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[英]How do i get the response back from server after issuing a request object(AJAX)
[英]how do i get a response from a jquery ajax request
我正在我的网站上工作,并向服务器发送了jQuery请求
$.ajax(
// do an ajax to send value to the database...
{
url:"pages/welcome_get.php",
type: "POST",
dataType: "json",
cache: false,
data: { wm_val: wel}
})
如何获取不是html数据的json数据作为响应,以及如何将json响应从服务器解析为html文件?
您编写PHP以发出JSON。
<?php
# Some code to populate $some_associative_or_non_associative_array
header("Content-Type: application/json");
echo json_encode($some_associative_or_non_associative_array);
?>
You need to use the parseJSON function in the js.
Here is the Php code:
function send_reply(){
echo json_encode(array('reply'=>'here is my reply'));
exit;
}
Here is the js code:
$.ajax({
url:'myajax.php',
data:{'func':send_reply},
type:'POST',
dateType:'json'
}).success(function(data){
data=$.parseJSON(data);
alert(data.reply);
}).error(function(jqxhr,error,status){
alert('Error sending reply');
});
正如@Quentin所说,您需要在PHP结果中输出JSON。 然后,您需要使用done()在客户端接收数据并从那里进行处理:
$.ajax({
url:"pages/welcome_get.php",
type: "POST",
dataType: "json",
cache: false,
data: { wm_val: wel}
}).done(function( json ) {
// do something with json here
});
您需要将“ JSON”标头添加到“ pages / welcome_get.php”文件中:
header("Content-Type: application/json");
还要在AJAX调用中记住添加“成功”和“错误”回调:
jQuery.ajax({
url:"pages/welcome_get.php",
type: "POST",
dataType: "json",
cache: false,
data: { wm_val: wel}
success: function(response) {
//Do stuff with response here
}
error: function(){
//Display error msg or something like that
}
});
可以说您正在用welcome_get.php检查用户
然后在您的welcome_get.php中
用这个
if(isset($_GET['wm_val'])){
$wm_val = $mysqli->real_escape_string($_GET['wm_val']);
$check_user = $mysqli->prepare("SELECT email FROM members WHERE username = ? LIMIT 1 ");
$check_user->bind_param('s', $wm_val);
$check_user->execute();
$check_user->store_result();
$check_user->bind_result( $email);
$check_user->fetch() ;
if ($check_user->num_rows == 1) { $datas['msg']= "failed" ;}
else{$datas['msg']= "success" ;}
$check_user->close() ;
echo json_encode($datas);
}
和你的ajax
$.ajax(
// do an ajax to send value to the database...
{
url:"pages/welcome_get.php",
type: "POST",
dataType: "json",
cache: false,
data: { wm_val: wel},
success: function(msg) {
if (data.msg == 'success'){
// do what you like here example
$('#mydata').html("<span >you welcome ! </span>").delay(4000).fadeOut('fast');
}else{
//do something else example
$('#mydata').html("<span >failed ! </span>").delay(4000).fadeOut('fast');
}
})
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