KISS

Question

我猜mytrue和myfalse被视为字符串而不是命令：

myfalse () {
    return 0
}

mytrue () {
    return 1
}

test-conditional ( ) {
    SHOW="dont-show"
    [[ ${SHOW} == "show" || mytrue ]] && echo "ok" || echo "wrong"
    [[ ${SHOW} == "show" && mytrue ]] && echo "err" || echo "ok"
    [[ ${SHOW} == "show" || myfalse ]] && echo "err" || echo "ok"
    [[ ${SHOW} == "show" && myfalse ]] && echo "err" || echo "ok"
    SHOW="show"
    [[ ${SHOW} == "show" || mytrue ]] && echo "ok" || echo "err"
    [[ ${SHOW} == "show" && mytrue ]] && echo "ok" || echo "err"
    [[ ${SHOW} == "show" || myfalse ]] && echo "ok" || echo "err"
    [[ ${SHOW} == "show" && mytrue ]] && echo "err" || echo "ok"

}

test-conditional

所以这显示了错误的输出：

ok
ok
err
ok
ok
ok
ok
err

（到处都是ok）

如何评估双括号内的命令？

Answer 1

您有true和false个函数返回相反的返回代码，您的上一个条件需要更正。

您可以使用：

myfalse () {
   return 1
}

mytrue () {
   return 0
}

test-conditional ( ) {
    SHOW="dont-show"
    [[ ${SHOW} == "show" ]] || mytrue && echo "ok" || echo "wrong"
    [[ ${SHOW} == "show" ]] && mytrue && echo "err" || echo "ok"
    [[ ${SHOW} == "show" ]] || myfalse && echo "err" || echo "ok"
    [[ ${SHOW} == "show" ]] && myfalse && echo "err" || echo "ok"
    SHOW="show"
    [[ ${SHOW} == "show" ]] || mytrue && echo "ok" || echo "err"
    [[ ${SHOW} == "show" ]] && mytrue && echo "ok" || echo "err"
    [[ ${SHOW} == "show" ]] || myfalse && echo "ok" || echo "err"
    [[ ${SHOW} == "show" ]] && myfalse && echo "err" || echo "ok"
}

test-conditional

另请注意函数调用如何放在[[ ... ]]之外，否则它不会调用函数。

在评估这些表达式时适用以下规则：

如果第一个条件失败，则&&之后的条件不会被评估。
如果第一个条件成功，则||之后的条件不会被评估。

Answer 2

在bash [[…]]内部（以及大多数shell中较旧的[…]内部）重要的是，如果测试的值有某些字符：

$ [[ somevalue ]] && echo "yes" || echo "no"

如果测试的值为空，则测试失败：

$ [[ "" ]] && echo "yes" || echo "no"
no

变量也是如此：

$ somevalue="a false string"
$ [[ $somevalue ]] && echo "yes" || echo "no"
yes

$ somevalue=""
$ [[ $somevalue ]] && echo "yes" || echo "no"
no

因此，[[…]]仅限于测试字符串而不是＆＃34;退出代码＆＃34; 您的函数定义＆＃34;退出代码＆＃34;不是字符串。
你可以

将mytrue和myfalse的值定义为代表这些值的变量：

mytrue=A_True_Value
myfalse=""

并使用它们：

$ show="dont-show"
$ [[ ${SHOW} == "show" || $mytrue ]] && echo "ok" || echo "wrong"
ok

$ [[ ${SHOW} == "show" || $myfalse ]] && echo "ok" || echo "wrong"
wrong

或者，实际测试[[…]]成语之外的退出代码：

$ mytrue(){ return 0; }
$ if mytrue; then echo "ok"; else echo "wrong"; fi
ok

$ mytrue && echo "ok" || echo "wrong"
ok

一个（更复杂的）替代方法是让函数发出一个值，并从[[…]]成语中调用函数中代码的执行：

$ mytrue(){ echo "A_True_Value"; }
$ myfalse(){ echo ""; }
$ show="dont-show"
$ [[ ${SHOW} == "show" || $(mytrue) ]] && echo "ok" || echo "wrong"
ok

KISS

但（可能）最简单的解决方案是最佳解决方案：

myfalse=""
mytrue="true"

test-conditional () {
    show="dont-show"
    [[ ${show} == "show" || $mytrue  ]] && echo "ok"  || echo "wrong"
    [[ ${show} == "show" && $mytrue  ]] && echo "err" || echo "ok"
    [[ ${show} == "show" || $myfalse ]] && echo "err" || echo "ok"
    [[ ${show} == "show" && $myfalse ]] && echo "err" || echo "ok"
    show="show"
    [[ ${show} == "show" || $mytrue  ]] && echo "ok"  || echo "err"
    [[ ${show} == "show" && $mytrue  ]] && echo "ok"  || echo "err"
    [[ ${show} == "show" || $myfalse ]] && echo "ok"  || echo "err"
    [[ ${show} == "show" && $myfalse ]] && echo "err" || echo "ok"
}

test-conditional

Bash：如何评估双括号内的命令结果

2 个答案:

KISS