如何在 bash 中使用 mod 运算符？

发布于 2024-11-01 09:52:02 字数 466 浏览 6 评论 0原文

我正在尝试这样的行：

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

我想要得到的输出是：

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

但我实际得到的是：

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

原文

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

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爱情眠于流年 2024-11-08 09:52:02

尝试以下操作：

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

$(( )) 语法执行内容的算术评估。

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.

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雨的味道风的声音 2024-11-08 09:52:02

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

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

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你如我软肋 2024-11-08 09:52:02

您必须将数学表达式放在 $(( )) 内。

一行：

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

多行：

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

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

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长亭外，古道边 2024-11-08 09:52:02

这可能是题外话。但是对于 for 循环中的 wget ，你当然可以这样做

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

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]

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叫思念不要吵 2024-11-08 09:52:02

bash 中的数学：如何使用所有 bash 运算符（`+`、`-`、`/`、`*`、`**`、`&`、`&&`、`<<` 等），以及算术扩展，在到目前为止，在这个问题的 346,000 名访客中

，我敢打赌，其中 344,900 名访客只是想回答这个问题的标题

Math in bash: how to use all bash operators (`+`, `-`, `/`, `*`, `**`, `&`, `&&`, `<<`, etc.), and arithmetic expansion, in bash

Of the 346k visitors to this question thus far, I'd be willing to bet 344.9k of them just want the title of this question answered ????:

How to use mod operator in bash?

Even I googled "bash modulus" looking for that answer, and landed here. So, now that I've figured it out, let's just jump straight to it:

How to use the modulus (`%`) operator in bash

Just do this, for instance:

# 7 mod 4 (answer is 3, but to print the output you must use one of the cmds
# below)
$((7 % 4))

# [PREFERRED: no quotes]
# print the result (double quotes are not required)
echo $((7 % 4))

# print the result (with double quotes if you like)
echo "$((7 % 4))"

Example with variables:

num1="7"
num2="4"

# [PREFERRED: no $ signs nor extraneous quotes] result is 3
echo $((num1 % num2))

# Also ok: with $ signs
echo $(($num1 % $num2))

# Also ok: with $ signs and extra quotes
echo "$(("$num1" % "$num2"))"

Store the result into a variable:

mod=$((num1 % num2))
echo "$mod"  # result is 3

The main links to study for these concepts are these, from the official GNU bash user manual:

More on bash "arithmetic expansion"

I learned the above from @Mark Longair's answer (although it took me some effort to comprehend it all), and that's where I got the link just below. I then did more research.

The $(( )) part is called "Arithmetic Expansion", and is described in the official GNU bash user manual here: https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Arithmetic-Expansion.

Basic examples (place echo in front of each one to see the result print to the screen):

# general form
$((mathematical_expression))

# addition
$((7 + 4))  # 11

# subtraction
$((7 - 4))  # 3

# modulus (remainder)
$((7 % 4))  # 3

# logical AND
$((7 && 4))  # 1

# bitwise AND
$((7 & 4))  # 4

# etc.
# See the full operator list below for more

Double quotes around the arithmetic expansion are not needed. From the manual above (emphasis added):

The expression is treated as if it were within double quotes, but a double quote inside the parentheses is not treated specially. All tokens in the expression undergo parameter and variable expansion, command substitution, and quote removal. The result is treated as the arithmetic expression to be evaluated. Arithmetic expansions may be nested.

For all shell arithmetic operators, see the "Shell Arithmetic" section of the GNU bash manual here: https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Shell-Arithmetic

You essentially have all of the C language mathematical operators at your disposal. The arithmetic is done "in fixed-width integers with no check for overflow", so if you do echo $((11/10)) or echo $((19/10)) you'll get 1 in both cases since the fractional part is truncated for integers.

From the manual link just above (emphasis added):

Evaluation is done in fixed-width integers with no check for overflow, though division by 0 is trapped and flagged as an error. The operators and their precedence, associativity, and values are the same as in the C language.

Since the arithmetic operators in bash have the same precedence as in C, as it states above, you can also reference the C Operator Precedence from the cppreference community wiki here: https://en.cppreference.com/w/c/language/operator_precedence <-- put that in your toolbag.

