如何在PHP中使用curl GET发送原始数据? [英] How to send raw data with curl GET in PHP?

问题描述

我正在开发REST API，尽管很容易为cURL中的POST请求设置原始JSON数据

I am developing REST API and while it is easy to set raw JSON data for request in cURL for POST

$payload = json_encode(array("user" => $data));

//attach encoded JSON string to the POST fields
curl_setopt($ch, CURLOPT_POSTFIELDS, $payload);

我不知道如何通过GET请求发送此类数据.

I cannot figure out how to send such data with GET requests.

是否有类似CURLOPT_GETFIELDS或CURLOPT_RAWDATA的东西?通过GET请求发送JSON的目的是传递一些参数.

Is there something like CURLOPT_GETFIELDS or CURLOPT_RAWDATA? The purpose of sending JSON with GET request is to pass in some params.

我不希望将formdata添加到请求中，我希望发布JSON，以便可以在接收方上对其进行解析.

I do not wish to add formdata to the request, I wish to post JSON so that it can be parsed on the receiver.

谢谢！

基于注释，我想避免造成混淆，因此生成的请求应如下所示:

based on comments I want to avoid confusion, so the resulting request should look like:

GET / HTTP/1.1
Host: 127.0.0.1:3000
Content-Type: application/json
Accept: application/json
Host: 127.0.0.1:3000
content-length: 13
Connection: keep-alive
cache-control: no-cache

{
    "a": "b"
}

如您所见，这里的GET请求包含数据，它已被解析并由Web服务器完美地工作.如何使用cURL做到这一点?

as you can see, GET request here has data and it is parsed and works perfectly by web server. How do I achieve this with cURL?

推荐答案

GET请求没有主体，这就是整个想法:您只是从服务器获取内容，而不是向其发布内容.来自 RFC 7231 :

GET requests do not have a body, that's the whole idea: you're just getting something from the server, as opposed to posting something to it. From RFC 7231:

GET请求消息中的有效载荷没有定义的语义； 在GET请求上发送有效内容正文可能会导致一些现有内容 拒绝请求的实现.

A payload within a GET request message has no defined semantics; sending a payload body on a GET request might cause some existing implementations to reject the request.

换句话说，一个GET请求可以有数据，但没有.摘自规范的早期版本，其中GET被定义为一种安全方法:

In other words, a GET request can have data, but it should not. From earlier in the spec, where GET is defined as a safe method:

如果请求方法定义的语义是

，则认为它们是安全的" 本质上是只读的；即客户不要求，而是要求 没想到，由于以下原因，原始服务器上的任何状态更改 将安全方法应用于目标资源.

Request methods are considered "safe" if their defined semantics are essentially read-only; i.e., the client does not request, and does not expect, any state change on the origin server as a result of applying a safe method to a target resource.

...

在此规范定义的请求方法中，GET，HEAD， OPTIONS和TRACE方法被定义为安全的.

Of the request methods defined by this specification, the GET, HEAD, OPTIONS, and TRACE methods are defined to be safe.

如果您确实希望在GET请求中包含JSON(并将其发送到合理实施的服务器资源)，那么它唯一可以放在的地方就是URI中作为查询字符串的一部分.对于GET请求，我发现使用 file_get_contents 来比处理cURL容易得多.

If you really want to have JSON in your GET request (and send it to a reasonably implemented server resource) the only place it can go is in the URI as part of the query string. For GET requests I find using file_get_contents to be much easier than dealing with cURL.

<?php
$payload = json_encode(["user" => $data]);
$url_data = http_build_query([
    "json" => $payload
]);
$url = "https://some.example/endpoint.php?" . $url_data;

$result = file_get_contents($url);

---

如果要将其发送到不合理实现的服务器资源，并且违反了HTTP RFC的精神，则可以执行以下操作:

---

If you want to send it to an unreasonably implemented server resource, and violate the spirit of the HTTP RFCs, you could do this:

<?php
$url = "https://some.example/endpoint.php";
$payload = json_encode(["user" => $data]);
$ctx = stream_context_create(["http" => [
    "header"=>"Content-Type: application/json",
    "content"=>$payload
]]);
$result = file_get_contents($url, false, $ctx);

如果确定要专门使用cURL进行此操作，则可能很幸运，将CURLOPT_CUSTOMREQUEST选项设置为"GET"，而将CURLOPT_POSTDATA设置为数据.

If you're determined to do this specifically with cURL, you might have luck with the CURLOPT_CUSTOMREQUEST option set to "GET" and CURLOPT_POSTDATA with your data.

这篇关于如何在PHP中使用curl GET发送原始数据?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！