Question

我创建了一个客户名称文本。已经注册的客户将在单击该文本字段时显示。只有该注册客户才能输入数据。我使用ajax来阻止新用户的添加。

这是ajax：

function validateForm()
    {
        var customerName    =   document.getElementById('customerName').value;
        var customerId      =   document.getElementById('customerId').value;
        var getResponse=0;

        var xmlhttp = new XMLHttpRequest();
            xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
           // document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
          getResponse= xmlhttp.responseText;alert(getResponse);

            if(getResponse=="0")
            {
                return false;
            }


          }

        };

        xmlhttp.open("GET", "customer_check_ajax.php?name="+customerName+"&id="+customerId, true);
        xmlhttp.send();

    }

这是ajax页面customer_check_ajax.php：

<?php
require("../../config/config.inc.php"); 
require("../../config/Database.class.php");
require("../../config/Application.class.php");
if($_SESSION['travelType']=='Admin')
{
    $check  =   1;
}
else
{
    $check  =   '';
    $logId  =   $_SESSION['travelId'];
    $proId  =   $_SESSION['proId'];
    $check  =   "proId='$proId'";
}

$cusName = $_REQUEST['name'];
$cusId   = $_REQUEST['id'];

$connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_DATABASE);
$qry="select * FROM ".TABLE_ACCOUNTS." WHERE ID='$cusId' and accountName='$cusName' and proId='$proId'";
$res = mysqli_query($connection, $qry);//echo $qry;
$num = mysqli_num_rows($res);
if($num>0)
echo '1';
else
echo '0';
?>

这里我会得到新用户的响应0和已注册用户的响应1。但我想在新用户添加时停止保存数据。请帮帮我。

Answer 1

您正在进行异步请求，并且在函数结束后会出现ajax的答案。

要做你想做的事，你必须做同步ajax请求

xmlhttp.open("GET", "customer_check_ajax.php?name="+customerName+"&id="+customerId, false); // `false` makes the request synchronous

当ajax响应为false时，如何防止保存数据？

1 个答案: