Find position of a character in a file and store its position bash

Question

I need to find all the positions of a specific character ## in a file using Bash script and also store all positions for future use.

This is how the file looks like:

assert property ( ( req1 == 0 ) ##1 ( req1 == 1 ) ##1 !( req2 == 1 ) || (gnt1 == 0 ) );
assert property ( ( req1 == 0 ) ##1 !( req1 == 1 ) || ( gnt1 == 1 ) );
assert property ( ( req1 == 1 && req2 == 0 ) ##1 !( req2 == 1 ) || ( gnt1 == 0 ) );

In the above file I want to get all locations for this character (##) and store them as well. How can this be done using shell script?

What do you mean by "locations"? Maybe zero-indexed positions from the front of the lines? — Ruslan Osmanov, Dec 07 '16 at 13:38
Or "locatoin" = 1 based from top of file thru the bottom? AND as we're about helping people **fix** their code, **what have you tried?** (update your Q, don't reply in comments please). Good luck. — shellter, Dec 07 '16 at 13:54
There will only be this character in the file ## not three # — mii9, Dec 07 '16 at 14:16

score 1 · Answer 1 · answered Dec 07 '16 at 14:45

You could use awk for this:

awk -F'##' '
{
   for(i=1;i<NF;i++){
     printf "%d ",length($i)+offset+1
     offset+=length($i)+length(FS)
   }
   printf "\n"
   offset=0
}' file

The parameter delimiter -F is set as your pattern.

Loop through all portions of the line ($1,$2...) and print the length of each portion that actually indicates the position of the pattern (adding the offset for line that would contain more than one match).

score 0 · Answer 2 · edited May 23 '17 at 12:01

Try this

#!/bin/bash

str1="assert property ( ( req1 == 0 ) ##1 ( req1 == 1 ) ##1 !( req2 == 1 ) || (gnt1 == 0 ) );"
str2="assert property ( ( req1 == 0 ) ##1 !( req1 == 1 ) || ( gnt1 == 1 ) );"
str3="assert property ( ( req1 == 1 && req2 == 0 ) ##1 !( req2 == 1 ) || ( gnt1 == 0 ) );"

char="##"

awk -v a="$str1" -v b="$char" 'BEGIN{print index(a,b)}' | xargs expr -1 +
awk -v a="$str2" -v b="$char" 'BEGIN{print index(a,b)}' | xargs expr -1 +
awk -v a="$str3" -v b="$char" 'BEGIN{print index(a,b)}' | xargs expr -1 +

The positions are 32, 32 and 45 for these examples. But there is only the first appearance of the ## char for each one. Not sure if you need to get more appearances.

Solution extracted from Dennis Williamsons comment on this post

Thank you, but I need to know the positions of all occurrences in the file for this character. — mii9, Dec 07 '16 at 14:14

Find position of a character in a file and store its position bash

2 Answers2