How to return with jquery an php json array object?

Question

Hello i have an ajax form submit and i want to return json data. For some reason it doesnt work as it should. When data.error is return it should give me the message Email is incorect. Same for the other responses. What did i do wrong? My php has json header and also datatype is json.

  $(function() {

    $("form#login").on('submit', function(e){
    e.preventDefault();
            $.ajax({
            type: "POST",
            url: "log.php",
            data: $('form#login').serialize(),
            dataType:"json",
            success: function(data){

                    if(data.error == "yes")
                        {
                        $("#msg").html('Email is incorect.')
                        }
                    else if (data.mandatory == "yes")
                        {
                        $("#msg").html('please complete email and pass')
                        }
                        else if (data.tip =='user')
                        {

                   alert('it works'+ data.id);
                   }
                                   },

        error: function(){
            alert("failure");
            }
              });
    });
});

my php

<?php
header('Content-Type: application/json');
session_start();
include ('core/dbconfig.php');  
$password=$_POST['password'];
$usernume=$_POST['email'];
$hash = hash('sha512', $password);

if ($password=='' or $usernume=='')
{
     $arr[] = array('mandatory'  => 'yes');
     echo json_encode($arr);
}


else
{
$stmt = $dbh->prepare("SELECT * FROM Users where Email=:username and Password= :hashed");   
                $stmt->bindParam(':username', $usernume);
                $stmt->bindParam(':hashed', $hash);
                $stmt->execute();
            if ($row = $stmt->fetch())
            {
                $_SESSION['id_user']=$row['ID_User'];
                 $arr[] = array(     
                'tip'  => 'user',
                'id'   => '3'     
                                );
                echo json_encode($arr);
            }

            else

            {   
                 $arr[] = array('error'  => 'yes',);
                echo json_encode($arr);
            }

}           
?>

Do a console.log(data); and have a look at the structure of your data. It is not what you expect (hint: data is an array). See Access / process (nested) objects, arrays or JSON for more help. — Felix Kling
– Felix Kling, Commented Apr 2, 2015 at 15:23

cmorrissey · Accepted Answer · 2015-04-02 15:25:41Z

2

turn all your php instances of $arr[] = to $arr =

answered Apr 2, 2015 at 15:25

cmorrissey

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CG_DEV · Accepted Answer · 2015-04-02 15:54:34Z

               if(data.error != undefined)  ///i think this is the right way
                {
                            $("#msg").html('Email is incorect.')


                }else if(data.length == 0){

                         alert("No users available");

                }else {

                       /* 
                         you will have to do an iteration here of your
                        "data" parent object through your child objects                             
                       */

                      for(var x in data){

                          if (data[x].mandatory == "yes")
                          {
                               $("#msg").html('please complete email and pass')
                          }
                          else if (data[x].tip =='user')
                          {

                              alert('it works'+ data[x].id);

                          }

                      } //close for                        


               } //close else

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