CODEWITHSUNDEEP
javascriptphpjqueryajaxjson
Venelin
Venelin

Reputation: 3308

jQuery - Ajax is not returning json response

Here is my JavaScript:

                    $.ajax({
                        url: 'CheckColorPrice.php',
                        type: 'POST',
                        data: {
                            url: '<?php echo $LINK;?>',
                            ColorId: ColorNumber
                        },
                        dataType: 'json',
                        success: function (data) {
                            $('#LoadingImage').hide();
                                $("#PRICE").text("£ " + data["price"]);                             
                        }
                    });

Here is CheckColorPrice.php:

<?PHP
$url = $_POST['url'];
$ColorId = $_POST['ColorId'];       
if(isset($_POST['url']))
{   
    libxml_use_internal_errors(true); 
    $doc = new DOMDocument();
    $doc->loadHTMLFile($url);

    $xpath = new DOMXpath($doc);

    $DataVariants = $xpath->query('//span[@class="ImgButWrap"]/@data-variants')->item(0)->nodeValue;

    $jsonStart = strpos($DataVariants, '[');
    $jsonEnd = strrpos($DataVariants, ']');

    $collections = json_decode(substr($DataVariants, $jsonStart, $jsonEnd - $jsonStart + 1));   

    foreach ($collections as $item) {
        $ColVarId = $item->ColVarId;

        $SizeNames = [];
        $SellPrice = [];
        foreach ($item->SizeVariants as $size) {
            $SizeNames[] = $size->SizeName;
            $SellPrice[0] = $size->ProdSizePrices->SellPrice;
        }
        $names = implode(',', $SizeNames);
        $price = implode('', $SellPrice);

    if($ColVarId == $ColorId){
                $healthy2 = array('£',' ','Â');
                $yummy2   = array('','','');
                $price = str_replace($healthy2, $yummy2, $price);
                $PRICE = $price;
            echo "price: ",  json_encode($PRICE), "\n";

    }
}

}   

?>

The result from CheckColorPrice.php is looking just like this:

price: "37.99"

Where is my mistake, why it is not receiving the response properly. I don't get it at all... Can you help me out ?

Thanks in advance!

Upvotes: 1

Views: 2873

Answers (3)

QuentinG
QuentinG

Reputation: 101

First thing, in your ajax request, add this parameter :

dataType : 'json'

Then, your response is not a correct json. You can return this :

echo json_encode(array("price"=>$PRICE));

Upvotes: 0

Marc B
Marc B

Reputation: 360572

You're not returning json. You're returning plain text which contains some json:

        echo "price: ",  json_encode($PRICE), "\n";
             ^^^^^^^^^^

That'd look like

    price: "$9.99"

which is NOT valid json.

You need to return an array for your JS code to work:

echo json_encode(array('price' => $PRICE));

which would output:

{"price":"$9.99"}

Upvotes: 4

Nik Chankov
Nik Chankov

Reputation: 6047

Add the following header to your script:

header('Content-type: application/json'); header("Content-Disposition: inline; filename=ajax.json");

Also change the line

echo "price: ",  json_encode($PRICE), "\n";

to

echo json_encode(array('price'=>$PRICE));

Hope that helps

Upvotes: 2

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