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 $.parseJSON(result); statement seems to be giving me this error:
Uncaught SyntaxError: Unexpected end of JSON input
I also tried to use serialize(), but I get an error message:
Uncaught exception 'PDOException' with message 'You cannot serialize or unserialize PDO instances'
Basically, My database is in my object, I'm guessing I can't serialize it...
What am I doing wrong here?
Thank you
3 Answers 3
In getPackage.php page :
echo json_encode($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);
?>
10 Comments
getPackage.php directly not help jquery just for test. and let me know can you get any data?Ajax expects the JSON inside the request body.
Use
die(json_encode($SOME_ARRAY_OR_OBJECT));
not "return" json_encode($SOME_ARRAY_OR_OBJECT)
Comments
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 );
2 Comments
mysql_ extensions. Also, this answer really misses the point of the question and makes a lot of assumptions about OP's use-case.
$packageobject?getInstanceByPackageSerial()function is returning either isn't what you think it is, or the way you're trying to do this isn't the right way. It would be helpful to know what your end objective is here. What do you end up trying to do withresin the javascript code? Third, you are usingasync: falsein your ajax setup. This is generally not considered a good thing to do..successfunction, why don't you log theresultto see what the actual return object is. Whatever PHP is returning is not correctly formatted for JSON.