json object in php not being read

Question

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,

there are no values in $usrname and $usrpass. Hence, the mysqli query executes but without any result. — Arjun Bhandari
– Arjun Bhandari, Commented May 11, 2015 at 6:44
for the incorrect password also it returns a success output on execution of the script — Arjun Bhandari
– Arjun Bhandari, Commented May 11, 2015 at 6:47
Warning: log() expects parameter 1 to be double, string given in /home/arjunmb/airtec-intl.com/checkpass.php on line 5 Success — Arjun Bhandari
– Arjun Bhandari, Commented May 11, 2015 at 6:58

chapskev · Accepted Answer · 2015-05-11 11:18:13Z

2

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();
}

edited May 11, 2015 at 11:18

answered May 11, 2015 at 6:43

chapskev

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6 Comments

Arjun Bhandari Over a year ago

did not work. although i enter the wrong password it still returns $output as success.

Maarten Peels Over a year ago

But $result, doesn't contain the actual number of count, if the query runs succesfull it wil return true, so $result=1 and is higher then 0, to get the count use mysql_num_rows

chapskev Over a year ago

It will return success since when using $result we are checking if the query wa successfull if you want to count the returned number of rows check my edited answer

Arjun Bhandari Over a year ago

thanks. got that error. but still it persists as the json object is not being parsed for the correct inputs as well

chapskev Over a year ago

are you passing the text field from a form, i edit my answer on how to pass the fields @ArjunBhandari

|

Mahesh Jagdale · Accepted Answer · 2015-05-11 07:12:04Z

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;
}

madalinivascu · Accepted Answer · 2015-05-11 06:59:25Z

0

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();

answered May 11, 2015 at 6:59

madalinivascu

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1 Comment

Arjun Bhandari Over a year ago

same problem, whether i use the correct or incorrect password. also i amended this to be $result = $result->num_rows and then checking if $result > 0

Shijin TR · Accepted Answer · 2015-05-11 07:18:59Z

0

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 );
});

answered May 11, 2015 at 7:18

Shijin TR

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Comments

Mudassar Ali · Accepted Answer · 2015-05-11 07:34:06Z

0

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.

answered May 11, 2015 at 7:34

Mudassar Ali

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