Bash Script - Will not completely execute - bash

Question

I am writing a script that will take in 3 outputs and then search all files within a predefined path. However, my grep command seems to be breaking the script with error code 123. I have been staring at it for a while and cannot really seem the error so I was hoping someone could point out my error. Here is the code:
#! /bin/bash -e
#Check if path exists
if [ -z $ARCHIVE ]; then
echo "ARCHIVE NOT SET, PLEASE SET TO PROCEED."
echo "EXITING...."
exit 1
elif [ $# -ne 3 ]; then
echo "Illegal number of arguments"
echo "Please enter the date in yyyy mm dd"
echo "EXITING..."
exit 1
fi
filename=output.txt
#Simple signal handler
signal_handler()
{
echo ""
echo "Process killed or interrupted"
echo "Cleaning up files..."
rm -f out
echo "Finsihed"
exit 1
}
trap 'signal_handler' KILL
trap 'signal_handler' TERM
trap 'signal_handler' INT
echo "line 32"
echo $1 $2 $3
#Search for the TimeStamp field and replace the / and : characters
find $ARCHIVE | xargs grep -l "TimeStamp: $2/$3/$1"
echo "line 35"
fileSize=`wc -c out.txt | cut -f 1 -d ' '`
echo $fileSize
if [ $fileSize -ge 1 ]; then
echo "no"
xargs -n1 basename < $filename
else
echo "NO FILES EXIST"
fi
I added the echo's to know where it was breaking. My program prints out line 32 and the args but never line 35. When I check the exit code I get 123.
Thanks!
Notes:
ARCHIVE is set to a test directory, i.e. /home/'uname'/testDir
$1 $2 $3 == yyyy mm dd (ie a date)
In testDir there are N number of directories. Inside these directories there are data files that have contain data as well as a time tag. The time tag is of the following format: TimeStamp: 02/02/2004 at 20:38:01
The scripts goal is to find all files that have the date tag you are searching for.

Answer 1

Here's a simpler test case that demonstrates your problem:
#!/bin/bash -e
echo "This prints"
true | xargs false
echo "This does not"
The snippet exits with code 123.
The problem is that xargs exits with code 123 if any command fails. When xargs exits with non-zero status, -e causes the script to exit.
The quickest fix is to use || true to effectively ignore xargs' status:
#!/bin/bash -e
echo "This prints"
true | xargs false || true
echo "This now prints too"
The better fix is to not rely on -e, since this option is misleading and unpredictable.

Answer 2

xargs makes the error code 123 when grep returns a nonzero code even just once. Since you're using -e (#!/bin/bash -e), bash would exit the script when one of its commands return a nonzero exit code. Not using -e would allow your code to continue. Just disabling it on that part can be a solution too:
set +e ## Disable
find "$ARCHIVE" | xargs grep -l "TimeStamp: $2/$1/$3" ## If one of the files doesn't match the pattern, `grep` would return a nonzero code.
set -e ## Enable again.
Consider placing your variables around quotes to prevent word splitting as well like "$ARCHIVE".
-d '\n' may also be required if one of your files' filename contain spaces.
find "$ARCHIVE" | xargs -d '\n' grep -l "TimeStamp: $2/$1/$3"