how to use fscanf to read from ( to )?

Question

How can you use scanf in c to read string between brackets? for example how can I get "myString" if text in a file is "text a 123 (myString) text"?

This question should be asked in a programming site like stackoverflow or [programmers](http://programmers.stackexchange.com) — Albert, Dec 31 '14 at 10:15
This question appears to be off-topic because it is about programming in C (and really has nothing to do with Unix or Linux). Please consider [so]. — derobert, Dec 31 '14 at 10:27
Instead of trying to squeeze `scanf()` to the limit of what it can do, and risk creating a bug, consider using a regexp library. — potrzebie, Dec 31 '14 at 14:43

score 1 · Answer 1 · answered Dec 31 '14 at 14:20

1

You can use %[] as a poor man's regexp :

#include <stdio.h>

int main(){
   const char data[] = "text a 123 (myString) text";
   char str[20];
   int cr = sscanf( data, "%*[^(](%[^)])", str ); /* skip until '(' and read up to ')' */
   if (cr == 1) {
        printf(">%s<\n", str);
   }
   else {
       printf("NOT FOUND\n");
   }
   return 0;
}

above program correctly outputs :

>myString<

answered Dec 31 '14 at 14:20

Serge Ballesta

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Likely good enough for OP, but this does not insure a final `')'` existed nor the subsequent "text". – chux - Reinstate Monica Dec 31 '14 at 22:47

Gopi · Answer 2 · 2014-12-31T14:46:58.960

0

Just to read your line you can have something like

char a[50];
fscanf(stdin,"%*s %*s %*d (%[^)]) %*s",a);

edited Dec 31 '14 at 14:46

answered Dec 31 '14 at 13:15

Gopi

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This answer is wrong. `a` is a char array. The `fscanf()` call should take `a` as argument, not `&a`. – user12205 Dec 31 '14 at 13:27
Quoting from [this SO answer](http://stackoverflow.com/a/5407121/3488231), "it's wrong, and it could fail on some platforms". They are pointers to different types. If you compile with gcc, this shows up: `warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[50]’ [-Wformat=]` – user12205 Dec 31 '14 at 13:32
@ace Very hard why this might fail and as the answer says who knows on which platform this fails – Gopi Dec 31 '14 at 13:42
1

I suspect there is an error: `s` after `[` is not a format specifier, but expected next character. – kestasx Dec 31 '14 at 13:54
@kestasx Sorry I didn't get ? – Gopi Dec 31 '14 at 13:55
@Gopi Lose the `s` after the `]`. – potrzebie Dec 31 '14 at 14:34
@Gopi, in Your format string `...(%[^)]s...` character `s` is not a format specifier for string. Format string as You wrote matches string until `)`, then instead of consuming this `)` it expects `s`, and only after it `)`. – kestasx Dec 31 '14 at 14:38

kestasx · Answer 3 · 2014-12-31T14:44:29.747

Full example:

#include <stdio.h>

int main(){
   const char data[] = "text a 123 (myString) text";
   char str[20];
   sscanf( data, "%*s %*c %*d (%[^)]", str );
   puts( str );
   return 0;
}

Updated:

You can replace sscanf line with the following (ignore avrytihing until ( and then read until ):

sscanf( data, "%*[^(](%[^)]", str );

Note: I'm not sure how portable such programming practice is, but it works at least on Ubuntu 12.04.

score 0 · Answer 4 · answered Dec 31 '14 at 14:11

Example :

#include <stdio.h>
#include <stdlib.h>
int main(){
        char mystring[1025];
        FILE * fp = NULL;
        fp= fopen("dd.txt","r");
        if(!fp){

        }else{
                fscanf(fp, "%*s %*c %*d (%[^)] %*s",mystring);
                printf("%s\n",mystring);
        }

        fclose(fp);

}

how to use fscanf to read from ( to )?

4 Answers4