如何使用 PHP 为 JQuery .ajax() 返回正确的成功/错误消息？
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How do I return a proper success/error message for JQuery .ajax() using PHP?
提问by Pieter
I keep getting the error alert. There is nothing wrong with the MYSQL part, the query gets executed and I can see the email addresses in the db.
我不断收到错误警报。MYSQL 部分没有任何问题，查询被执行，我可以在数据库中看到电子邮件地址。
The client side:
客户端：
<script type="text/javascript">
$(function() {
$("form#subsribe_form").submit(function() {
var email = $("#email").val();
$.ajax({
url: "subscribe.php",
type: "POST",
data: {email: email},
dataType: "json",
success: function() {
alert("Thank you for subscribing!");
},
error: function() {
alert("There was an error. Try again please!");
}
});
return false;
});
});
</script>
The server side:
服务器端：
<?php
$user="username";
$password="password";
$database="database";
mysql_connect(localhost,$user,$password);
mysql_select_db($database) or die( "Unable to select database");
$senderEmail = isset( $_POST['email'] ) ? preg_replace( "/[^\.\-\_\@a-zA-Z0-9]/", "", $_POST['email'] ) : "";
if($senderEmail != "")
$query = "INSERT INTO participants(col1 , col2) VALUES (CURDATE(),'".$senderEmail."')";
mysql_query($query);
mysql_close();
$response_array['status'] = 'success';
echo json_encode($response_array);
?>
回答by Muhammad Abrar
You need to provide the right content type if you're using JSON dataType. Before echo-ing the json, put the correct header.
如果您使用 JSON 数据类型，则需要提供正确的内容类型。在回显 json 之前，放置正确的标题。
<?php
header('Content-type: application/json');
echo json_encode($response_array);
?>
Additional fix, you should check whether the query succeed or not.
附加修复，您应该检查查询是否成功。
if(mysql_query($query)){
$response_array['status'] = 'success';
}else {
$response_array['status'] = 'error';
}
On the client side:
在客户端：
success: function(data) {
if(data.status == 'success'){
alert("Thank you for subscribing!");
}else if(data.status == 'error'){
alert("Error on query!");
}
},
Hope it helps.
希望能帮助到你。
回答by Alex
Just so you know, you can use this for debugging. It helped me a lot, and still does
正如您所知，您可以使用它进行调试。它对我有很大帮助，现在仍然如此
error:function(x,e) {
if (x.status==0) {
alert('You are offline!!\n Please Check Your Network.');
} else if(x.status==404) {
alert('Requested URL not found.');
} else if(x.status==500) {
alert('Internel Server Error.');
} else if(e=='parsererror') {
alert('Error.\nParsing JSON Request failed.');
} else if(e=='timeout'){
alert('Request Time out.');
} else {
alert('Unknow Error.\n'+x.responseText);
}
}
回答by Marc B
Some people recommend using HTTP status codes, but I rather despise that practice. e.g. If you're doing a search engine and the provided keywords have no results, the suggestion would be to return a 404 error.
有些人建议使用 HTTP 状态代码，但我很鄙视这种做法。例如，如果您正在使用搜索引擎并且提供的关键字没有结果，则建议返回 404 错误。
However, I consider that wrong. HTTP status codes apply to the actual browser<->server connection. Everything about the connect went perfectly. The browser made a request, the server invoked your handler script. The script returned 'no rows'. Nothing in that signifies "404 page not found" - the page WASfound.
然而，我认为这是错误的。HTTP 状态代码适用于实际的浏览器<->服务器连接。关于连接的一切都很顺利。浏览器发出请求，服务器调用您的处理程序脚本。脚本返回“无行”。没有在表示“404页找不到” -页WAS找到。
Instead, I favor divorcing the HTTP layer from the status of your server-side operations. Instead of simply returning some text in a json string, I always return a JSON data structure which encapsulates request status and request results.
相反，我倾向于将 HTTP 层与服务器端操作的状态分开。我不是简单地在 json 字符串中返回一些文本，而是总是返回一个 JSON 数据结构，它封装了请求状态和请求结果。
e.g. in PHP you'd have
例如在 PHP 中，你会拥有
$results = array(
'error' => false,
'error_msg' => 'Everything A-OK',
'data' => array(....results of request here ...)
);
echo json_encode($results);
Then in your client-side code you'd have
然后在您的客户端代码中，您将拥有
if (!data.error) {
... got data, do something with it ...
} else {
... invoke error handler ...
