shell built-in command
How do you run a "shell built-in command" inside a Unix shell script?

I'm trying to source a startup file but when I execute the file an error occurs

#!/bin/sh
echo "start"
source .cshrc
echo "finish"

colt 27% temp
start
temp[3]: source: not found
finish
colt 28%

Is there anyway to do this?

Thanks

Marty

I notice that you are trying to source a c shell source while starting the script in sh. any reason for this? try /bin/csh and use

source filenale (without the .cshrc - this is automatic)

Can you expand on what you are trying to do. Are you looking to access the parent shell or the child shell?

I'm writting a script that sources the .cshrc so the user does not have to do it manually.

The .cshrc contains many setenv. Is there away in a script to run set these?

Thanks

Marty

Two ways :-

1. Use #!/bin/sh on the first line of the script and do a
. /xyz/.profile

2. Use #!/bin/csh on the first line of the script and do a
source /xyz/.cshrc