Setting bash command-line variable to start my application

Question

I'm pretty new to programming for Linux environments, so I don't exactly know what to search for in order to answer this question for myself. I need to understand how applications set the shell to accept a certain command to start them. For example, you can start Firefox from the command line by executing the command: firefox.

I don't know where this is defined. Makefile? Configure script? In the source code itself?

Any resources / reading on Linux programming tidbits like these would be greatly appreciated!

Thank you.

Answer 1

Firefox由命令“firefox”启动，因为$ PATH环境变量中的一个文件夹中有一个名为“firefox”的可执行文件。

Answer 2

Are you talking about the PATH variable? It seems like you are.

In linux, you should be able to type: "echo $PATH" (without quotes) and get a ":"-separated list of locations where programs are located (like firefox).

If you need to add something to your path, you should be able to do:

export PATH=$PATH:/another/directory

In your shell (which is most likely bash)

You can also type:

which firefox

To display the location of the firefox executable.

Answer 3

Typically the shell is going to have an environment variable called $PATH set. This is just an ordered list of all the directories to look when somebody types in a command. As soon as it finds an executable file (by which I mean a file for which you have execute permissions, not a file ending in .exe) with the same name as whatever was typed, it will run that file. Common directories in $PATH might be /bin, /usr/local/bin, ~/bin, etc.

So, when you type 'firefox', the shell looks through all the directories in $PATH until it finds /usr/local/bin/firefox, which it then runs. To make your own programs run the same way, you'll either need to put them (or a symbolic link to them) in a directory that is likely to be in every user's path (/usr/local/bin/ is a good choice), or you'll need to get your users to add your program's directory to their $PATH.

For a more complete description, see the Wikipedia article about the $PATH variable.

Answer 4

As an alternative to the modification of $PATH mentioned earlier, you could also copy or link your executable in one of the directories already in your $PATH. more specifically, /usr/local/bin/ is available on most UNIX system for pretty much this purpose (installing software outside the default package management of the operating system).

Answer 5

It has to be in the path as everyone else mentioned, but you might also need to make it executable with something like this:

chmod +x /path/to/file

And if it's a script there's usually a shebang at the top that tells the os what to use to execute it:

#! /usr/bin/python

Answer 6

Often, large packages are installed in /opt with a wrapper script or link somewhere in the PATH . For example, on my system, Google Picasa is installed in /opt/google/picasa and there is a symlink at /usr/bin/picasa to /opt/google/picasa/3.0/picasa

Firefox is at /usr/bin/firefox on my system and that's a symlink to /usr/bin/firefox-3.0 which is itself a symlink to /usr/lib/firefox-3.0.11/firefox.sh - That shell file fumbles around until it finally runs /usr/lib/firefox-3.0.11/firefox (unless it finds a reason to do something else). That, finally, is a binary executable. So /usr/lib is where firefox is installed, for me.

You can use this command to find out where a program is:

type -a firefox

Then, you can find out what kind of file it is using this:

file /usr/bin/firefox

Also see the Filesystem Heirarchy Standard for more information about recommended locations for files and programs.