How I can run application from anywhere in the terminal without being in the full path

I have an application that I can run while I am in the file location as the below:

-$ cd home/myApps/app
-$ ./run.sh

I want to call the run.sh from anywhere in my terminal without being in the full path, I tried to used alais and add the below code to bashrc file.

alias mylovleyapp=/home/myApps/app/run.sh

the problem is whenever I want to run the app using mylovleyapp, the program return error because the run.sh is depending on another files in myApps folder

How I can achieve that?

Because the OP didn't provide a sample script, I'm going to make one. It will do one thing. Run ls in the directory of the script.

Yes, I know you can just do ls without the ./. But for this example, I'm pretending it is required, to demonstrate a script that uses the current directory of the script.

#!/usr/bin/env bash
ls ./

Save that script somewhere. I'll save it as /home/ubuntu/test/myscript. Make it executable with chmod +x /home/ubuntu/test/myscript. Now cd to the directory where the script is (e.g., cd /home/ubuntu/test), and run the script with ./myscript:

ubuntu@computer:~/test$ ./myscript
myscript someRandomFile

So far, so good. But now lets run it from a different directory. For this example, I'm going to run it from the root directory (/).

ubuntu@computer:~/test$ cd /
ubuntu@computer:/$ /home/ubuntu/test/myscript
bin  dev home lib   lib64  lost+found mnt proc run  snap sys usr
boot etc init lib32 libx32 media      opt root sbin srv  tmp var

Oops. We wanted it to print the files in the location of the script, not the location of the user. Before we figure out the solution, think about why it does this for a moment. The answer is simple. ./ is relative to the user's current directory, not the one of the script.

To get the directory of the script from inside of the script, we will look at this Stack Overflow question.

The gist of it is simple: script_dir="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )" will get the current directory of the script, and store it in the variable script_dir.

Let's update the script accordingly. Our script is now this:

#!/usr/bin/env bash
script_dir="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )"
ls "$script_dir"

Let's run it, and see if it works:

ubuntu@computer:/$ /home/ubuntu/test/myscript
myscript someRandomFile

It works! It prints the files in the directory of the script, even when you are not in the same directory as the script, We're done. The overall idea is simple. In the script, ./ refers to the user's directory. "$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )" refers to the directory the script is in. Update your script accordingly, and it should work.

You need to add your app path to PATH variable. To do that temporarily, go to the directory where you app is and then:

PATH=.:$PATH

You can now call your app from anywhere without having to type its full path. If you close the terminal this setting will be lost.

To set it permanently, you need to edit your .bashrc file and add :/home/myApps/app to the end of your PATH. Save the new .bashrc file and then:

source .bashrc

So you need to change your working directory to the one the target application is in, then execute the application so that it can find the dependent files.

I think you can do this by giving two separate commands in the alias: one to change directory and the other to execute the program, separated by a semi-colon.

So your .bashrc addition should read:

alias mylovleyapp='cd /home/myApps/app; ./run.sh'

That assumes the dependent files are in /home/myApps/app ... if not then you can change the working directory to the one with the dependent files, then run the app using an absolute path as the second component.