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c get請求json數據

張吉惟2年前7瀏覽0評論

在進行網絡編程的過程中,我們經常會需要向服務器請求一些數據,而獲取json數據是非常常見的一種請求。使用C語言實現get請求獲取json數據的方法如下:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <curl/curl.h>
struct MemoryStruct {
char *memory;
size_t size;
};
static size_t WriteMemoryCallback(void *contents, size_t size, 
size_t nmemb, void *userp) {
size_t realsize = size * nmemb;
struct MemoryStruct *mem = (struct MemoryStruct *)userp;
mem->memory = realloc(mem->memory, mem->size + realsize + 1);
if (mem->memory == NULL) {
printf("not enough memory (realloc returned NULL)\n");
return 0;
}
memcpy(&(mem->memory[mem->size]), contents, realsize);
mem->size += realsize;
mem->memory[mem->size] = '\0';
return realsize;
}
int main(void) {
CURL *curl;
CURLcode res;
struct MemoryStruct chunk;
chunk.memory = malloc(1);
chunk.size = 0;
curl_global_init(CURL_GLOBAL_ALL);
curl = curl_easy_init();
if(curl) {
curl_easy_setopt(curl, CURLOPT_URL, "http://example.com/json");
curl_easy_setopt(curl, CURLOPT_WRITEFUNCTION, WriteMemoryCallback);
curl_easy_setopt(curl, CURLOPT_WRITEDATA, (void *)&chunk);
res = curl_easy_perform(curl);
if(res != CURLE_OK) {
fprintf(stderr, "curl_easy_perform() failed: %s\n",
curl_easy_strerror(res));
}
else {
printf("%lu bytes retrieved\n", (unsigned long)chunk.size);
printf("%s", chunk.memory);
}
curl_easy_cleanup(curl);
}
free(chunk.memory);
curl_global_cleanup();
return 0;
}

代碼中使用了libcurl庫進行GET請求,并將返回的數據存儲在MemoryStruct結構體中。最后打印出返回的字節數以及請求的json數據。