How to pass parameters to a Linux Bash script?

Question

I have a Linux bash script 'myshell'. I want it to read two dates as parameters, for example: myshell date1 date2. I am a Java programmer, but don't know how to write a script to get this done.

The rest of the script is like this:

sed "s/$date1/$date2/g" wlacd_stat.xml >tmp.xml
mv tmp.xml wlacd_stat.xml

Answer 1

you use $1, $2 in your script eg

date1="$1"
date2="$2"
sed "s/$date1/$date2/g" wlacd_stat.xml >tmp.xml mv tmp.xml wlacd_stat.xml

Answer 2

$0 $1 $2

And so on will contain the script name, then the first and the second line argument.

Answer 3

Bash arguments are named after their position.

Moreover, if you need to handle one argument after the other, you can shift them and always use $1:

while [ $# -gt 0 ]
do
    echo $1
    shift
done

Answer 4

To iterate over the parameters, you can use this shorthand:

#!/bin/bash
for a
do
    echo $a
done

This form is the same as for a in "$@".