douque8861 2014-06-20 11:07
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将结果从pdo发送到ajax时遇到问题

I want to do is if the user successfully registered the pdo will provide an information and send it to ajax and the ajax will message if the user is registered or not. It was working properly after i put this condition in my pdo and now it wont insert no more and ajax tells "error registering user!" all the time.

script:

<script type="text/javascript">
$(document).ready(function() {

   $('#submit').click(function (e) {
       e.preventDefault();

        var data = {};
        data.name = $('#name').val();
        data.age = $('#age').val();
        data.gender = $('#gender').val();
        data.address = $('#address').val();
        data.image = $('#imgInp').val();


        $.ajax({
            type: "POST",
            url: "user.php",
            data: data,
            cache: false,
            success: function (response) {
            if (Number(response) == 1)
                {
                alert("User successfully registered");
                }
                      else
                {
                alert("Error registering user!");
                }
            }
        });
            return false;
    });

});
</script>

user.php:

<?php
$host = "localhost";
$user = "root";
$pass = "";
$db = "test";

$dbc = new PDO("mysql:host=" . $host . ";dbname=" . $db, $user, $pass);
$dbc->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

$name = @$_POST['name'];
$age = @$_POST['age'];
$address = @$_POST['address'];
$gender = @$_POST['gender'];
$imageName = @$_FILES['image']['name'];


 $q = "INSERT INTO students(name, age, address, gender, imageName ) VALUES(:name, :age, :address, :gender, :image)";

    $query = $dbc->prepare($q);
    $query->bindParam(':name', $name);
    $query->bindParam(':age', $age);
    $query->bindParam(':address', $address);
    $query->bindParam(':gender', $gender);
    $query->bindParam(':image', $imageName);


    $results = $query->execute();
    $results ? echo "1"; : echo "2"; ;
?>
  • 写回答

1条回答 默认 最新

  • dongzongzi0379 2014-06-20 11:15
    关注

    It seems that you have error in :

    $results ? echo "1"; : echo "2"; ;
    

    yours demo

    try like this :

    echo  $results ? "1" : "2";
    

    <kbd>working demo </kbd>

    you can see here a tutorial.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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