script path 1

[codemut]

How should a script in another directory be run, such as /another/directory/script ? My shell is bash.

 

[Annihilannic]

Exactly like you said, just /another/directory/script.

If you want to be able to run it from anywhere by just typing script, add that directory to your PATH:

PATH=$PATH:/another/directory

You can add this to your .profile or .bash_profile to set it automatically every time you log in.

Annihilannic.

 

[codemut]

Thanks Annihilannic,

With: /another/directory/script, returned is: bash: /another/directory/script: No such file or directory

With: ./another/directory/script, the terminal hangs.

While the following seems to work:
cmd="cd another/directory; ./script; cd .."
eval $cmd
unset cmd
... it seems overkill.

 

[Annihilannic]

I agree that's overkill. What is the the first line of the script? Does it contain a shebang line (e.g. #!/path/to/bash)? If so, is it pointing to the correct location of your bash shell?

What directory are you in when you run the command? Is another/directory a subdirectory of your current directory, or is it located in /?

Annihilannic.

 

[codemut]

The first line is #!/usr/bin/gawk -f, which works fine executing within the current directory. Hope to execute files that are in either a subdirectory or outside the current directory (which may require the full path from / ).

 

[Annihilannic]

I think it must be something related to the contents of the script itself, without seeing them it's difficult to say what's going wrong here.

Annihilannic.

 

[Mike042]

Hi zedlan,

It looks like there is a bit of confusion with absolute addressing and relative addressing.

/another/directory/script
is absolute addressing (where script is in directory which is in another - a directory on /)

./another/directory/script is the same as
another/directory/script which is the same as
$PWD/another/directory/script
is relative addressing, ie relative to the current directory (and can be several directories 'below' /)

To find the script script in a sub-directory somewhere 'below' the current working directory and give it its full path, you can use:
my_script=`find $PWD -name "script" -print`


I hope that helps.

Mike

 

[PHV]

Seems like the "script" reads something in another/directory ...
What about this ?

(cd another/directory; ./script)

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[Edcrosbys]

What comes after the "#!/usr/bin/gawk -f ". Is the filename specified given in absolute path?

 

[codemut]

PHV, your suggestion is closest to my ideal. Given that parentheses around a command starts a subshell, maybe one ought to include ";exit" near the end.

Edcrosbys, after the first line in the script it is just a typical gawk script with: BEGIN { ...

Mike042 and Annihilannic, you raise valid points.

 

[Annihilannic]

No need for exit, the subshell ends automatically.

Annihilannic.