not able to use local variable outside the function after exporting

Question

This is my script. Even after using the export command not able to use variable outside of the block. Below is the code that i have tried. i have tried other option like declare -x var but that is also not working. Can someone please please comment on this, am i doing right?

#!/bin/bash
{
    var="123"  
    export var   # exporting the variable so that i can access from anywhere   
    echo "var is "$var     # able to get the value of this variable
} | tee   log.txt

echo "var is "$var   # not able to get the value of this variable

still solution is not clear for me , so what the changes i have to done in code to access that variable . and thanks for ur reply and for ur valuable time . — brajkishore dubey, Jul 02 '17 at 08:47
The change you need to make is **DON'T set the variable in a pipeline** (except for the last segment in some non-bash shells and recent versions of bash with shopt lastpipe turned on, but you don't have it in the last segment anyway). — dave_thompson_085, Jul 02 '17 at 09:00

Jaleks · Answer 1 · 2017-07-03T10:22:34.823

According to this accepted SOF answer:

In a pipeline, all commands run concurrently (with their stdout/stdin connected by pipes) so in different processes.

So although the curly braces by themselves do not create a subshell, the pipe then does this (in bash); see e.g.

#!/bin/bash
var=256

ps -u | grep $0
{
ps -u | grep $0
{
var="123"  
export var
echo "var is "$var

ps -u | grep $0

} | tee log.txt
echo "var is "$var
}

echo "var is "$var

So what we want is avoiding the pipe, while still having output on screen and to the logfile. Luckily bash got the feature <(...) for creating temporary FIFOs. Example below shows a possibility where one still can use a code block, transfer its whole output to a log (here stdout and stderr are appended to different log files) and have the changed variables accessible later, as we do not get into a subshell.

#!/bin/bash

VAR=123
echo "VAR first: $VAR"

{
        VAR=256
        echo "VAR in block: $VAR"
        norealcommand
        # show open files to see FIFOs
        #lsof -p $$
        date
        otherbadcommand
}> >(tee -a stdout.log) 2> >(tee -a stderr.log >&2)

# dummy output to not mess up output order
echo | cat 

#lsof -p $$
echo "VAR after block: $VAR"

cat -n stdout.log 
cat -n stderr.log

which should result in something like this:

$ ./fifo                                                                            /dev/shm |myself@mydesk|0|12:04:10 
VAR first: 123
VAR in block: 256
./fifo: line 9: norealcommand: command not found
Mon Jul  3 12:04:37 CEST 2017
./fifo: line 13: otherbadcommand: command not found

VAR after block: 256
     1  VAR in block: 256
     2  Mon Jul  3 12:04:10 CEST 2017
     3  VAR in block: 256
     4  Mon Jul  3 12:04:37 CEST 2017
     1  ./fifo: line 9: norealcommand: command not found
     2  ./fifo: line 13: otherbadcommand: command not found
     3  ./fifo: line 9: norealcommand: command not found
     4  ./fifo: line 13: otherbadcommand: command not found

Hope this makes you happy :-)

awesome , thanks ! , thanks for ur valuable time , this solution is working . — brajkishore dubey, Jul 02 '17 at 18:24
how to get the value of var="123" , outside of curly braces . I am getting output 123 , 256 ,256 .. but i want 123 123 123 . — brajkishore dubey, Jul 02 '17 at 18:40

not able to use local variable outside the function after exporting

1 Answers1