CODEWITHSUNDEEP
curllibcurl
shadow dk
shadow dk

Reputation: 503

How to convert cURL to libcurl(c)

I have the following cmd and its work. - using digest auth - using json data

curl -v --digest -u user:passwd -X POST http://192.168.139.65:80/digest/getMode -H "Content-Type: application/json" -d '{"Data":{}}'

I would like to convert cURL to libcurl. 

curl_easy_setopt(curl, CURLOPT_URL, "http://192.168.139.65:80/digest/getMode");
headerlist = curl_slist_append(headerlist, "Content-Type: application/json");
curl_easy_setopt(curl, CURLOPT_HTTPHEADER, headerlist);
curl_easy_setopt(curl, CURLOPT_HTTPAUTH, CURLAUTH_DIGEST);
curl_easy_setopt(curl, CURLOPT_USERPWD, "user:passwd ");
curl_easy_setopt(curl, CURLOPT_CUSTOMREQUEST, "POST");
curl_easy_setopt(curl, CURLOPT_POSTFIELDS, "{\"Data\":\t{}\r\n}");

curl_easy_setopt(curl, CURLOPT_VERBOSE, 1L);

But I got fail, It return 401. Is there any missing? Thanks for help

Upvotes: 1

Views: 382

Answers (2)

Jodh Singh
Jodh Singh

Reputation: 383

Just add --libcurl -filename.c after your curl command. This will create a C source file performing the exact same operation as your curl command.

Upvotes: 4

shadow dk
shadow dk

Reputation: 503

Finally, I check the message different between two method. I found that TCP_NODELAY and USERAGENT used in cURL. I need to add these two command.

Upvotes: -2

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