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How to export a specify full directory by shell file

Requirement

  • project files

    myprj
    build
      env.sh
      Makefile
      submake
        MakeFlagTest1
        MakeFlagTest2
    doc
      readme.doc
    src
      a.h
      a.c
      b.h
      b.c
      main.c
    test
      test.c

  • In env.sh, need to export some environment variable, how to implement?
    If add hard code, it will looks like : export PRJ_HOME=/home/dennis/myprj

  • Using case

    • cd /home/dennis; source myprj/build/env.sh
    • cd /home/dennis; source ./myprj/build/env.sh
    • cd /home/dennis; source ~/myprj/build/env.sh
    • cd /home/dennis/myprj/build; source env.sh
    • cd /home/dennis/myprj/build; source ./env.sh
    • cd /home/dennis/myUT/src; source ../../myprj/build/env.sh
    • cd /home/dennis/myUT/src; source ~/myprj/build/env.sh

Test 1, using export PRJ_HOME=$(dirname $(dirname $0))

  • Failure case, when on suse release version, will failed to do dirname $0,
    you will find that echo $0 output -bash, not bash expected.

Test 2, using export PRJ_HOME=$(dirname $(dirname ${BASH_SOURCE[0]))

  • Failure case, when using relative directory.

Other test

echo "working directory is " $(pwd)
echo "BASH_SOURCE is " ${BASH_SOURCE}
echo "BASH_SOURCE[0] is " ${BASH_SOURCE[0]}
echo "dirname BASH_SOURCE[0] is " $(dirname ${BASH_SOURCE[0]})
echo "2 dirname BASH_SOURCE[0] is " $(dirname $(dirname ${BASH_SOURCE[0]}))
PARENT_DIR=$(dirname $(dirname ${BASH_SOURCE[0]}))
echo "parent dir is " ${PARENT_DIR}
export GMDB_HOME=$(pwd ${PARENT_DIR})

TODO, find the effective method

  • If we can get the bash file name?
  • How to translate relative directory to be full directory?

Reference