Shell Programming and Scripting

Hi,

I am checking for the existence of a file in my script and if the file is not found script should fail.

Below is what I have put together. But the script is not returning correct retrurn code. I need the script to fail if the file is not found. what should i do below for the script to fail?

Code:

if  [ -e $work_FILE ]
then echo "File is found"
errorcode=$?
else
if [[ $errorCode -ne 0 ]]
then
return $errorCode
fi
fi

Code:

errorcode=$?

is returning the exit code from the echo statement, not from the test.

Code:

[ -e "$work_FILE" ] || return 1

the script is not failing. i want the script to fail but its not happening

if [ -f "test.txt" ] || return 1
then
return 0
fi

with the code above

return value is 1 but script is not failing

Code:

if [ ! -f "test.txt" ]; then
 exit 1
fi
exit 0

Use always "$variable", not $variable.
Why ?
Ex. test command need some argument(s) and if value of variable is null, it's syntax error. But if you write "$var", it's always string, which length is 0 or more.
Read man test. So, it's more safety to use "$var" always as only $var. Ex. when using filenames and your filesystem filenames can include ex. spaces.

return = used in function, exit the function like exit do for whole script

Code:

#cmd1 || cmd2    = if cmd1 exit status is not 0 then do cmd2
#cmd1 && cmd2    = if cmd1 exit status is 0 then do cmd2
[ -e "$work_FILE" ] || exit 1

or same answer using if

Code:

if [ -e "$work_FILE" ] ; then
    exit 1
fi

Both "scottn" and "kshij" have the correct solution.
The "return" command is only relevant to a script called from another script.