Shell – What does ‘//’ mean in return from `which`

pathshellslashwhich

I have an executable script test under the full path /home/sawa/foo/bar/test. The directory /home/sawa/foo/bar is within $PATH, and has priority over the default ones including /usr/bin. When I do

`which test`

to see whether this command is correctly recognized, it returns

/home/sawa/foo/bar//test

with the double slash //. I know that there is a built in command with the same name test, and when I remove mine, this one under /usr/bin/test is returned by which, so I think it's interfering in some way.

What does this double slash mean here, and why is it appearing here?
My executable test does not seem to work correctly. Why is that?

Best Answer

I would guess that you have /home/sawa/foo/bar/ on your path - i.e. a path with a trailing slash.

which is iterating over each element of $PATH and appending /argv[1] and checking for the existence of that file. That causes a double-slash - one from the $PATH part, and one from /argv[1].

A double-slash is no problem. It is collapsed to a single slash by the kernel. Only at the beginning of a path may a double-slash have special meaning, and not always then.

As for test not working, ensure you are not using the shell built-in when calling test. You usually do this by using a full path, but with bash you can also use enable -n test to disable the built-in test command.

Related Solutions

Shell – Why does the “which” command give duplicate results

Check your path. It's not that hard to end up with duplicates in it. Example:

»echo $PATH
/usr/local/bin:/bin:/usr/bin:/usr/local/sbin:/usr/sbin:
»which -a bash
/bin/bash
/usr/bin/bash

This is because my /bin is a symlink to /usr/bin. Now:

»export PATH=$PATH:/usr/bin
»echo $PATH
/usr/local/bin:/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/usr/bin
»which -a bash
/bin/bash
/usr/bin/bash
/usr/bin/bash

Since /usr/bin is now in my $PATH twice, which -a finds the same bash twice.

linux – Why Does Variable Assigned ZERO Have Value TWO?

It is clear from you edit that the second script is called enigma2.sh, and that you call it by writing enigma2.sh on the command line. The script is somewhere in your path (you declare that it is at /usr/bin/enigma2.sh, and I shall assume that /usr/bin is in your PATH variable). Combining that the script could be found (in the PATH) by the shell and that the first line of the script is #!/bin/sh, the kernel will start a new process that has a command line of:

/bin/sh enigma2.sh

You can confirm this by simply making the first lines of that script as this:

#!/bin/sh 
ps ax|grep enigma2
exit 0
...

You don't need to erase the script, just place the lines shown above at the start of the script and leave all the rest below.

Run it to get:

$ enigma2.sh
 3109 pts/6    S+     0:00 /bin/sh enigma2.sh
 3111 pts/6    S+     0:00 grep enigma2

There are two processes that include the word enigma2.

That explains part of the problem. Now, if you change the script to:

#!/bin/sh 
enigma_pid=$(ps ax|grep enigma2)
echo "$enigma_pid"
exit 0
...                     # leave the old script below (no need to erase it).

And run it:

$ enigma2.sh
 3509 pts/6    S+     0:00 /bin/sh ./enigma2.sh
 3510 pts/6    S+     0:00 /bin/sh ./enigma2.sh
 3512 pts/6    S+     0:00 grep enigma2

When you use a $(…) one additional process appear.

Removing the grep process with grep -v grep leaves 2 processes.

That is the number you get 2.

Note: Please, don't use uppercase variable names, those are reserved for environment variables.
Don't use ~~ENIGMA_PID~~

Solutions

First: If you are going to identify a process by its name, don't use the exact same name in other programs or scripts. You may call the script enigma33.sh.
Second: In this case, you can restrict grep to match a name that ends at 2. The script ends in .sh and won't be counted. Use
```
 grep '[e]nigma2$'
```

And, since all you need is the count of matching lines use (-c):

enigma_pid=$(ps ax | grep -c '[e]nigma2$')

Yes, no need for grep -v grep since the regex [e]nigma2$ can not be equal (as an string) to enigma2.

Edited script:

#!/bin/sh


# Detect If enigma2 is running.
enigma_pid=$(ps ax|grep -c '[e]nigma2$')
echo "$enigma_pid"    

# Deetct if minisatip is running.
mini_pid=$(ps ax|grep -c '[m]inisatip$' )


# Check if the no_enigma file exists
if [ -f /home/root/no_enigma ]; then
    if [ -z "${enigma_pid##*[!0-9]*}" ] ; then
       echo "error: Not a number" >&2; exit 1
    else
        echo "It IS a number"
    fi
    
    if [ "$enigma_pid" -gt 0 ]; then
        echo "$enigma_pid"
        echo "enigma2 is running"
        killall -r '[e]nigma2$'           # make killing more restrictive
    else
        echo "enigma2 is not running"
    fi
    # Check if minisatip is running already, if not then start it
    if [ "$mini_pid" -gt 0 ]; then
        echo "minisatip is already running"
    else
        echo "minisatip is not running"
        echo "starting minisatip"
        /usr/bin/minisatip --satip-xml http://127.0.0.1:8554 -R /usr/share/minisatip/html       
    fi
else    
    # The no_enigma file does not exist
    # Check if minisatip is running, if yes then kill it
    if [ "$mini_pid" -gt 0 ]; then
        echo "minisatip is running, killing it now"
        killall minisatip
    else
        echo "minisatip is not running"
    fi  
    # Check if enigma2 is running already, if not then start it
    if [ "$enigma_pid" -gt 0 ]; then
        echo "enigma2 is already running"
    else
        echo "enigma2 is not running"
        echo "starting enigma2"
        /home/root/run_enigma.sh            
    fi
fi