如何在 bash 中使用 mod 运算符？
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How to use mod operator in bash?
提问by Eric
I'm trying a line like this:
我正在尝试这样的一行：
for i in {1..600}; do wget http://example.com/search/link $i % 5; done;
What I'm trying to get as output is:
我试图获得的输出是：
wget http://example.com/search/link0
wget http://example.com/search/link1
wget http://example.com/search/link2
wget http://example.com/search/link3
wget http://example.com/search/link4
wget http://example.com/search/link0
But what I'm actually getting is just:
但我实际得到的只是：
wget http://example.com/search/link
回答by Mark Longair
Try the following:
请尝试以下操作：
for i in {1..600}; do echo wget http://example.com/search/link$(($i % 5)); done
The $(( ))syntax does an arithmetic evaluationof the contents.
该$(( ))语法做一个算术评估的内容。
回答by Chris Eberle
for i in {1..600}
do
n=$(($i%5))
wget http://example.com/search/link$n
done
回答by Higor E.
You must put your mathematical expressions inside $(( )).
您必须将数学表达式放在 $(( )) 中。
One-liner:
单线：
for i in {1..600}; do wget http://example.com/search/link$(($i % 5)); done;
Multiple lines:
多行：
for i in {1..600}; do
wget http://example.com/search/link$(($i % 5))
done
回答by h__
This might be off-topic. But for the wget in for loop, you can certainly do
这可能是题外话。但是对于 for 循环中的 wget，你当然可以做到
curl -O http://example.com/search/link[1-600]
