# Unix Shell Script

1. May 24, 2007

### Surrealist

I have never written a shell script, but I am trying to learn. I want to make a generic Unix shell script that will allow me to run four commands in a row... something like this...

latex $filename bibtex$filename
latex $filename latex$filename

I would like to call the script "ltxprc.sh".

How would I make this file work for any filename?

For instance, if my file name were "paper", would I run a command like...

sh ltxprc.sh

where do I input the filename?

2. May 24, 2007

### D H

Staff Emeritus
First, you need to modify your script so it accesses the command line arguments sent to the script:

filename=\$1

You pass the filename on the command line:

sh ltxprc.sh file.tex

Even better is to make the shell script executable via chmod. You will need to add a "shebang" line to the very front of your script:

#!/bin/sh

When you do that, all you need to say is

ltxprc.sh file.tex

3. May 25, 2007

### Surrealist

Thanks for the help. That was probably the most useful response I have ever received from a message board.

I gotta ask... why is it that the #!/bin/sh works? I thought that the # symbol meant "comment out" that which follows.

4. May 25, 2007

### D H

Staff Emeritus
The # symbol is indeed a comment to the shell. The shell, however, is not what runs runs programs (directly, that is). The exec family of functions are what run programs on a Unix machine. The first thing exec does when asked to run a program is read the first two bytes of the program. When those first two bytes are "#!", exec does something quite special: It uses the line to indicate what program should be executed to run the script file. For example, you used #!/bin/sh . The #! tells exec this is an executable script. The exec invokes /bin/sh to run the script. Now /bin/sh reads lines from the script. Now we get back to your question: the first line is just a comment; /bin/sh does nothing with it.