Comparing php value with javascript variable?

Question

I am working on php and javascript project. I just want to compare echo return value of php with javascript variable.

php backend code which returns 'no' using echo

 if($connection){
    $sql ="SELECT secondname FROM account WHERE email = '$email' && password = '$password'";
    $searchquery = mysqli_query($connection, $sql);
    if(!(mysqli_num_rows($searchquery) == 0)) {
        $row = mysqli_fetch_array($searchquery);
        $secondname = $row['secondname'];
        echo $secondname ;
    } else {
       echo 'no';

    }

Now comparing with javascript variable

$.post("signnin.php",{
               email: email,
               password: password,
           },
        function(data, status){
        if(data == 'no'){
          console.log('same');

        }else{
          console.log('not same');

        }
        });

it give same result if value are same or not. i also JSON.stringify but it still not working

So check what the value of data actually contains..? I would assume that the problem is because you're returning a plain text response and there's some whitespace you need to remove with trim(), eg if (data.trim() === 'no'). This is why it's always a better idea to return a serialised response. — Rory McCrossan
– Rory McCrossan, Commented Apr 17, 2020 at 11:46
You should use standard SQL AND, not &&. && is non-standard SQL and is deprecated in MySQL 8 and will be removed in a future version. — Nick
– Nick, Commented Apr 17, 2020 at 11:50
Danger: You are vulnerable to SQL injection attacks that you need to defend yourself from. — Quentin
– Quentin, Commented Apr 17, 2020 at 11:50

Quentin · Accepted Answer · 2020-04-17 11:50:30Z

1

The output from the URL probably includes white space.

JSON is a good way to normalize that, but you need to apply it at the PHP end, not the JavaScript end.

    $secondname = $row['secondname'];
    header("Content-Type: application/json");
    echo json_encode([ "secondname" => $secondname ]);
 } else {
    header("Content-Type: application/json");
    echo json_encode([ "failure" => "Login failed" ]);
 }

and then:

    $.post(
        "signnin.php",
        { email, password },
        function(data, status){
            if(data.failure){
                console.log('same');
            } else {
                console.log('not same');
            }
        }
    );

answered Apr 17, 2020 at 11:50

Quentin

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Comments

chetan · Accepted Answer · 2020-04-17 15:44:52Z

0

You are probably sending whitespaces in your PHP script. There are 2 ways to solve this -

Ensure that you put the starting php tag from the very first line and don't put the ending php tag (?>).
Instead of comparing texts, compare numbers. In php script echo 1 and in js compare data==1. This way whitespaces are automatically ignored.

answered Apr 17, 2020 at 15:44

chetan

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