CODEWITHSUNDEEP
javascriptphpjqueryajax
user3009442
user3009442

Reputation: 21

Simple PHP to Ajax 'not logged in' error - how?

Currently I am using this to send success or failure to ajax:

$response = array( 'success' => true );
wp_send_json_success($response);

However, say the user is not logged in and I would like to end the php function right there, and send the error message that would prompt a login modal to popup. How would I do that? Like:

if(!is_user_logged_in)) {
wp_send_json_error('not logged in');
die()
}

or I could use the more common json_encode($response);

if(!is_user_logged_in)) {
echo json_encode($response);
die()
}

but how would this handler even know an error occurred? If I put all my options for success and errors into an array, how do I die() the function while still returning the json response? and how do I tell php which response to send?

Once it reached my ajax call, I am confident I can figure it out. Something like this is what I read:

var dataArray = jQuery.parseJSON(data);

Some examples would be of great help. Thank you in advance!

Upvotes: 1

Views: 281

Answers (2)

Razvan Pocaznoi
Razvan Pocaznoi

Reputation: 409

you can use json_encode and make your own protocol like this:

if(isset($_POST['operation']) && $_POST['operation'] == 'is_user_logged_in'){
    if(!is_user_logged_in()){
        $response->success = false;
        $response->msg = "not logged in";
    }else{
        $response->success = true;
    }
    echo json_encode($response);
    exit;
}

Then in your ajax:

$.post('AJAX-URL-HERE',{"operation":"is_user_logged_in"},function(response){
    if(response.success){
         //do whatever
    }else{
         var message = response.msg
         //pop up the login modal
    }
},'json').error(function(){alert('Call error')});

Upvotes: 0

Daniel W.
Daniel W.

Reputation: 32340

This should work:

if(!is_user_logged_in()) {
    $response = array(
               'loggedIn' => false,
               'alertMsg' => 'not logged in :,-(',
    );
} else {
    $response = array(
               'loggedIn' => true,
               'success'  => true,
               'alertMsg' => 'test!',
    );

}
wp_send_json_success($response);
die();

The Ajax call is something like:

jQuery.get('pathToYourJsonFunctionURL', function(data) {
    if (data.loggedIn == true) {
        jQuery('#divForLoggedInStatus').text('You are logged in');
    } else {
        jQuery('#divForLoggedInStatus').text('You are NOT logged in');
    }

    window.alert(data.alertMsg);
});

When you return JSON data, you do not use parseJSON (the data is already (parsed) json). data.loggedIn is the same as in PHP returned array('loggedIn' => value);

Upvotes: 1

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