The source of my bash script is printed instead of what was intended

[nolsen01]

I'm following along with a book. This is by bash script.

#!/bin/sh

# first.sh
# this file searches the current directory for the 
# string POSIX, then displays those files to the standard output

for file in *
do
 if grep -q POSIX $file
 then
  more $file
 fi
done

exit 0

For some reason, all it does is display the source code when I call it. I've tried:

/bin/sh ./first.sh

/bin/bash ./first.sh

and

chmod +x first.sh
./first.sh

What gives?

[Garius]

Hi nolsen01,

If you need to search all files containg string "POSIX" within current folder and its sub-folders you can try with:

1)
find . | xargs grep -n "POSIX"
This will print the file/s where the string "POSIX" is founded, the line number and 
the text within line itself as follow:

file name:line number: This is the line containing string "POSIX"

2)
find . | xargs grep -n "POSIX" | awk -F":" '{print $1,$2}'
This script does the same, but only printing the file name and line number where the 
string "POSIX" is founded, without text within line matched as follow.

inputdile 25

*Only run it in the folder you want

Hope it helps

[Cabhan]

So to answer your question, let's look at what the script actually does.

For every file in the current directory, it checks if the file contains the string "POSIX". Every such file is printed with the "more" command.

Well, your script is obviously in the current directory. Does it contain the string "POSIX"? Yes it does. Therefore, the script will be printed.

One way of getting around this problem would be to change the loop to the following:

for file in *
do
    if [ "$(readlink -m "$0")" == "$(readlink -m "$file")" ]
    then
        continue
    fi

    if grep -q POSIX "$file"
...

$0 contains the name of the script itself, so this would check if the file is the current script and skip over it.

Of course, it may be that you want the file to be printed, since it obviously does contain the string "POSIX".