Question

我创建了一个提交到php页面的AJAX表单。此表单已成功提交，我收到了我的数据库中的电子邮件地址，并且我收到了一封确认电子邮件。

但是，我收到一条警告信息“Error |”，显然它来自于：

error: function (xhr, textStatus, errorThrown) {
            alert(textStatus + "|" + errorThrown);

我不确定为什么错误会在它发生时抛出。另一件事，这个表单正在重新加载页面。我有event.preventDefault();，那么为什么页面会重新加载？

我感谢任何帮助。

<form action="" method="POST">
  <input type="email" id="footer-grid1-newsletter-input" placeholder="Your Email Address">
  <input type="submit" id="footer-grid1-newsletter-submit" name="submit">
</form>

$(document).ready(function(){ 
    $("#footer-grid1-newsletter-submit").on("click", function () {
        event.preventDefault();

        var newsletter_email = $("#footer-grid1-newsletter-input").val();

        $.ajax({ 
            url: "newsletterSend.php", 
            type: "POST",
            data: {
                "newsletter_email": newsletter_email
            },
            success: function (data) {
            //  console.log(data); // data object will return the response when status code is 200
                if (data == "Error!") {
                    alert("Unable to insert email!");
                    alert(data);
                } else {
                    /*$(".announcement_success").fadeIn();
                    $(".announcement_success").show();
                    $('.announcement_success').html('Announcement Successfully Added!');
                    $('.announcement_success').delay(5000).fadeOut(400);*/
                }
            },
            error: function (xhr, textStatus, errorThrown) {
                alert(textStatus + "|" + errorThrown);
                //console.log("error"); //otherwise error if status code is other than 200.
            }
        });
    });
});

ini_set('display_errors', 1);
error_reporting(E_ALL);

$newsletter_email = $_POST['newsletter_email'];

try {

    $con = mysqli_connect("localhost", "", "", "");

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    $stmt = $con->prepare("INSERT INTO newsletter (email, subscribed) VALUES (?, NOW())");
        if ( false===$stmt ) {
            die('Newsletter email prepare() failed: ' . htmlspecialchars($con->error));
        }
    $stmt->bind_param('s', $newsletter_email);
        if ( false===$stmt ) {
            die('Newsletter email bind_param() failed: ' . htmlspecialchars($stmt->error));
        }
    $stmt->execute();
        if ( false===$stmt ) {
            die('Newsletter email execute() failed: ' . htmlspecialchars($stmt->error));
        }
 } catch(Exception $e) {
    die($e->getMessage());
}

Answer 1

您的<EditText android:id="@+id/edittext" android:layout_width="match_parent" android:layout_height="wrap_content" android:digits="1234567890-"/>没有任何参考

event

应该是

.on("click", function () {

然后清除表单。你可以做到

.on("click", function (event) {

Ajax表单成功提交但出现警报错误

1 个答案: