如果不使用&b;& amp ;，则使用bash的正确方法是什么?||捷径? [英] What is the correct way to use the bash if else && || shortcut?

问题描述

我遇到了一些奇怪的行为(也许并不奇怪，但我不理解！)

I have come across some strange behavious (well probably not strange but I don't understand it !)

我想使用bash速记语法编写一些if/e，se语句:

I want to write some if / e,se statements using the bash shorthand syntax:

[[ 1 -eq 1 ]] && echo "true" || echo "false"

以上代码的输出为:

true

现在，上面的代码实际上可以正常工作.但是下面的代码没有:

Now the code above actually works fine. But this next code does not:

[[ 1 -eq 1 ]] && infMsg "true" || infMsg "false"

infMsg 只是一个函数.上面的代码输出为:

infMsg is just a function. The output from the above code gives:

true
false

我只希望它说"true".

I only want it to say 'true'.

在&&中，我是否缺少某些东西?||语法处理函数的退出代码?

Is there something I am missing in the way the && || syntax handles exit codes from functions ?

推荐答案

我怀疑您的'infMsg'退出代码不为0(成功).如果我们分解第一段代码:

I suspect your exit code for 'infMsg' is not 0 (success). If we break down the first piece of code:

[[ 1 -eq 1 ]] && echo "true" || echo "false"

发生了什么事

• [[1 -eq 1]]&&因为&& 的优先级高于 || (更多信息[此处语句.(&&->如果第一部分是正确的，请运行第二部分)
• 现在我们有了 ||回显"false" .由于上半部分( echo"true" )返回true，因为回声已成功打印((||->如果第一个部分为false，请运行第二个部分)，所以实际上该部分没有执行.部分).).

• The [[ 1 -eq 1 ]] && echo "true" portion of code is evaluated first since && has a higher precedence than || (more info[here1)
• The first half of that returns true. Then, since the first half was true, the second half executes the echo "true" statement. (&& -> If part one is true, run the second part)
• Now we have the || echo "false". This part actually doesn't get executed since the first half (echo "true") returned true since the echo printed successfully ((||-> If part one is false, run the second part).).

现在，如果我们分解第二段代码:

Now if we break down the second piece of code:

[[ 1 -eq 1 ]] && infMsg "true" || infMsg "false"

最有可能发生的事情 是:

What's most likely happening is:

• [[1 -eq 1]]&&infMsg"true" 被评估.第一面是正确的.然后，由于前半部分为true ，因此后半部分执行 infMsg"true" 语句.(&&->如果第一部分是正确的，请运行第二部分).
• 现在我们有 ||infMsg"false" .如果前一个命令的 infMsg"true" 没有返回解释为成功的状态代码(0)，则前半部分将被解释为false，从而触发 infMsg"false"((||->如果第一部分为假，请运行第二部分).)..

• [[ 1 -eq 1 ]] && infMsg "true" is evaluated. The first side is true. Then, since the first half was true, the second half executes the infMsg "true" statement. (&& -> If part one is true, run the second part).
• Now we have || infMsg "false". If the infMsg "true" from the previous command didn't return a status code that is interpreted as success (0), the first half is interpreted as false, triggering the infMsg "false"((||-> If part one is false, run the second part).)..

要移至此处的重要信息:

The important info to pull away here:

• ||仅在前半部分为FALSE时才运行后半部分.
• &&如果前半部分为TRUE，则仅运行后半部分.

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