Php 如果没有firebug断点,jquery ajax将无法工作
我使用以下方法调用php:Php 如果没有firebug断点,jquery ajax将无法工作,php,jquery,mysql,ajax,firebug,Php,Jquery,Mysql,Ajax,Firebug,我使用以下方法调用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 == ".") re
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 ".";
?>
为什么会这样
else {
return false;
$("#warning").show();
}
$(“#警告”).show()代码>将永远不会执行
编辑:好了:
function validateEmaiAjax(email){
var URL = 'https://localhost/Continental%20Tourism/register_ajax.php';
var Args = {email: email}
$('#warning').load(URL, Args, function(html){
if(html == '.'){
return true;
} else {
$('#warning').show();
return false;
}
});
return false;
}
或者您也可以尝试以下方法:
function validateEmaiAjax(email){
var URL = 'https://localhost/Continental%20Tourism/register_ajax.php';
var Args = {email: email}
$.ajax({
url: URL,
type: 'GET'
data: Args,
success: function(html){
if(html == '.'){
return true;
} else {
$('#warning').show();
return false;
}
}
});
return false;
}
这是因为函数一直在运行,并在返回ajax响应之前点击if(val==”)
。您需要在ajax回调函数中插入整个if语句
function validateEmaiAjax(email){
var success;
$("#warning").load("https://localhost/Continental%20Tourism/register_ajax.php",{email: email}, function(rspns, stat, xml){
if(rspns == ".")
success = true;
else {
$("#warning").show();
success = false;
}
});
return success;
}
还交换了警告显示和返回,因此它将执行,如果要使代码正常工作,应使用$.ajax方法,而不是load,并将async设置为false,使其等待响应
出现此问题的原因是您执行了一个异步请求。这意味着在从服务器收到响应之前,将到达if(rspns==”)
,并且结果将始终为false
为了将此代码封装在函数中,返回布尔值且不需要回调函数(阻塞过程),您需要使用同步请求:
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;
}
}
这仍然不能解决问题validateEmaiAjax()
现在将返回undefined
@idrumgood在AJAX响应实际设置变量之前,您仍在返回该变量。@idrumgood我按您所说的编辑代码。但还是不行。但是,如果我设置一个断点,它仍然可以工作。由于.load
函数不能直接返回值,因此您可能需要稍微重新构造其他代码以进行调整。不过,我的代码将解决您在帖子中提出的问题。这篇文章很棒。谢谢你的解释。现在我明白了。再次感谢。
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;
}
}