CODEWITHSUNDEEP
phpjqueryjsonmysqli
Arjun Bhandari
Arjun Bhandari

Reputation: 299

json object in php not being read

I want to pass username and password to a php script and check in the database. On the client side I use the following script to make a json object and post it to the php file.

var myobj = {};
myobj["usrname"]= $( "#customer option:selected" ).text();
myobj["usrpass"]= $("#psw").val();

var myjson = JSON.stringify(myobj);

$.ajax({
method: "POST",
url: "checkpass.php",
data: myjson
})
.done(function( msg ) {
    alert( msg );
});

On the server side, when I see in firebug, the post is passed as

Parametersapplication/x-www-form-urlencodedDo not sort {"usrname":"XXXXXXX...
JSON

usrname "XX"

usrpass "justdoit" Source {"usrname":"XXX","usrpass":"justdoit"}

however when i run the php script to check the query the it returns an error

$usrname = $_POST['usrname'];
$usrpass = $_POST['usrpass'];

$sql = "select count(*) from glusers where EmpName='$usrname' and EmpPass='$usrpass'";
$result = $conn->query($sql);

if($result >0){
$output = 'Success';
} else
{
$output = 'fail';
}

I have tried through all the posts but cannot get this to work.

Thanks in advance.

Regards,

Upvotes: 3

Views: 151

Answers (5)

chapskev
chapskev

Reputation: 982

Echo and die the statement in order for ajax to have a success event

Js File

 var myobj = {};

        myobj["usrname"] = 'myUsername';

        myobj["usrpass"] = 'myPassword';



        $.ajax({
            type: "post",
            url: "url",
            dataType: "json",
            data: {post_data: myobj},
            contentType: "application/x-www-form-urlencoded",
            success: function (responseData) {
                console.log(responseData);
            },
            error: function (errorThrown) {
                console.log(errorThrown);
            }
        });

PHP action File

           /** if we print post we will get the following array * */
//print_r($_Post);
//die()
//Array
//(
//    [post_data] => Array
//        (
//            [usrname] => myUsername
//            [usrpass] => myPassword
//        )
//
//)

if (isset($_Post['post_data'])) {
    $myPost = $_Post['post_data'];
    $usrname = $myPost['usrname'];
    $usrpass = $myPost['usrpass'];

    $sql = "select count(*) from glusers where EmpName='$usrname' and EmpPass='$usrpass'";
    $result = $conn->query($sql);
    $num_row = $result->num_rows;

    if ($num_row > 0) {
        $output = 'Success';
    } else {
        $output = 'fail';
    }
    echo json_encode($output);
    die();
}

Upvotes: 2

Mudassar Ali
Mudassar Ali

Reputation: 126

Well actually you php code is invalid because you pass json day not name value pair so you cant get it from $_POST['username']. You need to get the whole post data and decode it like this.

$data = json_decode(file_get_contents('php://input'), true);

Now $data is dictionary array of username and password. Also sanitize your data before passing to query to avoid sql injection.

Upvotes: 0

Shijin TR
Shijin TR

Reputation: 7768

Use Json2 library as follows,

var myobj = {};
myobj["usrname"]= $( "#customer option:selected" ).text();
myobj["usrpass"]= $("#psw").val();

var myjson = JSON2.stringify(myobj);

$.ajax({
 method: "POST",
 url: "checkpass.php",
 data: myjson
})
.done(function( msg ) {
  alert( msg );
});

Upvotes: 0

Mahesh Jagdale
Mahesh Jagdale

Reputation: 174

Try This : in js file :

$(document).on("ready", function(){

            // Create an object using an object literal.
            var ourObj = {};

            // Create a string member called "data" and give it a string.
            // Also create an array of simple object literals for our object.
            ourObj.data = "Some Data Points";
            ourObj.arPoints = [{'x':1, 'y': 2},{'x': 2.3, 'y': 3.3},{'x': -1, 'y': -4}];


            var savedata = JSON.stringify(ourObj)
            $.ajax({
                type:"POST",
                url:"Users.php",
                data: {"points" : JSON.stringify(ourObj)},
               success: function(data) {
                    // Do something with data that came back. 
                    alert(data);
               }

            })
        });

In PHP File :

if (isset($_POST["points"])) {
$points = json_decode($_POST["points"]);


echo "Data is: " . $points->data . "<br>";
echo "Point 1: " . $points->arPoints[0]->x . ", " . $points->arPoints[0]->y;
}

Upvotes: 1

madalinivascu
madalinivascu

Reputation: 32354

Try this:

var myobj = '{
usrname:'+$( "#customer option:selected" ).text()+',
usrpass:'+$("#psw").val()+'
}';

or 

var myobj = {};
myobj.usrname= $( "#customer option:selected" ).text();
myobj.usrpass= $("#psw").val();

Upvotes: 0

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