Question

我们编写了一个简单的shell脚本

cd location\address

var1='grep -w -c exception server.log
var2='grep -w -c exception server.log.1

var3= $var1 + $var2

echo $var3
echo $var3
echo 'addition of exception : ' $var3

输出：

240
82
240+82

如何正确地得到总和

Answer 1

在UNIX shell中有多种方法可以执行算法，但是，在我进入之前，你需要意识到除非你的shell 与大多数shell不同，否则你需要定义变量变量名称，=或值之间没有空格：var1='abc'

为此，我假设你 使用bash

echo $((( 1 + 10 )))
echo $[ 1 + 10 ]
# 11

var=0
((++var))
echo $var
# 1 

((++var))
echo $var
# 2

((--var))
echo $var
# 2

((var = 1 + 2))
echo $var
# 3

let "var = 3 + 1"  # let is another way of writing `(( ... ))`
echo $var
# 4

echo $(( 1 + 2 ))
# 3

(((var = 20 + 1)))
echo $var
# 21

Answer 2

bc命令应该可以正常工作

echo "4+10" | bc

Answer 3

为了帮助您入门，这里有一个带注释，语法正确的代码版本，它应该适用于所有与POSIX兼容的shell（例如bash）：< / p>

#!/bin/sh

# Note that '/' (not '\') must be used as the path separator.    
cd location/address

# Capture command output using *command substitution*, `$(...)`
# Note: NO spaces around `=` are allowed.
var1=$(grep -w -c exception server.log)
var2=$(grep -w -c exception server.log.1)

# Use *arithmetic expansion*, `$((...))`, to perform integer calculations.
# Note: Inside `$((...))`, the prefix `$` is optional when referencing variables.
var3=$(( var1 + var2 ))

# Generally, it's better to *double*-quote variable references to
# make sure they're output unmodified - unquoted, they are subject to
# interpretation by the shell (so-called *shell expansions*).
echo "$var1"
echo "$var2"

# Output the result of the calculation. Note the use of just one, double-quoted
# string with an embedded variable reference.
# (By contrast, a *single*-quoted string would be used verbatim - no expansions.)
echo "sum of exceptions: $var3"

一般提示：

http://shellcheck.net是用于语法检查shell代码的优秀工具。
http://explainshell.com是一个很棒的网站，根据man页面解释给定的命令行。
学习资源bash ，可能是使用最广泛的POSIX兼容shell：
- 简介：http://www.faqs.org/docs/Linux-HOWTO/Bash-Prog-Intro-HOWTO.html
- 指南：http://mywiki.wooledge.org/BashGuide
- 备忘单：http://mywiki.wooledge.org/BashSheet

shell中程序的基本添加

3 个答案: