bash的逻辑运算符如何工作(或为什么它们在这里不工作)? [英] How do bash's logical operators work (or why don't they work here)?

问题描述

我有以下包装器脚本:

#!/bin/bash

$local_time=`date +%H%M`
if [ $local_time -ge 1430 ] && [ $local_time -le 2230 ] ||
   [ $local_time -ge 0300 ] && [ $local_time -le 0400 ]
then
   echo 'pass'
else
   echo 'skip'
fi

这个想法是创建一个包装器脚本，该脚本在两个不相交的时间间隔之间运行一个脚本.每个要进行或"运算的表达式都可以正确求值，但是，当我对两个表达式进行或"运算时，测试将失败.

The idea being to create a wrapper script that runs a script between two disjoint time intervals. Each of the expressions being OR'd is evaluated correctly, however, when I OR both expressions, the test fails.

我该如何解决?

推荐答案

您有两个问题

1. 分配变量时，不要在变量前加上$前缀，只有在扩展变量时才使用
2. 您需要使用括号( )
强制执行运算符优先级

1. You don't prefix your variables with a $ when you assign to them, only when you expand them
2. You need to enforce operator precedence with parens ( )

此外，由于您已经在使用bash，因此最好将其更好的语法与(( ))构造一起使用，该构造允许您使用比较运算符< > == !=，并且可以使用$()代替命令反引号/刻痕

Also, since you're already using bash, might as well use its better syntax with the (( )) construct that allows you to use the comparison operators < > == != and you can use $() for command substitution instead of the backticks/gravemarks

#!/bin/bash

local_time=$(( 10#$( date +'%H%M' ) ))
if (( ( local_time >=    1430 && local_time <=    2230 ) ||
      ( local_time >= 10#0300 && local_time <= 10#0400 )    ))
then
  echo 'pass'
else
  echo 'skip'
fi

以零开头的数字被解释为八进制.
同样，任何以0开头且包含8或9的数字都会显示错误.
前缀(10#)可以解决该问题.

Numbers that start with a zero are interpreted as octal.
Also, any number that start with 0 and contains an 8 or 9 will print an error.
Prefixing (10#) solves that.

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