Question

你好，我有一个ajax表单提交，我想返回json数据。由于某种原因，它不能正常工作。当data.error返回时，它应该给我消息Email is incorect。其他回复也一样。我做错了什么？我的php有json头，数据类型也是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");
            }
              });
    });
});

我的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);
            }

}           
?>

Answer 1

将$arr[] =的所有php实例转换为$arr =

Answer 2

               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

如何用jquery返回一个php json数组对象？

2 个答案: