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我想獲取網站的HTML代碼。 我使用的代碼是使用https獲取網站的源代碼,使用cURL
static size_t write_data(void *ptr, size_t size, size_t nmemb, void *stream)
{
int written = fwrite(ptr, size, nmemb, (FILE *)stream);
return written;
}
int main(void)
{
CURL *curl_handle;
static const char *headerfilename = "head.txt";
FILE *headerfile;
static const char *bodyfilename = "body.txt";
FILE *bodyfile;
curl_global_init(CURL_GLOBAL_ALL);
curl_handle = curl_easy_init();
curl_easy_setopt(curl_handle, CURLOPT_URL, "http://www.chess.com");
curl_easy_setopt(curl_handle, CURLOPT_NOPROGRESS, 1L);
curl_easy_setopt(curl_handle, CURLOPT_WRITEFUNCTION, write_data);
curl_easy_setopt(curl_handle,CURLOPT_FOLLOWLOCATION,1);
headerfile = fopen(headerfilename,"w");
if (headerfile == NULL) {
curl_easy_cleanup(curl_handle);
return -1;
}
bodyfile = fopen(bodyfilename,"w");
if (bodyfile == NULL) {
curl_easy_cleanup(curl_handle);
return -1;
}
curl_easy_setopt(curl_handle, CURLOPT_WRITEDATA, headerfile);
curl_easy_perform(curl_handle);
fclose(headerfile);
curl_easy_cleanup(curl_handle);
return 0;
}
這個代碼適用於谷歌(使用HTTP),而不是chess.com(使用HTTPS)。 我怎樣才能讓它爲兩者工作?
我想如果沒有仔細考慮先做這是一個徹頭徹尾的適得其反的建議,並打開該用戶的MITM攻擊等等。 –
嗨@DanielStenberg,謝謝你指出,你會建議什麼呢? –