使用特定的多字节定界符标记字符串

Question

I need to parse a string like this:

link:a link:blink:c link:d lkjh

The output should be a , blink:c , d

But the output using strtok understably is a , b , c , d , jh

How do I make sure only link: explicitly splits the string (avoiding the case of blink:c getting split. Also how do I ensure the last kjh don't appear (k seems to be the delimiter here).

Answer 1

Firstly, passing a delimiter string to strtok does not do what you seem to think. If you pass "link:" as the delim field, it will use any of those characters as a delimiter. This is why lkjh is being split and returning jh .

You are better off splitting by spaces, and then checking the beginning matches "link:" .

const char * delim = " ";
const char * prefix = "link:";
const size_t len_prefix = strlen( prefix );

char * token = strtok( input_string, delim );
while( token != NULL ) {
    if( 0 == strncmp( token, prefix, len_prefix )
    {
        printf( "%s\n", token + len_prefix );
    }
    token = strtok( NULL, delim );
}

If you need something more complicated than this, roll your own or use a regular expression.

Answer 2

Answering my own question.

Basically find the first link: and then keep moving forward till the next space/null, and repeat on the string remnant.

Better approach is to simply use strstr:-

char* parseSubFn(char *string)

{

    if(string==NULL || *string=='\0')
    return;

    else{
    //printf("Parsing on %s \n",string);
    const char* needle = "link:";
    char ret[128];//large buffer character for return token
    char *location=strstr(string,needle);
    char *start=location+5;
    int i=0; 
    while(*start!='\0' && *start!='\n' && *start!=' ')
    {
    ret[i]=*start;
    start++;i++;
    }
    start++;
    printf("%s\n",ret);
    parseSubFn(start);
    }

}