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It's probably a lame question. But I am getting 3 arguments from command line [ bash script ]. Then I am trying to use these in a for loop.
for i in {$1..$2} do action1 done This doesn't seem to work though and if $1 is "0" and $2 is 2 it prints {0..2}' and callsaction1` only once. I referred to various examples and this appears to be the correct usage. Can someone please tell me what needs to be fixed here?
Thanks in advance.
638
You can handle command-line arguments in a bash script in two ways. One is by using argument variables, and another is by using the getopts function.
$# is the number of positional parameters passed to the script, shell, or shell function. This is because, while a shell function is running, the positional parameters are temporarily replaced with the arguments to the function. This lets functions accept and use their own positional parameters.
bash [filename] runs the commands saved in a file. $@ refers to all of a shell script's command-line arguments. $1 , $2 , etc., refer to the first command-line argument, the second command-line argument, etc. Place variables in quotes if the values might have spaces in them.
Use the for loop to iterate through a list of items to perform the instructed commands.
You can slice the input using ${@:3} or ${@:3:8} and then loop over it
For eg., to print arguments starting from 3
for i in ${@:3} ; do echo $i; done or to print 8 arguments starting from 3 (so, arguments 3 through 10)
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