Passing JSON object built from SQL from PHP to JavaScript

Question

I am trying to pass a JSON object from PHP to Javascript. the object is being filled from an SQL Database here is the PHP code I am using.

<?php
    $conn = mysql_connect("localhost","root","");
    if(! $conn )
    {
         die('Could not connect: ' . mysql_error());
    }
    mysql_select_db('db') or die( 'Error'. mysql_error() );
    $query = "SELECT * FROM products;";
    $results = mysql_query($query, $conn);
    $return = array();

    while($result = mysql_fetch_assoc($results))
    {
        $mount = array('product_code' => $results['product_code'], 'colour'  =>    $results['colour'], 'price' => $results['price']);
        array_push($return, $mount);
    }

    return json_encode($return);
?>

I have changed a few of the variable names but the same functionality is there.

now when I try to do an AJAX Get to this .php file it crashes at the JSON.Parse part code shown below:

$.get("JSON.php", function(data) {
    var JSONdata = JSON.parse(data);
    alert(JSONdata[0].colour);
});

My alert is there just for testing. I Understand the problem may lie in my building of the $return array. Rather new to JSON, any help would be greatly appreciated.

EDIT: taking all the information from below I have corrected my PHP code to look as such.

<?php
    $conn = mysql_connect("localhost","root","");
    $error = array("result" => false, "error" => mysql_error());

    if(! $conn )
    {
        die(json_encode($error));
    }
    mysql_select_db('db') or die(json_encode($error));
    $query = "SELECT * FROM products;";
    $results = mysql_query($query, $conn);
    $return = array();

    while($result = mysql_fetch_assoc($results))
    {
        $mount = array('product_code' => $result['product_code'], 'colour'  => $result['colour'], 'price' => $result['price']);
        array_push($return, $mount);
    }

    echo json_encode($return);
?>

I'm Looking into changing the mysql_* functions to new more compatible versions

put alert(JSONdata[0].colour); before you parse. And usually, we need to print response to JSON request, not return it. — user4035
– user4035, Commented Sep 17, 2013 at 9:11
Please, do not use the mysql_* functions anymore. With the release of PHP 6 they will be deprecated. Look into the PDO object in PHP! — stUrb
– stUrb, Commented Sep 17, 2013 at 9:15
thank you for the heads up and surely you need to parse the data before it can be used in the alert? — Phil
– Phil, Commented Sep 17, 2013 at 10:03

Jite · Accepted Answer · 2013-09-17 09:33:36Z

4

You are using 'return' at the end of the php script, echo the json encoded string :

echo json_encode($return);

Also, it might be a good idea to set the contenttype header to application/json.

Edit:

If you script fails, you use die('error'.mysql_error());
This will also be a response to the ajax call, but not json, and the parse function will throw an exception.
I would recommend returning a json object instead, like:

 $error = array("result" => false, "error" => mysql_error());
 die(json_encode($error));

Also as stated in comments, do not use mysql_* functions, they are deprecated in later php versions and will be gone from php all together in future releases.
Check out mysqli or even better, PDO.

edited Sep 17, 2013 at 9:33

answered Sep 17, 2013 at 9:14

Jite

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4 Comments

Phil Over a year ago

Thanks for the update on mysql_* functions, will change accordingly. i have taken your advice and replaced the die methods, also I must have been in .net mode for the return, that has been corrected to print_r now. I have commented out the parse and added alert(data) and now i am getting a parse error "Syntax error, unexpected T_STRING in filename at line 21" that being the new print_r line. ps. i also added the content-type now thank you for reminding me.

Phil Over a year ago

just noticed as well that I am using $results and not $result within the while loop but I still get the same error after correction.

Jite Over a year ago

How does line 21 look?

Phil Over a year ago

Print_R json_encode($return); i have not changed it to an echo and it works.

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