Need to check whether any extension is present before query string - c

Question

I have written the code to check the query string
Logic::if query string is "?" should remove all the characters from the query string and print the vailid URL.
char str[] = "http://john.org/test.mp4?iufjdlwle";
char *pch;
pch = strtok(str,"?");
printf("%s\n",pch);
Output::
bash-3.2$ ./querystring
http://john.com/test.mp4
But i have to check one more case
Need to get the URL only if there is any extensions present before query string?
if No extensions are present before the query string,need to skip.
I have tried this way,
continuation of the code
char *final;
final = pch+(strlen(pch)-3);
printf("%s\n",final);
if(strcasecmp(p,"mp4"))
printf("falure case\n");
else
printf("Success case\n");
It will work for .mp4 extension alone.
Incase if i'm getting *.mpeg or *.m3u8 or *.flv as an extensions,it will fail.
Can someone guide me how to solve this problem and make it working?

Answer 1

A query string is what starts after a question mark ?, fine.
You should try to define what an extension is. For me, it is what can happen after a dot (.) in the last component of the url, where the components are delimited with slashes (/)
So you should do:
first remove the possible query string including the initial ?
then locate the last /
then locate the last . that occurs after the last /
If you find one, it is the starting point of the extension.
So assuming pch contains the url without any query string, you can do:
char * ix = strrchr(pch, '/');
if (ix == NULL) {
// an URL without / is rather weird, better report and abort
...
}
ix = strrchr(ix, '.');
if (ix == NULL) {
// no extension here: ignore the url
...
}
else {
// found an URL containing an extension: process it
// ix+1 points to the extension
...
}