4

I noticed that this does not work: 

/tomcatDirectory/bin $ startup.sh //command not found

but this does work 

/tomcatDirectory $ bin/startup.sh

I am used to Windows. It seems counter-intuitive to me that I can not run a program from its working directory, only from the parent folder.

What's the bigger picture of what's happening here?

Rui F Ribeiro
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bernie2436
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    See the related [Why do you need the “./” when executing programs in the current directory?](http://unix.stackexchange.com/questions/16022/why-do-you-need-the-when-executing-programs-in-the-current-directory) too. – manatwork Feb 13 '13 at 14:03
  • You could put that bin in your path. Add `export PATH=$PATH:/tomcatDirectory/bin` to `~/.profile`. – goldilocks Feb 13 '13 at 14:11

2 Answers2

7

The current directory (i.e., .) is not in your path. Try with

./startup.sh

You can check your path with

echo ${PATH}

You could add the current directory (.) to your path but this is considered a risk (especially if . is before other directories): when typing a command the shell will first try to execute it in the current directory. This will execute what is there instead of the default one.

Summarizing: just start executables in the current directory with ./ in front of them.

Matteo
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0

Because "startup.sh" doesn't really look like a path to file and you don't have . in your PATH environment variable. But still you can start it as ./startup.sh

gelraen
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