Using Regex Group in C

Question

I need to get the groups that matches with my Regex in C to manipulate a Java program logs.

I have tested the Regex:

(Client:\s[a-zA-Z\s]+)|(Wallet:\s[a-zA-Z0-9]+)|(ID\s*:\s*[0-9]{3}.{0,1}[0-9]{3}.{0,1}[0-9]{3}-{0,1}[0-9]{2})

here and it works.

But in my C program, it doesn't work as well.

#include 
#include 
#include 
#include 

int main(void) {
  const char *source =
      "[com.example.app.JavaClass.JavaMethod(JavaClass.java:1)] (Thread-1) - "
      "Client: FirstName MiddleName AnotherName LastName, Wallet: WL01, "
      "Agency: 9999, ID: 06611486123, Ticket: TKR211";
  const char *regexString =
      "(Client:\s[a-zA-Z[:space:]]+)|(Wallet:\s[a-zA-Z0-9]+)|(ID\s*:\s*[0-"
      "9]{3}.{0,1}[0-9]{3}.{0,1}[0-9]{3}-{0,1}[0-9]{2})";

  regex_t regexCompiled;

  regcomp(®exCompiled, regexString, REG_ICASE | REG_EXTENDED);

  size_t ngroups = regexCompiled.re_nsub + 1;
  regmatch_t *groups = malloc(ngroups * sizeof(regmatch_t));

  regexec(®exCompiled, source, ngroups, groups, 0);

  char cursorCopy[strlen(source) + 1];
  strcpy(cursorCopy, source);
  size_t nmatched;
  for (nmatched = 0; nmatched < ngroups; nmatched++) {
    if (groups[nmatched].rm_so == (size_t)(-1)) {
      break;
    }

    char *match =
        calloc(groups[nmatched].rm_eo - groups[nmatched].rm_so, sizeof(char));
    memcpy(match, &source[groups[nmatched].rm_so],
           groups[nmatched].rm_eo - groups[nmatched].rm_so);
    printf("Match: [%2u-%2u]: \"%s\"
", groups[nmatched].rm_so,
           groups[nmatched].rm_eo, match);
  }
  regfree(®exCompiled);

  return 0;
}

Executing:

$ gcc -Wall -Wextra -Wwrite-strings reg.c && ./a.out

Generates the output:

Match: [70-119]: "Client: FirstName MiddleName AnotherName LastName"
Match: [70-119]: "Client: FirstName MiddleName AnotherName LastName"

But what I want is:

Match: [xx-xx]: "Client: FirstName MiddleName AnotherName LastName"
Match: [xx-xx]: "Wallet: WL01"
Match: [xx-xx]: "ID: 06611486123"

Can someone tell me if it is possible to do using C or I need another approach?

edit:

In my case is possible that some fields ("Client", "Wallet" or "ID") won't coming in log.

user_3791842 · Answer

Your regex is composed like this: (a)|(b)|(c) where a, b, and c correspond to the Client regex, the Wallet regex, and the ID regex.

This is not what you want - you can see in your own RegExr that you are not getting one match, but three different matches. In C, you are only matching once.

What you're really trying to accomplish is matching your source string only once, and having each of the groups contain their string. In other words, we want to change your regex:

(a)|(b)|(c) -> (a),(b),(c) - one single match matching the entirety of the string.

This does the trick:

const char *regexString =
    "(Client:\s[a-zA-Z[:space:]]+), (Wallet:\s[a-zA-Z0-9]+).*(ID\s*:\s*[0-"
    "9]{3}.{0,1}[0-9]{3}.{0,1}[0-9]{3}-{0,1}[0-9]{2})";

I changed the first | to a , which is what separates the Client and Wallet substrings, and I changed the second | to a .* which encapsulates everything between the Wallet and ID substrings.

Running this now gives:

Match: [70-164]: "Client: FirstName MiddleName AnotherName LastName, Wallet: WL01, Agency: 9999, ID: 06611486123"
Match: [70-119]: "Client: FirstName MiddleName AnotherName LastName"
Match: [121-133]: "Wallet: WL01"
Match: [149-164]: "ID: 06611486123"

The first line gives you the entire match, while the following lines give you the contents of each separate group.

A more visual way of looking at this is going from:

to:

Using Regex Group in C

1 Answers