how to get the working dir of the program-bin?
lazarus / 1-Linux

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Amnon82 · 2007-05-09 16:27

I coded an installer for a linux-distro. I'm using external sh-scripts
to do additional stuff. Atm the paths to the sh-scripts are hardcoded.

Now I've a situation where in two branches the path differ from
eachother.

Now I need a variable to do it proper:

execute sh-script from <program-bin dir>/scripts/script.sh

Code: [Select]Processandwait('sudo
'+ProgramDIR+'/installer/14_reboot.sh '+TargetPart);

ProgramDIR should be the working dir of the running program-bin.

How can I do that?

In delphi it is:

Code: [Select]ProgramDIR:=ExtractFilePath(Application.Exename);

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theo · 2007-05-09 17:07

What's the Problem with
ProgramDIR:=ExtractFilePath(Application.Exename); in Lazarus ?

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felipemdc · 2007-05-09 19:55

On unixes there is no application directory:

http://wiki.lazarus.freepascal.org/Multiplatform_Programming_Guide#On_
Unix_there_is_no_.22application_directory.22

Even if sometimes it works it's not at all guaranteed or a correct way
to do this.

The best way to deploy applications on Linux is using the default
package managers. Tutorials here:

http://wiki.lazarus.freepascal.org/Deploying_Your_Application

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eric · 2007-05-10 09:23

Good heavens! I didn't know that. I've been using Application.ExeName
on Linux for years and never had a problem with it.

Eric