mjhjmtu

PROBLEM:

I can't get the $@ variable to do what I want in a for loop, the loop only finds one name in the file while looping, it should loop through all the arguments and write them to the file USERS.txt each on its own line.

Here is the file:

something78
something79
something7
dagny
oli
bjarni
toti
stefan_hlynur
jessie

Here is the test code:

#!/bin/bash

prepare_USERS()


prepare_USERS "$@"

#for user in "$@"
#do
# echo "$user" >> USERS.txt
#done

for user in USERS.txt
do
 printf "%s" $user
done

Here are the arguments I pass:

./somethingDELETEme.sh jessie henry allison jason

CURRENT output:

$./somethingDELETEme.sh jessie henry allison jason
jessie henry allison jason
jessie

EXPECTED output:

The loop loops through all names from the argument list and writes it to the file USERS.txt.

QUESTION:

I have used this variable ($@) before and never had this problem.

Why is the loop not iterating through all names in the argument list ($@) and how is the right way coding this?

HERE IS THE REAL CODE:

prepare_USERS()
 echo "writing to USERS.txt failed"; exit 127
 done

The problem is with the incorrect usage of exit 127 in your for loop, which is exiting after the first for-loop iteration. You need to group the echo statement and the exit as a compound block under .. to prevent this.

echo "$user" >> USERS.txt || echo "writing to USERS.txt failed"; exit 127; 

Without this grouping what happens is the || defined applies only for the echo command and always run the exit command no matter if write to fail passed or failed.