我正在尝试使用 libcurl(在 C 中)将文件上传到网络服务器。服务器要求我使用 POST,而不是 PUT(这是 CURLOPT_UPLOAD 将使用的)。我还需要在标头中发送一个特定的令牌。这是我到目前为止所拥有的:
int upload(char *filepath, char *filename, char *token) {
CURL *curl;
CURLcode res;
FILE *fd;
struct stat file_info;
char *file;
asprintf(&file, "%s/%s", filepath, filename);
fd = fopen(file, "rb");
if (!fd) {
fprintf(stderr, "Could not open file.\n");
return 1;
}
if (fstat(fileno(fd), &file_info) != 0) {
fprintf(stderr, "Could not get file information.\n");
return 1;
}
curl = curl_easy_init()
curl_easy_setopt(curl, CURLOPT_URL, "https://127.0.0.1/upload");
curl_easy_setopt(curl, CURLOPT_READDATA, fd);
curl_easy_setopt(curl, CURLOPT_POST, 1L);
struct curl_slist *headers = NULL;
headers = curl_slist_append(headers, "Content-Type:application/octet-stream");
char *my_token = malloc(snprintf(NULL, 0, "X-Token:%s", token) + 1);
sprintf(my_token, "X-Token:%s", token);
headers = curl_slist_append(headers, my_token);
curl_easy_setopt(curl, CURLOPT_HTTPHEADER, headers);
curl_easy_setopt(curl, CURLOPT_VERBOSE, 1L);
res = curl_easy_perform(curl);
if (res != CURLE_OK) {
fprintf(stderr, "curl_easy_perform() failed: %s\n", curl_easy_strerror(res));
return 1;
}
fprintf(stdout, "Upload OK.\n");
return 0;
}
我从服务器得到的唯一响应是“400 Bad Request”。
我已经用 Python 编写了执行相同操作的代码,并且此文件上传有效:
import urllib, urllib2, json, os
def upload(self, filepath, filename, token):
# Open file
f = open(os.path.join(filepath, filename), 'rb')
filedata = f.read()
f.close()
request = urllib2.Request("https://127.0.0.1/upload" % filedata, {"Content-Type":"application/octet-stream", "X-Token":token})
res = self.opener.open(request)
return json.loads(res.read())
更新:以详细模式运行后:
> POST /upload HTTP/1.1
Host: 127.0.0.1
Accept: */*
Content-Type:application/octet-stream
X-Token: 17f684-b98c
Content-Length: -1
Expect: 100-continue
< HTTP/1.1 Bad Request
....