Question

尝试获取正确的regexp语法以匹配标记的行（使用*）：

#32 0 0 A [2] *
#13 3 2 A [1]
#44 1 2 A [2]
#44 0 0 A [1]
#44 0 0 A [2] *
#44 0 0 B [2] *

我想要的是包含“0 0”的行和在“[2]”之后的某处。任何字符都可以出现在这两个规则之前，之间或之后。

这就是我今天所拥有的，它找到了第一部分：

#.. 0 0

我如何完成它？

此外，如果有人想像我这样的正则表达式新秀分享一个好的网址，我会更加感激！

Answer 1

`我想要的是，包含“0 0”的行和“[2]”之后的某个地方。任何角色都可以在之前，之间发生

您可以使用此正则表达式：

^#[0-9]{2} 0 0.*?\[2\]

<强>解释

^ assert position at start of the string
# matches the character # literally
[0-9]{2} match a single character present in the list below
Quantifier: Exactly 2 times
0-9 a single character in the range between 0 and 9
 0 0 matches the characters  0 0 literally
.*? matches any character (except newline)
Quantifier: Between zero and unlimited times, as few times as possible, expanding as needed [lazy]
\[ matches the character [ literally
2 matches the character 2 literally
\] matches the character ] literally

Answer 2

以下内容如何：

echo '#32 0 0 A [2]
#13 3 2 A [1]
#44 1 2 A [2]
#44 0 0 A [1]
#44 0 0 A [2]
#44 0 0 B [2]' | grep -E '.*0 0.*\[2\].*'

Answer 3

您可以使用：

#.. 0 0.*\[2\].*

另一个简单的RegEx比赛

3 个答案: