directory path as command line argument in bash

Question

The following bash script finds a .txt file from the given directory path, then changes one word (change mountain to sea) from the .txt file

#!/bin/bash
FILE=`find /home/abc/Documents/2011.11.* -type f -name "abc.txt"`
sed -e 's/mountain/sea/g' $FILE 

The output I am getting is ok in this case. My problem is if I want to give the directory path as command line argument then it is not working. Suppose, I modify my bash script to:

#!/bin/bash
FILE=`find $1 -type f -name "abc.txt"`
sed -e 's/mountain/sea/g' $FILE 

and invoke it like:

./test.sh /home/abc/Documents/2011.11.*

Error is:

./test.sh: line 2: /home/abc/Documents/2011.11.10/abc.txt: Permission denied

Can anybody suggest how to access directory path as command line argument?

Answer 1

Your first line should be:

FILE=`find "$@" -type f -name "abc.txt"`

The wildcard will be expanded before calling the script, so you need to use "$@" to get all the directories that it expands to and pass these as the arguments to find .

Answer 2

You don't need to pass .* to your script.

Have your script like this:

#!/bin/bash

# some sanity checks here
path="$1"

find "$path".* -type f -name "abc.txt" -exec sed -i.bak 's/mountain/sea/g' '{}' \;

And run it like:

./test.sh "/home/abc/Documents/2011.11"

PS: See how sed can be invoked directly from find itself using -exec option.