cut command behaving differently when its output is " * "

Question

I am trying to create a shell script, below is a part of which where i am trying to extract some value from a record in a file:

tgt_val=`cat $file_name | grep "$string_name" | cut -d"|" -f$column_no`

This work perfectly when the specified column has some numeric or string value however it fails when th column has value as " * "

eg:

102|Sam|*|USA

This command would work correctly for column 1,2 & 4

However in case of column 3 it gives me output as all the files in present directory.

Someone please help over this.

Quote your variables!. Use printf '%s\n' "$tgt_val", not echo $tgt_val (same for those $file_name and $column_no that you forgot to quote). See also Security implications of forgetting to quote a variable in bash/POSIX shells — Stéphane Chazelas, Jan 13 '16 at 12:37

score 1 · Answer 1 · answered Jan 13 '16 at 12:23

1

What is happening is that somehow the * is being interpreted by the shell. Make sure all extracted fields are always properly quoted.

answered Jan 13 '16 at 12:23

vonbrand

1 Answers1