regex - 使用正则表达式检查字符串是否以数字开头

Question

如果一行以 03:32:33 （时间戳）之类的数字开头，我在匹配时正在编写 filebeat 配置。我目前正在这样做——

\d

但它没有得到认可，还有什么我应该做的。我不是特别好/有正则表达式的经验。帮助将不胜感激。

Answer 1

真正的问题是filebeat 不支持\d.

替换\d为[0-9]，您的正则表达式将起作用。

我建议你看看 filebeat 的Supported Patterns。

另外，请确保您使用过^，它代表字符串的开头。

Answer 2

您可以使用此正则表达式：

^([0-9]{2}:?){3}

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演示

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Assert position at the beginning of the string «^»
Match the regex below and capture its match into backreference number 1 «([0-9]{2}:?){3}»
   Exactly 3 times «{3}»
      You repeated the capturing group itself.  The group will capture only the last iteration.  Put a capturing group around the repeated group to capture all iterations. «{3}»
      Or, if you don’t want to capture anything, replace the capturing group with a non-capturing group to make your regex more efficient.
   Match a single character in the range between “0” and “9” «[0-9]{2}»
      Exactly 2 times «{2}»
   Match the character “:” literally «:?»
      Between zero and one times, as many times as possible, giving back as needed (greedy) «?»

Answer 3

Regex: (^\d)

1st Capturing group (^\d)
    ^ Match at the start of the string
    \d match a digit [0-9] 

Answer 4

您可以使用：

^\d{2}:\d{2}:\d{2}

字符 ^ 匹配行首。