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Understanding $PATH better

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# 1  
Old 07-14-2015
Hammer & Screwdriver Understanding $PATH better
Dear all,

I have a script which triggers a C program in it after initialising parameters.

Below is the sample

Code:
#! /bin/sh

. /my_data/environment

HOME=/home/execs/testing/

cd $HOME

compare $1 $2 X

exit $?

When I tried to run the script as sh -x my_script1.sh file1 file2 it is throwing 127 error.
# 2  
Old 07-14-2015
Firstly, I would remove the space between #! and /bin/sh on the first line.

Can you post us the full error message? I'm assuming that compare is a function declared within /my_data/environment, but it would be good to know the content of that too.



I'm not sure where $PATH comes into your post though?


Thanks, in advance,
Robin
This User Gave Thanks to rbatte1 For This Post:
PikK45
# 3  
Old 07-14-2015
Robin,

That is a typo on the first line Smilie compare is the executable from a C program to compare files. environment file will have several variables defined and exported.

It does have export $PATH in it. And the $PATH variable has the "." in it.

I wonder why, the script is not looking in the current directory while I execute compare.

Thanks
# 4  
Old 07-14-2015
Well, if you display the value of PATH before your call to the program, we can discuss it.

Can you also paste the full error message as the number 127 is not unique as an error message.



Robin
# 5  
Old 07-14-2015
PATH=/my_data/execs:/universal/execs/:/newpath/configure/:.:/test_dir/:/bin/

This site has the message details
# 6  
Old 07-14-2015
Does compare reside in any of those directories? Does it have "execute" permission?
# 7  
Old 07-14-2015
the executable compare is present in HOME=/home/execs/testing/. I switch directory before I run compare as well
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