1

I am relatively new to Unix, so perhaps this is an extremely simple problem.
Anyways, the problem is such:

I am taking an online course and the professor typed:

echo $path

and got this result....

/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin

I typed the same command (echo $path)

and I got:

/usr/local/mysql/bin:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin 

because this class is related to web development (we will be using gitHub and Node.js) ... will this slightly more complicated path become an issue?
Should I change the location of mysql?

  • 1
    Welcome to U&L. Are you sure you typed a space between $ and path and that if you did not your return value did not start with a / (i.e. /usr/local/....) I formatted your code and removed non-relevant parts from your post. click on the edited ... link to see what changed. A better readable post, is more likely to be understood and answered. – Anthon Apr 30 '16 at 13:58
7

This is not a problem at all.

The $PATH variable simply states where executables reside in your filesystem.

When you want to run a program from your shell, normally you would have to type the complete path to this program to run - e.g. /bin/grep. But since the path /bin is included in your $PATH variable, you only have to type grep. In other words, when you type a single command in your shell, the directories in your $PATH variable are searched for that command. If it is found there, it is executed without you having to find the location of the program and typing the whole path.

  • Thanks, geruetzel. I thought it meant that all of the files below /usr/local had been moved to an SQL file system. If so, I was worried that the functionality of most programs might be impeded. Thanks for the tip... – Jacob Apr 30 '16 at 17:58
  • You are welcome. If my post helped you, I would appreciate if you marked my question as accepted solution ;) – geruetzel Apr 30 '16 at 18:04

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