On Thu, 10 Oct 2019, ka sc via curl-library wrote:
url_flags = "\"[\\\"INSERT INTO foo2 values(1,11)\\\"]\"";
curl_easy_setopt(curl, CURLOPT_POSTFIELDS, url_flags.c_str());
Is there a way to see the output url so I can try that in commandline?
The data you set with CURLOPT_POSTFIELDS is not part of the URL - it is part
of the request body - so seeing the URL won't help you.
You can see the URL with CURLOPT_VERBOSE set to 1L, but since you probably
rather see the request body so that you get to see what you post you want to
add a CURLOPT_DEBUGFUNCTION callback and show the data that way.
Possibly based on this example code: https://curl.haxx.se/libcurl/c/debug.html
Alternatively, you make a command line version of the request with curl that
works and then you convert that to libcurl code with the "--libcurl code.c"
option.
--
/ daniel.haxx.se | Get the best commercial curl support there is - from me
| Private help, bug fixes, support, ports, new features
| https://www.wolfssl.com/contact/
-------------------------------------------------------------------
Unsubscribe: https://cool.haxx.se/list/listinfo/curl-library
Etiquette: https://curl.haxx.se/mail/etiquette.html
