returning multiple variables in JSON/PHP

Question

Basically, i'm using an JSON response from a local PHP file to append data inside a div, currently it's only returning one variable, in which the code i'm using is shown here :

    $.ajax({
        url: "functions/ajax.php",
        data: "func=auto",
        type: "GET",
        dataType: "json",
        success: function(data){
            $.each(data.search, function (i, v) {
                    $div = $('.bs-docs-example');
                    $div.append('+ v +');
            });
        },
        error: function (jqXHR, textStatus, errorThrown){
            console.log('Error ' + jqXHR);
        }
    });

The code which is returning the JSON is currently

while($row = mysql_fetch_assoc($res)) {
     $search[] = $row['number'];
}
echo json_encode($search);

What i want to do is something like this

while($row = mysql_fetch_assoc($res)) {
     $search[] = $row['number'];
     $search[] = $row['name'];
}
echo json_encode($search);

But if i was to do the following, how would i access both number and name since in the success on the jQuery it's parsing using $.each(data.search, function (i, v) { which means the original $search[] = $row['number'] is now stored inside v

would i do something like v['name'] v['number'] or is that completely wrong ?

Adrian · Accepted Answer · 2013-03-13 20:56:50Z

2

You'll want to pass an associative array to json_encode:

$search['number'] = $row['number'];
$search['name'] = $row['name'];
json_encode($search);

Or, if there will be multiple rows of results, as an array of associative arrays:

while($row = mysql_fetch_assoc($res)) {
    $search[] = $row;
}

And then access the values in JavaScript by name:

var name = v['name'];

Or

var name = v[0]['name']

If you're using multiple rows of results.

answered Mar 13, 2013 at 20:56

Adrian

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3 Comments

jdp Over a year ago

You could just drop the while loop and do json_encode(mysql_fetch_assoc($res)), I believe, assuming there isn't a reason to only pass in those two columns specifically.

Adrian Over a year ago

Yes you could; I left the loop in case some preprocessing needed to be done to the rows, e.g. if not all returned columns are desired in the JSON output.

Curtis Crewe Over a year ago

Perfect exactly, with little but very exact explanations +1

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