Bash if [假] ; 返回真
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Bash if [ false ] ; returns true
提问by tenmiles
Been learning bash this week and ran into a snag.
本周一直在学习 bash 并遇到了障碍。
#!/bin/sh
if [ false ]; then
echo "True"
else
echo "False"
fi
This will always output True even though the condition would seem to indicate otherwise. If I remove the brackets []then it works, but I do not understand why.
即使条件似乎另有指示，这也将始终输出 True。如果我删除括号[]然后它工作，但我不明白为什么。
回答by chepner
You are running the [(aka test) command with the argument "false", not running the command false. Since "false" is a non-empty string, the testcommand always succeeds. To actually run the command, drop the [command.
您正在使用参数“false”运行[（又名test）命令，而不是运行命令false。由于“false”是一个非空字符串，所以test命令总是成功的。要实际运行该命令，请删除该[命令。
if false; then
echo "True"
else
echo "False"
fi
回答by Beejor
A Quick Boolean Primer for Bash
Bash 的快速布尔入门
The ifstatement takes a command as an argument(as do &&, ||, etc.). The integer result code of the command is interpreted as a boolean (0/null=true, 1/else=false).
该if声明需要一个命令作为参数（如做&&，||等）。命令的整数结果代码被解释为布尔值（0/null=true，1/else=false）。
The teststatement takes operators and operands as argumentsand returns a result code in the same format as if. An alias of the teststatement is [, which is often used with ifto perform more complex comparisons.
该test语句将运算符和操作数作为参数，并以与if. test语句的别名是[，通常用于if执行更复杂的比较。
The trueand falsestatements do nothing and return a result code(0 and 1, respectively). So they can be used as boolean literals in Bash. But if you put the statements in a place where they're interpreted as strings, you'll run into issues. In your case:
该true和false语句什么也不做，返回的结果代码（0和1，分别）。所以它们可以在 Bash 中用作布尔文字。但是，如果将语句放在将它们解释为字符串的位置，则会遇到问题。在你的情况下：
if [ foo ]; then ... # "if the string 'foo' is non-empty, return true"
if foo; then ... # "if the command foo succeeds, return true"
So:
所以：
if [ true ] ; then echo "This text will always appear." ; fi;
if [ false ] ; then echo "This text will always appear." ; fi;
if true ; then echo "This text will always appear." ; fi;
if false ; then echo "This text will never appear." ; fi;
This is similar to doing something like echo '$foo'vs. echo "$foo".
这类似于做类似echo '$foo'vs. 的事情echo "$foo"。
When using the teststatement, the result depends on the operators used.
使用test语句时，结果取决于所使用的运算符。
if [ "$foo" = "$bar" ] # true if the string values of $foo and $bar are equal
if [ "$foo" -eq "$bar" ] # true if the integer values of $foo and $bar are equal
if [ -f "$foo" ] # true if $foo is a file that exists (by path)
if [ "$foo" ] # true if $foo evaluates to a non-empty string
if foo # true if foo, as a command/subroutine,
# evaluates to true/success (returns 0 or null)
In short, if you just want to test something as pass/fail (aka "true"/"false"), then pass a command to your ifor &&etc. statement, without brackets. For complex comparisons, use brackets with the proper operators.
简而言之，如果您只想测试通过/失败（又名“真”/“假”），则将命令传递给您的if或&&等语句，不带括号。对于复杂的比较，请使用带有正确运算符的括号。
And yes, I'm aware there's no such thing as a native boolean type in Bash, and that ifand [and trueare technically "commands" and not "statements"; this is just a very basic, functional explanation.
是的，我知道在 Bash 中没有本地布尔类型这样的东西，而且ifand[和 andtrue技术上是“命令”而不是“语句”；这只是一个非常基本的功能性解释。
回答by Rodrigo
I found that I can do some basic logic by running something like:
我发现我可以通过运行类似的东西来做一些基本的逻辑：
A=true
B=true
if ($A && $B); then
C=true
else
C=false
fi
echo $C
回答by psh
Using true/false removes some bracket clutter...
使用 true/false 消除了一些括号混乱...
#! /bin/bash
# true_or_false.bash
[ "$(basename ##代码##)" == "bash" ] && sourced=true || sourced=false
$sourced && echo "SOURCED"
$sourced || echo "CALLED"
# Just an alternate way:
! $sourced && echo "CALLED " || echo "SOURCED"
$sourced && return || exit
