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Bash [[ 模式匹配不起作用

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Bash [[ 模式匹配不起作用

这有关条件构造的 Bash 参考部分说：

When the ‘==’ and ‘!=’ operators are used, the string to the right of the 
operator is considered a pattern and matched according to the rules described 
below in Pattern Matching, as if the extglob shell option were enabled. ...
An additional binary operator, ‘=~’, is available, with the same precedence as 
‘==’ and ‘!=’. When it is used, the string to the right of the operator is 
considered an extended regular expression and matched accordingly (as in regex 
3)).

但后来我尝试这样做：

$ [[ -good == -* ]] ; echo $?
0
$ [[ -g == -* ]] ; echo $?
bash: syntax error in conditional expression
bash: syntax error near `-*'
$ [[ -g == -? ]] ; echo $?
bash: syntax error in conditional expression
bash: syntax error near `-?'

然后我尝试正则表达式匹配运算符=~：

$ [[ -good =~ -.* ]] ; echo $?
0
$ [[ -g =~ -.* ]] ; echo $?
bash: syntax error in conditional expression
bash: syntax error near `-.*'
$ [[ -g =~ -.? ]] ; echo $?
bash: syntax error in conditional expression
bash: syntax error near `-.?'

为什么会出现语法错误？

答案1

==好吧，你必须小心或左边的文本=~不是!=其中之一Bash 条件表达式中识别的运算符。

在您的情况下，-g测试位setgid集。如果您提供了在此上下文中不被识别为运算符的内容，那么它会起作用：

$ [[ -i == -* ]] ; echo $?
0
$ [[ -i == -? ]] ; echo $?
0
$ [[ -i =~ -.* ]] ; echo $?
0
$ [[ -i =~ -.? ]] ; echo $?
0

可靠地提供任何字符串的一种方法是引用它，因此它不会被识别为运算符，而只是被识别为字符串：

$ [[ "-good" == -* ]] ; echo $?
0
$ [[ "-g" == -* ]] ; echo $?
0
$ [[ "-g" == -? ]] ; echo $?
0
$ [[ "-good" =~ -.* ]] ; echo $?
0
$ [[ "-g" =~ -.* ]] ; echo $?
0
$ [[ "-g" =~ -.? ]] ; echo $?
0

但请注意不要引用右侧运算符，因为它会阻止将其识别为模式：

$ [[ "-good" == "-*" ]] ; echo $?
1
$ [[ "-g" == "-*" ]] ; echo $?
1
$ [[ "-g" == "-?" ]] ; echo $?
1
$ [[ "-good" =~ "-.*" ]] ; echo $?
1
$ [[ "-g" =~ "-.*" ]] ; echo $?
1
$ [[ "-g" =~ "-.?" ]] ; echo $?
1

答案2

我们应该将字符串放在引号内以将它们评估为字符串。如果我们尝试评估嵌入连字符（-）的字符串，那么 shell 会将其视为测试选项，而不是字符串。有关运营商的更多信息，您可以查看此这里

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