}
回答by philippe
In order to build an AJAX webservice, you need TWO files :
为了构建 AJAX 网络服务，您需要两个文件：
- A calling Javascript that sends data as POST (could be as GET) using JQuery AJAX
- A PHP webservice that returns a JSON object (this is convenient to return arrays or large amount of data)
- 使用 JQuery AJAX 将数据作为 POST（可以作为 GET）发送的调用 Javascript
- 一个返回 JSON 对象的 PHP 网络服务（这方便返回数组或大量数据）
So, first you call your webservice using this JQuery syntax, in the JavaScript file :
因此，首先您在 JavaScript 文件中使用此 JQuery 语法调用您的网络服务：
$.ajax({
url : 'mywebservice.php',
type : 'POST',
data : 'records_to_export=' + selected_ids, // On fait passer nos variables, exactement comme en GET, au script more_com.php
dataType : 'json',
success: function (data) {
alert("The file is "+data.fichierZIP);
},
error: function(data) {
//console.log(data);
var responseText=JSON.parse(data.responseText);
alert("Error(s) while building the ZIP file:\n"+responseText.messages);
}
});
Your PHP file (mywebservice.php, as written in the AJAX call) should include something like this in its end, to return a correct Success or Error status:
您的 PHP 文件（mywebservice.php，如 AJAX 调用中所写）应在其末尾包含类似内容，以返回正确的成功或错误状态：
<?php
//...
//I am processing the data that the calling Javascript just ordered (it is in the $_POST). In this example (details not shown), I built a ZIP file and have its filename in variable "$filename"
//$errors is a string that may contain an error message while preparing the ZIP file
//In the end, I check if there has been an error, and if so, I return an error object
//...
if ($errors==''){
//if there is no error, the header is normal, and you return your JSON object to the calling JavaScript
header('Content-Type: application/json; charset=UTF-8');
$result=array();
$result['ZIPFILENAME'] = basename($filename);
print json_encode($result);
} else {
//if there is an error, you should return a special header, followed by another JSON object
header('HTTP/1.1 500 Internal Server Booboo');
header('Content-Type: application/json; charset=UTF-8');
$result=array();
$result['messages'] = $errors;
//feel free to add other information like $result['errorcode']
die(json_encode($result));
}
?>
回答by Frank Violette
I had the same issue. My problem was that my header type wasn't set properly.
我遇到过同样的问题。我的问题是我的标题类型设置不正确。
I just added this before my json echo
我只是在我的 json echo 之前添加了这个
header('Content-type: application/json');
回答by larachiwnl
Server side:
服务器端：
if (mysql_query($query)) {
// ...
}
else {
ajaxError();
}
Client side:
客户端：
error: function() {
alert("There was an error. Try again please!");
},
success: function(){
alert("Thank you for subscribing!");
}
回答by Roland Roos
...you may also want to check for cross site scripting issues...if your html pages comes from a different domain/port combi then your rest service, your browser may block the call.
...您可能还想检查跨站点脚本问题...如果您的 html 页面来自不同的域/端口组合，那么您的休息服务，您的浏览器可能会阻止调用。
Typically, right mouse->inspect on your html page. Then look in the error console for errors like
通常，鼠标右键->检查您的 html 页面。然后在错误控制台中查看类似的错误
Access to XMLHttpRequest at '...:8080' from origin '...:8383' has been blocked by CORS policy: No 'Access-Control-Allow-Origin' header is present on the requested resource.
从源 '...:8383' 访问 XMLHttpRequest at '...:8080' 已被 CORS 策略阻止：请求的资源上不存在 'Access-Control-Allow-Origin' 标头。
回答by Francois Brand
adding to the top answer: here is some sample code from PHP and Jquery:
添加到最佳答案：这里是一些来自 PHP 和 Jquery 的示例代码：
$("#button").click(function () {
$.ajax({
type: "POST",
url: "handler.php",
data: dataString,
success: function(data) {
if(data.status == "success"){
/* alert("Thank you for subscribing!");*/
$(".title").html("");
$(".message").html(data.message)
.hide().fadeIn(1000, function() {
$(".message").append("");
}).delay(1000).fadeOut("fast");
/* setTimeout(function() {
window.location.href = "myhome.php";
}, 2500);*/
}
else if(data.status == "error"){
alert("Error on query!");
}
}
});
return false;
}
});
PHP - send custom message / status:
PHP - 发送自定义消息/状态：
$response_array['status'] = 'success'; /* match error string in jquery if/else */
$response_array['message'] = 'RFQ Sent!'; /* add custom message */
header('Content-type: application/json');
echo json_encode($response_array);
