正则表达式 - 匹配任意数量的数字

问题描述

  \ +稍后
  

在您的示例文件中，您也有一行：

  datafhj893724897290384later 
  

目前由于数据和数字之间存在字母，因此不会匹配。我们可以通过添加一个 [^ 0-9] * 来匹配数据之后的任何数字，直到数字。

我们的最终命令是：

  grep'^ [Dd] ata [^ 0-9] * \ d \ +稍后$'文件名
  

I've got a file that has lines in it that look similar as follows

data
datalater
983290842
Data387428later
datafhj893724897290384later
4329804928later

What I am looking to do is use regex to match any line that starts with data and ends with later AND has numbers in between. Here is what I've concocted so far:

^[D,d]ata[0-9]*later$

However the output includes all datalater lines. I suppose I could pipe the output and grep -v datalater, but I feel like a single expression should do the trick.

解决方案

Use + instead of *.

+ matches at least one or more of the preceding.
* matches zero or more.

^[Dd]ata[0-9]+later$

In grep you need to escape the +, and we can use \d which is a character class and matches single digits.

^[Dd]ata\d\+later$

In you example file you also have a line:

datafhj893724897290384later

This currently will not be matched due to there being letters in-between data and the numbers. We can fix this by adding a [^0-9]* to match anything after data until the digits.

Our final command will be:

grep '^[Dd]ata[^0-9]*\d\+later$' filename

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