我有这个jQuery Ajax脚本

$.ajax({
    type: "POST",
    url: "register.php",
    data: "name=" + name + "&email=" + email + "&password=" + password + "&confirm_password=" + confirm_password + "&captcha=" + captcha,
    success: function(data){
    if(data == "Success"){
      $("#tombol_submit").remove();
      $("#register_sukses").fadeIn(500);
    }else{
      $("#register_gagal").html(data).fadeIn(500);
      $("#submit").removeAttr("disabled").attr("value", "Submit");
    }
}


在register.php中,一旦数据成功添加到数据库中,它将回显“成功”一词,成功一词出现在表单页面中,但是按钮(tombol_submit)未被删除,否则它将返回到Submit按钮(只是就像在“其他”声明中一样)。如何删除按钮,使客户端无法再次单击提交按钮?

这是register.php脚本

<?php
    session_start();
    include "config/koneksi.php";

    if(!empty($_POST['captcha'])){
         if($_POST['captcha'] == $_SESSION['hasil']){
             $fullname = $_POST['name'];
             $email = $_POST['email'];
             $password = md5($_POST['password']);

             $queryform = "INSERT INTO register (fullname,email,pass)
                    VALUES('$fullname','$email','$password')";
             if ($hasilform = mysqli_query($konek2, $queryform)) {
                 echo "Success";
             } else {
                 echo "Failed";
             }
         } else {
             echo "The captcha code is wrong";
         }
     } else {
         echo "The captcha cannot be empty";
     }
?>

最佳答案

通常$().remove()将起作用。

如果您问为什么if...else...不能正常工作并且总是转到else,请检查后端的响应以确保响应只是一个“成功”字符串。

关于javascript - jQuery Ajax无法捕获数据,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40214217/

10-09 17:50
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