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Phil
Phil

Reputation: 23

Passing JSON object built from SQL from PHP to JavaScript

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

Upvotes: 0

Views: 2895

Answers (1)

Jite
Jite

Reputation: 5847

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.

Upvotes: 4

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