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Export shell variable

Posted on 2015-02-10
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Hi ,

In my script i use export variable . So i can use the variable in another script which was sourced from main script.

But in command line, if i use $variable i see an empty value returned.

Please help on this. My aim is if some problem happened the script will be started in middle so if they start script 2 it should aware of the variable which was set in main script.

Thanks
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Question by:magento
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Accepted Solution

by:
ozo earned 1200 total points
ID: 40601136
You would need to source the script that defined $variable from the command line in order to use $variable in the command line.
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Assisted Solution

by:woolmilkporc
woolmilkporc earned 800 total points
ID: 40601152
"export" makes a variable available to subshells of the shell where it was issued, but never to the parent of that shell.

That's a security measure to avoid that a script could modify the environment of your login shell.
As ozo said, you must source the script which exports the variable to make it available to the current shell (= the parent of the script containing "export"):

. scriptname # note it's  "dot space" on the left!)

or (only in bash)

source scriptname
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Author Comment

by:magento
ID: 40601167
say script1.sh has a variable named code_version . This is the input to script2.sh

Sometimes the system will fail for unknown causes ..at that time, script 1.sh completed so we will start script2.sh . At that time the variable input will not be available.

Can you advice me how to solve this issue ?

Script1.sh
export $(code_version=1.7)

script2.sh
if [ $code_version == 1.7 ] then ...

Thanks
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Expert Comment

by:woolmilkporc
ID: 40601187
. Script1.sh

Please note that the script must not contain "exit"! This would terminate the calling shell (probably your login shell)!

By the way, what's this strange syntax "$(code_version=1.7)" ??
This will definitely not work (at least not in the shells I use to use).

Correct syntax is simply:

export code_version="1.7"
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Expert Comment

by:simon3270
ID: 40601316
If you want variables to be seen in other scripts, your best bet is to write the valeus to a file, as:

script1.sh
  code_version=1.7
  echo $code_version > /tmp/code_ver.txt

script2.sh
   code_ver=$(cat /tmp/code_ver.txt)
   echo Code version is $code_ver

Note that the variable name does not have to be the same in both scripts.


An alternative is to write the definition of the variable to the file, as

script1.sh
  echo export code_version=$code_version > /tmp/code_ver.txt

then source that file in the second script to define the value

script2.sh
  . /tmp/code_ver.txt

In this case, the variable name in script2.sh will be whatever you wrote to /tmp/code_ver.txt.
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Expert Comment

by:ozo
ID: 40601420
If you want the input to script2.sh to come from script1.sh,
you can pipe the input;
script1.sh | script2.sh
with
cat > script1.sh <<< 'echo 1.7'
cat > script2.sh <<< 'read code_ver; echo $code_ver'
or pass an arg:
script2.sh `script1.sh`
with
cat > script2.sh <<< 'echo $1'
or call script2 from script1
with
cat > script1.sh <<< 'script2.sh 1.7'
or
cat > script1.sh <<< 'echo 1.7 | script2.sh
or
cat > script1.sh <<< 'export code_ver=1.7; script2.sh'
or call script1 from script2
with
cat > script2.sh <<< 'code_ver=`script1.sh`; echo $code_ver'
or with
cat > script1.sh <<< 'code_ver=1.7'
cat > script2.sh <<< '. script1.sh; echo $code_ver'
or with
cat > script1.sh <<< 'exit 17'
cat > script2.sh <<< 'script1.sh; echo $?'
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Expert Comment

by:simon3270
ID: 40614528
Just remember that sourcing a shell script updates your current shell environment, possibly affecting anything you do afterwards (e.g. adding entries to the PATH, changing current directory, creating function names that silently overlay commands and so on).

It will probbaly be fine for you, but is something to bear in mind.
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