"For" loop in bash script only run once - linux

Question

The script goal is simple.
I have many directory which contains some captured traffic files.
I want to run a command for each directory. So I came up with a script. But I don't know why the script is run only with the first match.
#!/bin/bash
# Collect throughput from a group of directory containing capture files
# Group of directory can be specify by pattern
# Usage: ./collectThroughputList [regex]
# [regex] is the name pattern of the group of directory
for DIR in $( ls -d $1 ); do
if test -d "$DIR"; then
echo Collecting throughputs from directory: "$DIR"
( sh collectThroughput.sh $DIR > $DIR.txt )
fi
done
echo Done\!
I try it with:
for DIR in $1; do
or
for DIR in `ls -d $1`; do
or
for DIR in $( ls -d "$1" ); do
or
for DIR in $( ls -d $1 ); do
But the result is the same. The for loop runs only one time.
Finally I found this one and did some tricks for it to work. However, I would like to know why my first script doesn't work.
find *Delay50ms* -type d -exec bash -c "cd '{}' && echo enter '{}' && ../collectThroughput.sh ../'{}' > ../'{}'.txt" \;
"*Delay*" is the directory pattern name that I want to run the command with.
Thanks for pointing out the issues.

Answer 1

Since you want to find all sub-directories under $1, use it like this:
for DIR in $(find $1 -type d)

Answer 2

Problem
Most probably the problem you are encountering is due to the fact that you are trying to use some kind of pattern like * as argument to your script.
Running it with something like:
my_script *
What's happening here is, that the shell will expand * prior to calling your script.
Thus after word splitting has been performed $1 in your script will just reference the first entry returned by ls.
Example
Given the following directory layout:
directory_a
directory_b
directory_c
Calling my_script * will result in:
my_script directory_a directory_b directory_c
being called thus your loop just iterating over $(ls -d directory_a) which in fact is nothing else but directory_a alone.
Solution
To have the program run with $1=* you would have to escape the * prior to calling your script.
Try running:
my_script \*
To see it effectively does what it is intended to do then. This way $1 in your script will contain * instead of directory_a which most probably is the way you wanted your script to work.

Answer 3

as mikyra has pointed out, the shell expands your argument * to all entries in your directory prior to passing it to your script.
if you want shell-expansion of your wildcards (e.g. * matches all but hidden files), you could simply leave the expansion to the shell and use the result, by iterating over all arguments, rather than just the first one:
for DIR in $#; do
# ...
done
if you want to do the expansion yourself (e.g. because the pattern should be applied only to a pre-filtered list or to files in a different directory, or because you want regex-expansion rather than shell globbing), you have to protect the argument from being expanded by the shell, either using backslash notation (like mikyra's \*) or by using quotes (which is often easier to use):
my_script "*"