php 带有 contentType 的 Ajax 调用：'application/json' 不起作用
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Ajax call with contentType: 'application/json' not working
提问by Sebsemillia
I have an ajax call, that sends form data to a php function. Since I read a lot that using contentType: 'application/json'is best practice I wanted to give it a try as well. But unfortunately my script doesn't return anything when I use it. If I remove it, the script does what it is supposed to do.
我有一个 ajax 调用，它将表单数据发送到一个 php 函数。因为我读了很多，使用contentType: 'application/json'是最佳实践，所以我也想尝试一下。但不幸的是，我的脚本在使用时没有返回任何内容。如果我删除它，脚本就会做它应该做的事情。
Do you have any idea what the reason might be and why? Thank you!
您知道原因可能是什么吗？谢谢！
$('#Form').submit(function(e) {
e.preventDefault();
var content = $(this).serialize() + "&ajax=1";
$.ajax('app/class/controller/contactForm.php', {
type: "POST",
//contentType: 'application/json',
dataType: 'json',
data: content,
success: function(result) {
console.log(result);
}
});
})
and my PHP:
和我的 PHP：
if(isset($_POST['ajax']) && $_POST['ajax'] === '1') {
echo json_encode(validateForm($_POST));
}
回答by Mike Brant
When using contentType: 'application/json'you will not be able to rely on $_POSTbeing populated. $_POSTis only populated for form-encoded content types.
使用时，contentType: 'application/json'您将无法依赖$_POST被填充。 $_POST仅针对表单编码的内容类型填充。
As such, you need to read your data from PHP raw input like this:
因此，您需要像这样从 PHP 原始输入读取数据：
$input = file_get_contents('php://input');
$object = json_decode($input);
Of course if you want to send application/jsonyou should actually send JSON, which you are not doing. You either need to build the object serialization to JSON directly, or you need to do something like this - Convert form data to JavaScript object with jQuery- to serialize the object from the form.
当然，如果您想发送，application/json您实际上应该发送 JSON，而您并没有这样做。您要么需要将对象序列化直接构建为 JSON，要么需要执行以下操作 -使用 jQuery 将表单数据转换为 JavaScript 对象- 从表单序列化对象。
Honestly in your case, since you are dealing with form data, I don't quite think the use case for using application/jsonis there.
老实说，在您的情况下，由于您正在处理表单数据，因此我不太认为 using 的用例在application/json那里。
回答by LSerni
The best practice you refer to is about the server scriptsetting the Content-Typefor JSON to "application/json":
您参考的最佳实践是关于将Content-TypeJSON设置为“application/json”的服务器脚本：
Header('Content-Type: application/json; charset=UTF8');
This is because otherwise a default Content-Typewill be sent, often a catch-all text/html, and this could lead to an incomprehension with the client.
这是因为否则Content-Type将发送一个默认值，通常是一个包罗万象的text/html，这可能导致客户端不理解。
If you do notspecify yourself a Content-Type in the jQuery request, jQuery will determine the most appropriate one. The problem here is that you were sending a POST form, for which the default Content-Typeset by jQueryis application/x-www-form-urlencoded, which tells PHP to decode the data as POST fields and populate $_POST. Your script would have then recovered its parameters from $_POST(or maybe $_REQUEST).
如果您没有在 jQuery 请求中为自己指定 Content-Type，jQuery 将确定最合适的一个。这里的问题是您发送了一个POST 表单，jQuery为它Content-Type设置了默认值，它告诉 PHP 将数据解码为 POST 字段并填充。然后，您的脚本将从（或可能）恢复其参数。application/x-www-form-urlencoded$_POST$_POST$_REQUEST
By changing it to application/json, $_POSTwill no longer be populated, the receiving script operation won't receive the parameters where it was expecting to, and the operation breaks.
通过将其更改为application/json，$_POST将不再填充，接收脚本操作将不会在它期望的位置接收参数，并且操作会中断。
So you either need to:
所以你要么需要：
- not specify the Content-Type yourself (better, IMHO)
- set a Content-Type of
application/x-www-form-urlencoded; charset=UTF-8 - set a Content-Type of
application/json; charset=UTF-8and modify the script to parse the POST stream and decode the JSON data; see this answer.
- 不要自己指定内容类型（更好，恕我直言）
- 设置一个内容类型
application/x-www-form-urlencoded; charset=UTF-8 - 设置 Content-Type
application/json; charset=UTF-8并修改脚本以解析 POST 流并解码 JSON 数据；看到这个答案。
The third option requires proper handlingof php://input.
第三个选项需要妥善处理的php://input。
回答by ceejayoz
The PHP script should be setting the Content-Type header.
PHP 脚本应该设置 Content-Type 标头。
if(isset($_POST['ajax']) && $_POST['ajax'] === '1') {
header('Content-Type: application/json');
echo json_encode(validateForm($_POST));
}
