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Bash shell script with mandatory and optional input

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# 1  
Old 05-18-2015
Bash shell script with mandatory and optional input
Hello
I would like to write a bash shell script which will need user to supply one variable which is mandatory and some other optional variables. If mandatory variable is not supplied by user, the script will exit. If optional values are not supplied by user, hard-coded value (in the script) will be used.

Please help
# 2  
Old 05-18-2015
1. You can start by learning what positional parameters are.
2. If you want something advanced, learn about getopts
3. And, please share with us what you've tried along with your OS/shell information.
# 3  
Old 05-18-2015
You can continue your code/script followed by this below if statement.

Code:
#!/bin/bash

if [ $# -lt "1" ]; then
        echo "Usage:  $0 <something>"
        exit 1;
fi

# 4  
Old 05-18-2015
Both answers that you have already are useful. I would use a combination of if [ $# -lt "1" ]; then and getopts. You should start by writing a test script that will take the parameters that you want and print out the parameter passed in, specifically the optional parameters. Once you know that the script is handling the parameters correctly, then you can write the rest of the script.
Last edited by Don Cragun; 05-18-2015 at 07:36 PM.. Reason: Add ICODE tags.
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