CODEWITHSUNDEEP
javascriptphpjqueryajax
Mathieu Turcotte
Mathieu Turcotte

Reputation: 354

returning a PHP object from an ajax file to my javascript code

So I'm still a newbie when it comes to javascript and php. I am having this issue:

from javascript, I scan a package's barcode using a barcode reader. I send this to an PHP file using ajax, Which builds an object, and needs to return it to my javascript code.

I'm doing this:

function LoadPackage(ScannedCode) {
    var res;

    console.time("Load package " + ScannedCode);
    $.ajax({
        type: "POST",
        url: "ajax/3gmodule_inventory_ajax/getPackage.php",
        data: "packageSerial=" + ScannedCode,
        cache: false,
        async: false //inline operation, cannot keep processing during the execution of the AJAX
    }).success(function(result) {
        res = $.parseJSON(result);
    });

    console.timeEnd("Load package " + ScannedCode);

    return res;
}

The php file:

    <?php
        include_once "../../init.php";

        $packageSerial = $_POST["packageSerial"];

        $package = tbProductPackage::getInstanceByPackageSerial($packageSerial, $db);
        return json_encode($package);
// edit: first part of the problem was here, I was supposed to ECHO here. not RETURN.
    ?>

I am 100% certain my object gets built properly. I did do a var_dump of my $package object, and everything is fine with it. However, when trying to get it back to javascript, I tried a bunch of different things, nothing works.

The $.parseJSON(result); statement seems to be giving me this error:

Uncaught SyntaxError: Unexpected end of JSON input

I also tried to use serialize(), but I get an error message:

Uncaught exception 'PDOException' with message 'You cannot serialize or unserialize PDO instances'

Basically, My database is in my object, I'm guessing I can't serialize it...

What am I doing wrong here?

Thank you

Upvotes: 4

Views: 6012

Answers (3)

Razib Al Mamun
Razib Al Mamun

Reputation: 2713

In getPackage.php page : 

echo json_encode($package);

not use return

In Jquery should be : 

data: {packageSerial:ScannedCode},

After success not need $.parseJSON( because getPackage.php already retrieve json encode

so, is should be : 

}).success(function(result) {
    res = result
});

also add dataType: 'json', after data: {packageSerial:ScannedCode},

So, Final correction code is :

Jquery : 

function LoadPackage(ScannedCode) {
    var res;

    console.time("Load package " + ScannedCode);

    $.ajax({
        context: this,
        type: "POST",
        url: "ajax/3gmodule_inventory_ajax/getPackage.php",
        data: {packageSerial:ScannedCode},
        dataType: 'json',
    }).success(function(result) {
        res = result;
    });

    console.timeEnd("Load package " + ScannedCode);

    return res;
}

PHP : 

<?php
    include_once "../../init.php";

    $packageSerial = $_POST["packageSerial"];

    $package = tbProductPackage::getInstanceByPackageSerial($packageSerial, $db);
    echo json_encode($package);
?>

Upvotes: 7

V&#237;tor Oliveira
V&#237;tor Oliveira

Reputation: 2091

Ajax expects the JSON inside the request body.

Use 

die(json_encode($SOME_ARRAY_OR_OBJECT));

not "return" json_encode($SOME_ARRAY_OR_OBJECT)

Upvotes: 0

Syl OR
Syl OR

Reputation: 127

I don't know how your database is done, but as a first step, you could make a SELECT query and fetch it as an array. Then, json_encode would work without problem. Something along the lines of:

$vec = array();
$query="SELECT * From YourTable
                    WHERE 1";

$result= mysql_query($query) or die(mysql_error() . $query)

while ($r=mysql_fetch_array($result)) {
   $vec [] = $r;
}

$toreturn=json_encode($vec );

Upvotes: -1

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