Shell Programming and Scripting

Hi,

I'm new to UNIX, at least shell programming and am having trouble figuring out a problem i'm having. In one section in my nested if statement, i want the program to test if the file does not exist, based on an argument supplied at the command line by the user. What i have is

and from there it goes down a long list of other conditions, but this is the only one giving me problems. It skips completely over it and doesnt print anything to screen.

Another problem i'm having is with error codes. I want certain error code values to display if the user should decide to ask for the error code level at the command line, using a logical AND in addition to the arguments supplied for the main part of hte program. I tried changing the last line in all of my nested ifs to

bu that threw an error saying an assignment was attempted on a nonvariable. OK fine, so how would this be done? Is it even possible to set up specified error codes in a shell script that you can call by using 'echo $?' ?

Any help is appreciated, thanks !!


Using BASH shell btw.

I don't think you can set "?"

For one thing, you do assignment without the "$", so even if you tried, it would be

Code:

$ set ?=127
$ echo $?
0

However, it doesn't achieve the desired result, as after every command (even "set"), "?" is set to the exist status of last command.

You could just use an argument to exit (or return if in a function), as in

Code:

exit 127 (or return 127)

And for your file test...try the ! at the left of the test

Code:

if [ ! -e $4 ]; then
  stuff;
  exit 127;
fi

Concerning your first problem, you need to move the ! in front of "-e". Otherwise you're just negating the command line argument.
About your second problem: Why don't you call "exit" with the respective error code? OK, I wasn't fast enough

Both worked - thanks a million guys !