How to use mod operator in bash?

问题

I'm trying a line like this:

for i in {1..600}; do wget http://example.com/search/link $i % 5; done;

What I'm trying to get as output is:

wget http://example.com/search/link0
wget http://example.com/search/link1
wget http://example.com/search/link2
wget http://example.com/search/link3
wget http://example.com/search/link4
wget http://example.com/search/link0

But what I'm actually getting is just:

    wget http://example.com/search/link

回答1:

Try the following:

 for i in {1..600}; do echo wget http://example.com/search/link$(($i % 5)); done

The $(( )) syntax does an arithmetic evaluation of the contents.

回答2:

for i in {1..600}
do
    n=$(($i%5))
    wget http://example.com/search/link$n
done

回答3:

You must put your mathematical expressions inside $(( )).

One-liner:

for i in {1..600}; do wget http://example.com/search/link$(($i % 5)); done;

Multiple lines:

for i in {1..600}; do
    wget http://example.com/search/link$(($i % 5))
done

回答4:

This might be off-topic. But for the wget in for loop, you can certainly do

curl -O http://example.com/search/link[1-600]

来源：https://stackoverflow.com/questions/5688576/how-to-use-mod-operator-in-bash