Bash run script from script using variable

Question

I am trying to run bash script from bash script using variable.

The variable is assigned before from parsed CSV, and when I echo it the path to another script looks ok. When I try to run the script path from the variable it fails.

Assiging the variable (2nd column from csv):

THE_VARIABLE=`echo ${CHOICE[${WHICH}]} | awk -F";" '{print $2}'`

echo $THE_VARIABLE (ok):

/fake_path/scripts/test.sh

Bash debug:

    + echo $'/fake_path/scripts/test.sh\r'

However after that when trying to run:

bash $THE_VARIABLE
: No such file or directoryripts/test.sh

It seems like taking only part of the string in variable(?).

Thanks for help!

Could be the \r for the carriage return causing the problem. — Raman Sailopal
– Raman Sailopal, Commented Aug 17, 2017 at 8:48
Yes, this cannot be fully reproduced without having all the code and the csv. I know only basics of bash and I am wondering why its cuts the string assigned to variable when trying to run it (echo is ok). Maybe some Bash parameter? — user8476799
– user8476799, Commented Aug 17, 2017 at 8:50
@grail echo "#$THE_VARIABLE#" ----> #/fakepath/scripts/test.sh — user8476799
– user8476799, Commented Aug 17, 2017 at 8:53

Nahuel Fouilleul · Accepted Answer · 2017-08-17 09:53:38Z

1

The string ends with ASCII character 13 (\x0D or \015 or \r) and not the sequence '\' 'r', with echo the trailing \r can't be seen because echo adds a \n, but appending a # character for example will show at begining of line.

echo "${THE_VARIABLE}#"
#fake_path/scripts/test.sh

to remove the trailing \r:

THE_VARIABLE=${THE_VARIABLE%$'\r'}

otherwise it means that input file has dos line ending \r\n, it could be modified with dos2unix.

edited Aug 17, 2017 at 9:53

answered Aug 17, 2017 at 9:46

Nahuel Fouilleul

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1 Comment

user8476799 Over a year ago

Perfect! I just run dos2unix on the csv file and everything works. Thanks!

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