在C中使用Curl发送HTTP Get请求 [英] send HTTP Get request using Curl in c

问题描述

请帮助我使用C语言中的curl实现HTTP Get请求.

Please help me to implement an HTTP Get request using curl in C.

我需要使用 https://这样的参数来访问URL www.googleapis.com/tasks/v1/users?name=pradeep&lastname=singla 我使用CURLOPT_HTTPHEADER设置带有Header的参数，但没有成功.我将其实现为

I need to hit URL with parameters like https://www.googleapis.com/tasks/v1/users?name=pradeep&lastname=singla I used CURLOPT_HTTPHEADER to set parameters with Header but without success. I implemented it like

struct curl_slist* contentheader = NULL;    

contentheader = curl_slist_append(contentheader, "name=pradeep");
contentheader = curl_slist_append(contentheader, "lastname=singla");

curl_easy_setopt(curl, CURLOPT_URL, "https://www.googleapis.com/tasks/v1/users");
curl_easy_setopt(curl, CURLOPT_HTTPHEADER, contentheader);
curl_easy_perform(curl); 

在这种情况下，发生了诸如无正确的APi请求"之类的错误.所以我认为我可以使用

In this case an error occured like "no correct APi request". So I thought that I can use

char *charff = "name=pradeep&lastname=singla";    
curl_easy_setopt(curl, CURLOPT_POSTFIELDS, charfff);
curl_easy_setopt(curl, CURLOPT_HTTPGET, 1L); curl_easy_perform(curl);

但发生相同的错误.

有人可以帮助我吗?我可以同时请求POST和GET方法，因为服务器方法可能随时更改.

Can anyone please help me ? I can put my request for both POST and GET method because server method may change anytime.

推荐答案

该URL只是具有"parameters"的URL，您可以通过

The URL is just a URL even with "parameters" and you set it in full with CURLOPT_URL.

它们不是标题，也不是后字段.

They're not headers and they're not postfields.

CURL *curl = curl_easy_init();
curl_easy_setopt(curl, CURLOPT_URL, 
    "https://www.googleapis.com/tasks/v1/users?name=pradeep&lastname=singla");
curl_easy_perform(curl);

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