If I run the bash script below as ./myscript.sh -i what will be the output to terminal?#!/bin/bashOPTERR="Invalid option"if [[ $# -gt 0 ]]; then while getopts "zpa" opt; do case $opt in z) echo "Apple";; p) echo "Banana";; a) echo "Orange";; *) echo "$OPTERR" && exit 1;; esac donefiexit 0
Question
If I run the bash script below as ./myscript.sh -i what will be the output to terminal?#!/bin/bashOPTERR="Invalid option"if [[ # -gt 0 ]]; then while getopts "zpa" opt; do case opt in z) echo "Apple";; p) echo "Banana";; a) echo "Orange";; *) echo "$OPTERR" && exit 1;; esac donefiexit 0
Solution 1
The output will be "Invalid option". This is because the script is designed to only accept the options "z", "p", and "a". When you run the script with the "-i" option, it doesn't match any of the specified options, so the script echoes the value of the variable OPTERR ("Invalid option") and then exits with a status of 1.
Solution 2
The output will be "Invalid option".
Here's the step by step explanation:
- The script is run with the -i option.
- The script checks if there are any arguments passed to it. In this case, there is one argument (-i).
- The script enters a while loop where it uses the getopts function to iterate over the options. The options it is looking for are z, p, and a.
- When it encounters the -i option, it doesn't match any of the specified options (z, p, a).
- Therefore, it goes to the *) case, which is the default case when no other cases match.
- It then echoes "Invalid option" to the terminal and exits the script with a status of 1.
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