Why am I unable to successfully export a path within a bash script?

Question

In order to get some C code to compile correctly, I need to export a folder so that are certain file is found.

On the command line, if I first run

export LD_LIBRARY_PATH=/bin/path../version/:$LD_LIBRARY_PATH

and then the executable from the C code works. If I do not export LD_LIBRARY_PATH, then the executable will fail, as the C package cannot find the necessary file, i.e.

exectuable: error while loading shared libraries: file1.so: cannot open shared object file: No such file or directory

I would now like to run this code via a bash script. Here is my bash script, run_stuff.sh:

#!/bin/bash

source ~/.bash_profile

export LD_LIBRARY_PATH=/bin/path../version/:$LD_LIBRARY_PATH

path/to/executable/executable 

This fails, with the error above. I have since tried

echo "export LD_LIBRARY_PATH=/bin/path../version/:$LD_LIBRARY_PATH"

within the bash script, and I have added

export LD_LIBRARY_PATH=/bin/path../version/:$LD_LIBRARY_PATH

within ~/.bash_profile. It still doesn't work---the C code cannot find this file.

How does one correctly export a folder within a bash script?

Answer 1

1

It can be exported locally.

#!/bin/bash

LD_LIBRARY_PATH=/bin/path../version/:$LD_LIBRARY_PATH path/to/executable/executable 

using VAR=value cmd will set and export VAR for cmd process.

2

export it in two step

#!/bin/bash

LD_LIBRARY_PATH=/bin/path../version/:$LD_LIBRARY_PATH 
export LD_LIBRARY_PATH

path/to/executable/executable 

this is the way I do in my scripts. (note that this contradict what man page says)

export [-fn] [name[=word]] ...

If a variable name is followed by =word, the value of the variable is set to word.