尝试使用 C 访问 Twitter Streaming API [英] Trying to Access Twitter Streaming API with C
问题描述
我从 C libcurl 得到输出到一个字符串.我修改了它,我想用它来访问 Twitter Stream.我添加了 curl_easy_setopt(curl, CURLOPT_URL, "http://stream.twitter.com/1/statuses/sample.json");
和 curl_easy_setopt(curl, CURLOPT_USERPWD, "neilmarion:my_password");
在代码中.但问题是每当我执行它时,都没有输出.一定是什么问题?谢谢.
I got the code from C libcurl get output into a string. I modified it and I want to use it to access the Twitter Stream. I added curl_easy_setopt(curl, CURLOPT_URL, "http://stream.twitter.com/1/statuses/sample.json");
and curl_easy_setopt(curl, CURLOPT_USERPWD, "neilmarion:my_password");
in the code. But the problem is whenever I execute it, there is no output. What must be the problem? Thanks.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <curl/curl.h>
struct string {
char *ptr;
size_t len;
};
void init_string(struct string *s) {
s->len = 0;
s->ptr = malloc(s->len+1);
if (s->ptr == NULL) {
fprintf(stderr, "malloc() failed
");
exit(EXIT_FAILURE);
}
s->ptr[0] = ' ';
}
size_t writefunc(void *ptr, size_t size, size_t nmemb, struct string *s)
{
size_t new_len = s->len + size*nmemb;
s->ptr = realloc(s->ptr, new_len+1);
if (s->ptr == NULL) {
fprintf(stderr, "realloc() failed
");
exit(EXIT_FAILURE);
}
memcpy(s->ptr+s->len, ptr, size*nmemb);
s->ptr[new_len] = ' ';
s->len = new_len;
return size*nmemb;
}
int main(void)
{
CURL *curl;
CURLcode res;
curl = curl_easy_init();
if(curl) {
struct string s;
init_string(&s);
curl_easy_setopt(curl, CURLOPT_URL, "http://stream.twitter.com/1/statuses/sample.json");
curl_easy_setopt(curl, CURLOPT_USERPWD, "neilmarion:my_password");
curl_easy_setopt(curl, CURLOPT_WRITEFUNCTION, writefunc);
curl_easy_setopt(curl, CURLOPT_WRITEDATA, &s);
res = curl_easy_perform(curl);
printf("%s
", s.ptr);
free(s.ptr);
/* always cleanup */
curl_easy_cleanup(curl);
}
return 0;
}
推荐答案
最快的下一步可能是设置 CURLOPT_HEADER,在正文输出中包含标题.最有可能的是,我猜它在安全性方面失败了,您会在标题中看到详细信息.
The quickest next step is probably to set CURLOPT_HEADER, to include headers in the body output. Most likely, I would guess it is failing on security, and you'll see the details in the headers.
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