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?

Solution

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.

Comments

[Marc B]

./test.sh *.txt will make the shell expand that wildcard before test.sh gets invoked, so it'd be functionally identical to ./test.sh file1.txt file2.txt file3.txt etc...

[Barmar]

The name of the directory isn't 2011.11, it's 2011.11.10. He needs the wildcard to make it find the directory with any suffix.