Question

我正在使用ajax与PHP进行通信。当获取的数据显示数据库中存在已键入的电子邮件时，我使用了echo的消息＆＃39;已经收到电子邮件，尝试另一封电子邮件＆＃39;。但是ajax responseText正在返回header file and echoed content。这是一张图片

enter image description here

如何将回显的消息作为ajax responseText？

这是我的代码

的Javascript

function signup(){
  var f=document.getElementById("firstname").value;
  var l=document.getElementById("lastname").value;
  var p=document.getElementById("password1").value;
  var p2=document.getElementById("password2").value;
  var e=document.getElementById("email").value;
  var status=document.getElementById("status");

  if(f=="" || l=="" || p=="" || p2=="" || e==""){
     status.innerHTML="All fields are required";
  }
  else if(p!=p2){
     status.innerHTML="passwords didn't match";
  }
  else{
    document.getElementById("submitX").style.display = "none";
    //status.innerHTML = 'please wait ...';
    var x=new XMLHttpRequest();
    x.open("POST","signup.php",true);
    x.setRequestHeader("content-type","application/x-www-form-urlencoded");
    x.onreadystatechange=function(){
    if(x.readyState == 4 && x.status == 200){
     if(x.responseText != "success"){  //this condition is not met at any circumstances
         document.getElementById("submitX").style.display = "block";
         status.innerHTML="";
         //alert(x.responseText);
         status.innerHTML = x.responseText;
        }
       else{
         window.scrollTo(0,0);
         status.innerHTML ="";
         document.getElementById("diff_for").innerHTML = "";
         document.getElementById("form1").innerHTML = "";
         document.getElementById("form1").innerHTML = "Thank you for creating an account.A Welcome email has been sent to your email.<br/><br/>Go to <a href='login.php'>Login</a>";

     }

    }
   }
  x.send("F="+f+"&L="+l+"&E="+e+"&P="+p+"&P2="+p2);
 }
}

PHP

if(isset($_POST['F'])){
  $firstname=$_POST['F'];
  $lastname=$_POST['L'];
  $password=$_POST['P'];
  $passmatch=$_POST['P2'];
  $pass=md5($password);
  $email=$_POST['E'];

 $sqlx1="SELECT id FROM user_det WHERE email=? LIMIT 1";

   mysqli_stmt_prepare($stmt30, $sqlx1);
   mysqli_stmt_bind_param($stmt30, "s", $email);
   mysqli_stmt_execute($stmt30);
   mysqli_stmt_store_result($stmt30);
   mysqli_stmt_bind_result($stmt30, $idx1);
   $num_row1x=mysqli_stmt_num_rows($stmt30);

     if($num_row1x>0){
       //ob_end_clean();
       //ob_start();
       $abul="<br>This email is already registered.Please use another email";
       echo $abul;
       exit();
     }
     else{
       $sql11x="INSERT INTO user_det(first_name,last_name,password,email,signup_date)
               VALUES(?,?,?,?,now())";

       mysqli_stmt_prepare($stmt30, $sql11x);
       mysqli_stmt_bind_param($stmt30, "ssss", $firstname, $lastname, $pass, $email);
       mysqli_stmt_execute($stmt30);                   

       echo "success";
     }

我真的很感谢你的帮助。

感谢您的时间。

Answer 1

随着你的额外评论，现在我看到发生了什么。

仅使用您发布的部分PHP代码不可见，如果不是太重，您应该发布整个代码。

但是非常确定您的错误位于PHP文件顶部附近：当您测试if(isset($_POST['F'])){时，您已经输出了原始页面的开头，该页面由同一个PHP文件生成。

顺便说一句，你应该在回复响应Ajax调用的消息后确保exit;。

ajax responseText返回带有echo内容的标题栏

1 个答案: