Getting particular data from json_encode response

Question

I have a file ajax.php which I am using to process data passed through jquery. I have this particular line of code called on successful form verification:

$.post("/ajax.php",{'request': 'emailLogin', 'loginmail': mail, 'loginpass': pass}, function(data) {}  );

data in my case is: {"valid":true}{"auth":false}which is returned as a response from ajax.php, but I can't seem to file the correct way of defining "auth" and a variable with value "false".

My ajax.php is just checking if login and password are in the database and than echo json_encode(array('auth' => false)); or echo json_encode(array('auth' => true)); depending on the result. But it has also contain these lines:

if( isset($_POST['loginmail'])) {
    $usermail = htmlspecialchars($_POST['loginmail']);

    if (!filter_var($usermail, FILTER_VALIDATE_EMAIL)) {
        $response = array('valid' => false, 'message' => 'You did not enter a correct email address.'); 
    } else {                      
        // All good
        $response = array('valid' => true);
    }
}
echo json_encode($response);

Your data seems to be returning two JSON objects instead of one. Once you fix that, have you set the dataType to json in ajaxSetup? api.jquery.com/jQuery.ajaxSetup — Zach Rattner
– Zach Rattner, Commented Feb 6, 2015 at 1:00
You should post your ajax.php since the problem seems to be there and not in your JS — Machavity
– Machavity ♦, Commented Feb 6, 2015 at 1:01
That is correct. One is used to confirm validation, another one authentication. — Ivan Venediktov
– Ivan Venediktov, Commented Feb 6, 2015 at 1:01
@IvanVenediktov You can only return one JSON object. You need to combine them into a single array, and then call json_encode() on that. — Barmar
– Barmar, Commented Feb 6, 2015 at 1:02
Can you please include your php code that generates the response data? — Jonathan Crowe
– Jonathan Crowe, Commented Feb 6, 2015 at 1:02

Barmar · Accepted Answer · 2015-02-06 01:29:27Z

1

Don't echo json_encode($response) separately from the authentication result, you need to combine them. After you do the authentication, do:

$response['auth'] = $result_of_authentication;

then do

echo json_encode($response);

Once you do this, you should be able to access data.auth in Javascript. You should tell $.post that it's returning JSON:

$.post("/ajax.php",{
        'request': 'emailLogin', 
        'loginmail': mail, 
        'loginpass': pass}, 
    function(data) {
        alert(data.auth);
    },
    "json");

edited Feb 6, 2015 at 1:29

answered Feb 6, 2015 at 1:17

Barmar

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1 Comment

Ivan Venediktov Over a year ago

I did that, good man, now my response is: {"valid":true,"auth":"false"}

Jonathan Crowe · Accepted Answer · 2015-02-06 01:26:23Z

0

Based on your PHP code you should be able to access the valid attribute like so:

$.post("/ajax.php",{'request': 'emailLogin', 'loginmail': mail, 'loginpass': pass}, function(data) {
    var auth = data.valid;
    if (auth) {
        // do something!
    } else {
        // do something else!
    }
});

Also there is a bug in your PHP code, you need to set up a default value for $response like so:

$response = array('valid' => false, 'message' => 'Email address is required');

if( isset($_POST['loginmail'])) {
    $usermail = htmlspecialchars($_POST['loginmail']);

    if (!filter_var($usermail, FILTER_VALIDATE_EMAIL)) {
        $response = array('valid' => false, 'message' => 'You did not enter a correct email address.'); 
    } else {                      
        // All good
        $response = array('valid' => true);
    }
}

echo json_encode($response);

Otherwise if $_POST['loginmail'] is not set your app with throw an undefined variable exception

EDIT:

As Barmar pointed out in his answer, you should only echo a response back once time to avoid creating an invalid response. Any data you need should be sent back in a single array. I don't see you doing that in your PHP code but you do make mention of echoing another array ['auth' => false] which will not work the way you want it to

edited Feb 6, 2015 at 1:26

answered Feb 6, 2015 at 1:16

Jonathan Crowe

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4 Comments

Barmar Over a year ago

This won't work until he fixes the problem with the invalid JSON.

Jonathan Crowe Over a year ago

I'm not sure from his code where invalid JSON would be coming from. Hopefully his example data was just incorrect

Jonathan Crowe Over a year ago

Oh I think I see what you are talking about, in his question he does mention echoing json_encodeded data twice...

Ivan Venediktov Over a year ago

@JonathanCrowe I did what Barmar said and tried your jquery code, but it doesn't seem to get data out of JSON.

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