Shell Arithmetic: here are all of the supported operators from the GNU Bash manual

They are listed in order of highest to lowest precedence:

id++ id--
variable post-increment and post-decrement
++id --id
variable pre-increment and pre-decrement
- +
unary minus and plus
! ~
logical and bitwise negation
**
exponentiation
* / %
multiplication, division, remainder
+ -
addition, subtraction
<< >>
left and right bitwise shifts
<= >= < >
comparison
== !=
equality and inequality
&
bitwise AND
^
bitwise exclusive OR
|
bitwise OR
&&
logical AND
||
logical OR
expr ? expr : expr
conditional operator
= *= /= %= += -= <<= >>= &= ^= |=
assignment
expr1 , expr2
comma

Using alternate bases in your arithmetic, such as binary (base-2), octal (base-8), and hex (base-16)

To learn about using different bases, such as base-2 (binary), base-8 (octal) or base-16 (hex) with the bash arithmetic operators, read the next couple paragraphs below the "Shell Arithmetic" list above in the manual.

Here are a few quick examples with input numbers which are decimal (base-10), octal (base-8), hex (base-16), and binary (base-2), used in the math:

# hex 0xa (decimal 10) + decimal 5 = decimal 15
echo $((0xa + 5))  # prints `15` (decimal 15)
# OR (same thing)
echo $((16#a + 5))  # prints `15` (decimal 15)

# octal 071 (decimal 57) + hex 0xaa (decimal 170) = decimal 227
echo $((071 + 0xaa))  # prints `227` (decimal 227)
# OR (same thing)
echo $((8#71 + 16#aa))  # prints `227` (decimal 227)

# binary 1011 (decimal 11) + decimal 2 = decimal 13
echo $((2#1011 + 2))  # prints `13` (decimal 13)

# binary 1111 (decimal 15) + binary 11111 (decimal 31) = decimal 46
echo $((2#1111 + 2#11111))  # prints `46` (decimal 46)

To print as hex, use printf "0x%X\n" number:

# prints `0x2E` (hex 2E, or decimal 46)
printf "0x%X\n" $((2#1111 + 2#11111))

To print as binary, use bc (see my answer here):

# prints `0b101110` (decimal 46)
printf "0b%s\n" "$(echo "obase=2; $((2#1111 + 2#11111))" | bc)"

[my answer] How do I use floating-point arithmetic in bash?
If you need a full floating point library in bash, that you can easily import (source), use my code here: bash/floating_point_math.sh in my eRCaGuy_hello_world repo.
1. See also my answer about using Bash libraries here: Detailed example: how do you write, import, use, and test libraries in Bash?

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昵称有卵用 2024-11-08 09:52:02

这篇文章相当老了，但我想我会做出贡献，因为我在尝试研究通过自动化设置键盘颜色的相同问题时偶然发现了它。

我创建了一个简单的 BASH 脚本，每分钟从我的 ROOT chrontab 调用该脚本，以随着时间的推移设置键盘颜色。您可以调整颜色模式和模数以满足您的需求。这只是一个好的起点。

#!/bin/bash
# must run as ROOT to work
# put in your root crontab to change the color at set times

sec=$(date +%s)
min=$(( $sec / 60 ))
col=$(( $min % 7 ))
colors=('0000FF' '00FF00' '00FFFF' 'FF0000' 'FF00FF' 'FFFF00' 'FFFFFF')
colorFile="/sys/class/leds/system76_acpi::kbd_backlight/color"

if [ -f "$colorFile" ]; then
    echo "Set keyboard to color $col ~ ${colors[$col]}"
    echo "${colors[$col]}" > "$colorFile"
fi

希望你喜欢它。

This post is rather old but I thought I would contribute since I stumbled upon it while trying to research the same issue of setting keyboard color through automation.

I created a simple BASH script that I call from my ROOT chrontab every minute to set the keyboard color as the day progresses. You can tweak the color patterns and the modulo to match your needs. This is just a good starting point.

#!/bin/bash
# must run as ROOT to work
# put in your root crontab to change the color at set times

sec=$(date +%s)
min=$(( $sec / 60 ))
col=$(( $min % 7 ))
colors=('0000FF' '00FF00' '00FFFF' 'FF0000' 'FF00FF' 'FFFF00' 'FFFFFF')
colorFile="/sys/class/leds/system76_acpi::kbd_backlight/color"

if [ -f "$colorFile" ]; then
    echo "Set keyboard to color $col ~ ${colors[$col]}"
    echo "${colors[$col]}" > "$colorFile"
fi

Hope you like it.

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~没有更多了~