LinuxQuestions.org - Reading a bash variable in bash scripting problem

- Programming (https://www.linuxquestions.org/questions/programming-9/)

- - Reading a bash variable in bash scripting problem (https://www.linuxquestions.org/questions/programming-9/reading-a-bash-variable-in-bash-scripting-problem-686390/)

Reading a bash variable in bash scripting problem

Hi,

I have a config file that contains:

my.config:

Code:

Section

INSTALLATION_PATH = $HOME

EndSection

Now in my bash script, I want to get the output /home/user instead of $HOME once read. So far, I have managed to get the $HOME variable but I can't get it to echo the variable. All I get is the output $HOME.

here is my parse_cmd script:

Code:

#Get the parameter from the section                                            

RESULT=`sed -n "/Section/,/EndSection/p" $1 | grep -w $3`                                                                            



#Ignore comment                                                                

RESULT=`echo $RESULT | awk -F# '{print $1}'`



#remove all beginning white spaces                                              

RESULT=`echo $RESULT | awk -F= '{print $NF}' | awk '{sub("^ ", "", $0); print $0}'





#return value                                                                  

echo $RESULT



exit 0

Any idea how to get around the problem.

Thanks,
Indy

Code:

eval echo $RESULT

Moin,

you can do the config parsing in one step:

Code:

jan@jack:~/tmp/config> cat cfg

Section

INSTALLATION_PATH = $HOME                # the installation path

EndSection

jan@jack:~/tmp/config> eval echo `sed -nr '/^Section/,/^EndSection/{/Section/d;s/.* *= *//;s/#.*//;p}' cfg`

/home/jan

Jan

thanks colucix, easy and neat.

Jan64,
I prefer not to do everything in one line. If it needs to be modified or improved, it's better policy in my opinion to have it separated into several commands than just one. Thanks for the tip though.

Indy