[SOLVED] Multiline + pipeline command assigns values to variables in bash

Issue

I have writen a bash file like this:

  JSESSIONID=$(curl -I 'http://17....212' \
  -H 'Accept: text/h...b3;q=0.7' \
  -H 'Accept-Language: zh...8' \
  -H 'Cache-Control: ma...0' \
  -H 'Connection: k...ve' \
  -H 'Upgrade-Insecure-Requests: 1' \
  -H 'User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/118.0.0.0 Safari/537.36' \
  --insecure|grep -o -e "JSESSIONID=\(.*\); P"|sed "s/JSESSIONID=//g"|sed "s/; P//g")
  echo $(JSESSIONID)

(Apostrophe is for privacy)

I want to get JSESSIONID, I can run it in the command-line, and it will return a pure string: The JSESSIONID. However, I dont know how to assign it to a variable. By using the code above, it will shows:

/etc/rc.common: line 37: JSESSIONID: not found

I know this question is very basic, but I have searched for a long time and have not found a workable answer.

have tried to switch "$()" to "``", But it is useless.

Solution

JSESSIONID already contain result of execution your command. You don't need to run it again with command substitution.

For getting value use

echo "${JSESSIONID}"

or

echo "$JSESSIONID"

Answered By - Neo Lip

Answer Checked By - Candace Johnson (WPSolving Volunteer)