' }

 ];

     data = _.filter(data, function(item) { return item.name.toLowerCase().indexOf(query.toLowerCase()) > -1 });
     callback.call(this, data);
    }
  });

<?php
}}
?>
});
</script>

Expected result:

{
    id: 295,
    name: 'Mike',
    'avatar': 'http: //cdn0.4dots.com/i/customavatars/avatar7112_1.gif',
    'type': 'contact'
},
{
    id: 296,
    name: 'John',
    'avatar': 'http: //cdn0.4dots.com/i/customavatars/avatar7112_1.gif',
    'type': 'contact'
}

Actual output:

{
    id: 295,
    name: 'Mike',
    'avatar': 'http: //cdn0.4dots.com/i/customavatars/avatar7112_1.gif',
    'type': 'contact'
}

It just returns 1 item.

Marc B · Accepted Answer · 2013-04-13 22:25:15Z

2

You database fetch loop isn't preserving each fetched value. You simply overwrite the previous fetch call with the current fetch. perhaps you meant something like this:

while ($row_info = mysql_fetch_asssoc($result)) {
    $username[] = $row_info['name'];
    $user_id[] = $row_info['profile_id'];
}

The [] notation on the vars tells PHP to treat the vars as array and push the new values into the array.

You'd then insert the arrays into javascript with:

var usernames = <?php echo json_encode($username); ?>;
var user_ids = <?php echo json_encode($user_id); ?>;

answered Apr 13, 2013 at 22:25

Marc B

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4 Comments

pilsetnieks Over a year ago

No, the loop isn't finished when they're outputting the values, so they should be both shown. Edit: on second thought, it's a javascript problemm as in they're overwriting the onDataRequest event every time.

Aaron Over a year ago

This doesn't seem to work, or maybe I'm not able to integrate your answer with my existing code, I don't know...

Marc B Over a year ago

well, since your onDataRequest is inside the loop, it'll be output on EVERY iteration. that means you end up with the equivalent of a many-element array all sharing the same key - last entry in the list "wins".

Aaron Over a year ago

I thought this could be done with a simple php to js array conversion, but I guess I was wrong. Anyway thanks. I still can't get it working though.

pilsetnieks · Accepted Answer · 2013-04-13 22:40:03Z

2

Every time you loop, you are assigning new values for your onDataRequest javascript function, however, it's the same function every time. You should think about execution order - first, you do PHP&MySQL server-side and what is generated as HTML or Javascript code there gets rendered and/or executed client-side later.

Basically, you have a PHP loop that goes over a set of values and those values are put inside a Javascript code that is within a HTML code block. So whatever you fetch from your database server last, is what is actually executed client-side.

The main idea here is that you shouldn't mix your server-side PHP code with client-side HTML or Javascript code.

answered Apr 13, 2013 at 22:40

pilsetnieks

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1 Comment

Aaron Over a year ago

I kinda knew that, but this is only for testing, good advice tho.

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