shell script source behaviour

Question

I am new to shell script. I have a shell script which sources another file. Another file is just a file with name value pair. I have a confusion here.

test.dat
TEST_VARIABLE_PATH=/usr/project/workspace

script file (test_script.sh)

TEST_VARIABLE_PATH=
source test.dat
echo ${TEST_VARIABLE_PATH}

The above echo statement prints different values which makes me confused. In line 1, if I make

TEST_VARIABLE_PATH=/somepath/

, even I have sourced test.dat which has TEST_VARIABLE_PATH variable already set, echo is printing only

TEST_VARIABLE_PATH=/somepath/

where sourcing the file did not overwrite the variable.

Suppose If I have the line 1 as

TEST_VARIABLE_PATH=\`dirname $0\`/../

, source command overwrites and echo prints

TEST_VARIABLE_PATH=/usr/project/workspace

Why is this source behaving differently ? Please help me to understand.

Gilles Quénot · Accepted Answer · 2012-11-21 08:24:53Z

2

It seems like you don't really source the file.

To do it properly, put the full path of the file, like :

source '/full/path/to/file'

edited Nov 21, 2012 at 8:24

answered Nov 21, 2012 at 8:18

Gilles Quénot

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Jite Over a year ago

Or just . /path/to/file. Notice that it is dot[space]/path/to/file

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