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RE: curl_easy_init not defined

From: Cristina Gil <cristina.gil_at_sumenor.com>
Date: Fri, 8 Aug 2008 12:30:47 +0200

Thanks a bunch! I gonna try. But sorry.. i have another questions im reading the examples and the curl api to find any function which can receive the stream the server (I sent the http request) is gonna send.

I try to program this in vb.net and it worked with a "GET" but here I don't see which option is the one I need. Could it be:

curl_easy_setopt(curl, CURLOPT_POST, xxx);

and xxx is a variable that contains a stream??

Thanks again

De: curl-users-bounces_at_cool.haxx.se [mailto:curl-users-bounces_at_cool.haxx.se] En nombre de Luigi Grilli
Enviado el: viernes, 08 de agosto de 2008 12:07
Para: the curl tool
Asunto: R: curl_easy_init not defined

You need to check in your system where is the library installed.
In my system is in /usr/lib
Maybe you have it in /usr/local/lib

It's seems that ld can't find the library in your system paths.
Try to add the path where you have the libcurl.so.4 to your LD_LIBRARY_PATH variable, like this:
export LD_LIBRARY_PATH=/usr/local/lib
substitute the path with the one where libcurl is installed.

This is a temporary (session) solution. You will need to change system configuration as root to make it permanent.

Da: curl-users-bounces_at_cool.haxx.se [mailto:curl-users-bounces_at_cool.haxx.se] Per conto di Cristina Gil
Inviato: venerd� 8 agosto 2008 11.57
A: the curl tool
Oggetto: RE: curl_easy_init not defined

Thanks for your answers, i did it and it compiled but when i try to execute with ./progam I have this error:

Error while loading shared libraries: libcurl.so.4: cannot shared object file: no such file or directory

Do I need to install another library??

De: curl-users-bounces_at_cool.haxx.se [mailto:curl-users-bounces_at_cool.haxx.se] En nombre de Luigi Grilli
Enviado el: viernes, 08 de agosto de 2008 11:32
Para: the curl tool
Asunto: R: curl_easy_init not defined

You need also to link to the library.
If the library is installed in the system you just need to have a -lcurl in your gcc command line like this:
gcc -lcurl -o program program.c

You also need to change void to int as main return value.

Da: curl-users-bounces_at_cool.haxx.se [mailto:curl-users-bounces_at_cool.haxx.se] Per conto di Cristina Gil
Inviato: venerd� 8 agosto 2008 11.24
A: curl-users_at_cool.haxx.se
Oggetto: curl_easy_init not defined

Hello!
Im a newbie in C programming, i have to program an application to use it as a daemon which sends periodically a http request to the server.

Im write this:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <curl/curl.h>

void main(){
CURL *curl;
char *url;

url="http://192.168.0.63:80/admin/serverreport.cgi";
curl = curl_easy_init();

if(curl) {
   curl_easy_setopt(curl,CURLOPT_URL, url);
   curl_easy_setopt(curl, CURLOPT_HTTPAUTH, CURLAUTH_BASIC);
   curl_easy_setopt(curl, CURLOPT_USERPWD, "user:password");
   curl_easy_cleanup(curl);

};
}

But when I try to compile, gcc -o program program.c,
The compiler says that
(.text+0x12) Reference to curl_easy_init not defined
(.text+0x36) Reference to curl_easy_setopt not defined
(.text+0x41) Reference to curl_easy_setopt not defined
(.text+0x64) Reference to curl_easy_setopt not defined
(.text+0x92) Reference to curl_easy_cleanup not defined

Thanks for your help!

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Received on 2008-08-08

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