Simple BASH script?

JayC89 · #1 (**permalink**) 10-02-2007

Hi guys, I'm new to the forum so forgive me if I'm sounding ... daft.

I currently work in a Tech Support role. Every day we have to generate data by running around 10 .sh scripts. I was thinking instead of having to ./filename 10 times is it possible to right a new script that will run these for me on the relevant order? I'm assuming one of you guys could do this in 2 seconds, but would anyone mind pointing me in the right direction? E.g a template of what the script should look like?

Obviously I still think myself to be a novice when it comes to UNIX, but anything to make the little jobs easier would be great!

Cheers
Jamie

porter · #2 (**permalink**) 10-02-2007

Put the following in a file call runall.sh then do

chmod +x runall.sh

to execute use

./runall.sh

Code:

#!/bin/sh
for d in    /path...../first-file.sh \
             /path..../second-file.sh \
             /path..../third-file.sh \
             /path..../fourth-file.sh
do
     $d
     RC=$?
     if test "$RC" != "0"
     then
          exit $RC
     fi
done

this will run the scripts and exit if there is an error returned.

JayC89 · #3 (**permalink**) 10-02-2007

Thanks porter, appreciated!

What does the +x specify after the chmod, isn't this normally numbers to specify read, write, execute permissions? Such as 660 or similar?

Could you have it so if an error occurred it would ask if you wanted to continue?

Thanks again!

porter · #4 (**permalink**) 10-02-2007

"chmod +x .... " sets the "executeable" flags which tells the operating system this is a runable program.

Yes you could,

Code:

if test "$RC" != "0"
then
     echo "continue? [y]/n"
     read N
     case "$N" in 
        Y* | y* ) 
              ;;
        * )
              exit $RC
              ;;
      esac
fi

JayC89 · #5 (**permalink**) 10-02-2007

So something along the lines of;

Code:

#!/bin/sh
for d in    /path...../first-file.sh \
             /path..../second-file.sh \
             /path..../third-file.sh \
             /path..../fourth-file.sh \
if test "$RC" != "0"
then
     echo "continue? [y]/n"
     read N
     case "$N" in 
        Y* | y* ) 
              ;;
        * )
              exit $RC
              ;;
      esac
fi
done

Do the .sh scripts not need to be told to "./"?

Thanks porter!

Jamie

porter · #6 (**permalink**) 10-02-2007

Quote:

Originally Posted by JayC89

Do the .sh scripts not need to be told to "./"?

Depends what is on the path, in UNIX (unlike windows) you normally don't execute programs from the current directory.

If the directory name is not in the command name then the operating system will look for programs in directories listed in the PATH variable.

The "./" basically tells the OS what directory "runall.sh" is in. Similarly you should replace the "/path...." to what ever directory your scripts are in.

JayC89 · #7 (**permalink**) 10-02-2007

If I had 4 scripts in;
/usr/local/production/temp/

And another 4 in;
/usr/local/production/temp/newfolder

I could use;

Code:

#!/bin/sh
for d in    /usr/local/production/temp/first-file.sh \
             /usr/local/production/temp/second-file.sh \
             /usr/local/production/temp/third-file.sh \
             /usr/local/production/temp/fourth-file.sh \
             /usr/local/production/temp/newfolder/first-file.sh \
             /usr/local/production/temp/newfolder/second-file.sh \
             /usr/local/production/temp/newfolder/third-file.sh \
             /usr/local/production/temp/newfolder/fourth-file.sh \
if test "$RC" != "0"
then
     echo "continue? [y]/n"
     read N
     case "$N" in 
        Y* | y* ) 
              ;;
        * )
              exit $RC
              ;;
      esac
fi
done

Where I currently work we generally go to the directory the .sh script is located and simply;

Code:

./script.sh

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