Reputation: 2491
I have a simple bash script, simple.sh, as follows:
#/usr/local/bin/bash
for i in $1
do
echo The current file is $i
done
When I run it with the following argument:
./simple.sh /home/test/*
it would only print and list out the first file located in the directory.
However, if I change my simple.sh to:
#/usr/local/bin/bash
DIR=/home/test/*
for i in $DIR
do
echo The current file is $i
done
it would correctly print out the files within the directory. Can someone help explain why the argument being passed is not showing the same result?
Upvotes: 11
Views: 15326
Reputation: 531055
Your script does not receive "/home/test/*" as an argument; the shell expands the patter to the list of files that match, and your shell receives multiple arguments, one per matching file. Quoting the argument will work:
./simple.sh "/home/test/*"
Your change to using DIR=/home/test/*
did what you expected because filename generation is not performed on the RHS of a variable assignment. When you left $DIR
unquoted in the for
loop, the pattern was expanded to the list of matching files.
Upvotes: 1
Reputation: 72639
This is as portable as it gets, has no useless forks to ls and runs with a minimum of CPU cycles wasted:
#!/bin/sh
cd $1
for i in *; do
echo The current file is "$i"
done
Run as ./simple.sh /home/test
Upvotes: 2
Reputation: 1505
If you take "$1", it is the first file/directory, which is possible! You should do it in this way:
for i in "$@"
do
echo The current file is ${i}
done
If you execute it with:
./simple.sh *
They list you all files of the actual dictionary
"$1" is alphabetical the first file/directory of your current directory, and in the for loop, the value of "i" would be e.g. a1.sh and then they would go out of the for loop! If you do:
DIR=/home/<s.th.>/*
you save the value of all files/directories in DIR!
Upvotes: 12
Reputation: 3202
How about list the file manully instead of using *:
#/usr/local/bin/bash
for i in $(ls $1)
do
echo The current file is $i
done
and type
./simple.sh /home/test/
Upvotes: -2