Issue
I have a shell script sample.sh. Inside the shell script, there are many commands, it looks like this for example:
#!/bin/bash
command 1 ......
command 2 ......
command 3 ......
txt1="/users/doc/folder1/sam.txt"
txt2="/users/doc/folder2/pam.txt"
txt3="/users/doc/folder3/ram.txt"
echo "run done"
First I gave a run with this script like sh sample.sh. After running this shell script I want to run a command in which I wanted to give txt3 which was the name I'm using instead of the path for the file ram.txt
The command looks like this for eg:
convert -i txt3 > sim.tsv
This gave me an error. Error: The requested txt file (txt3) could not be opened. Exiting!
May I know how this works without giving the path to the file in the command?
Solution
You need to source the script, not run it, and export your variables to have them persist after the script finished.
You have to refer to the variable as $txt3 after the sourcing.
Answered By - tink Answer Checked By - Gilberto Lyons (WPSolving Admin)