jquery ajax在没有firebug断点的情况下不工作

Question

我使用以下方法来调用php:

function validateEmaiAjax(email){
    val = null;
    $("#warning").load("https://localhost/Continental%20Tourism/register_ajax.php",{email: email}, function(rspns, stat, xml){
        val = rspns;
    });

    if(val == ".")
        return true;
    else {
        return false;
    }
}

我的PHP代码是:

<?php
    $dbc = mysqli_connect("localhost","root","pass","continental_tourism") OR die(mysqli_connect_error());

    $email = $_REQUEST['email'];

    $query = "SELECT email FROM customer_info WHERE email = '$email' ";

    $r = mysqli_query($dbc, $query) OR die(mysqli_error($dbc));

    if(mysqli_num_rows($r) > 0)
        echo "Email address exists!";
    else
        echo ".";   
?>

基本上这会检查数据库,如果存在电子邮件显示"存在电子邮件地址!" 如果不是我想return true(所以我回应"."并比较它).奇怪的是,如果我在if(val == ".")程序附近使用firebug设置断点并且返回true.如果我删除该断点函数总是返回false.我不明白为什么会这样.请帮忙!谢谢.

Answer 1

您遇到此问题的原因是您已执行异步请求.这意味着if(rspns == ".")将在从服务器收到响应之前到达,结果将始终为false.

为了在函数中包装此代码,返回一个布尔值,并且不需要回调函数(阻塞过程),您将需要使用同步请求:

function validateEmaiAjax(email) {

  // This is the correct way to initialise a variable with no value in a function
  var val;

  // Make a synchronous HTTP request
  $.ajax({
    url: "https://localhost/Continental%20Tourism/register_ajax.php",
    async: false,
    data: {
      email: email
    },
    success: function(response) {
      // Update the DOM and send response data back to parent function
      $("#warning").html(response);
      val = response;
    }
  });

  // Now this will work
  if(val == ".") {
    return true;
  } else {
    $("#warning").show();
    return false;
  }

}