使用grep regex匹配任意位数 [英] Matching arbitrary number of digits using grep regex

问题描述

我有一个文件，其中包含类似于以下内容的行

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

data
datalater
983290842
Data387428later
datafhj893724897290384later
4329804928later

我想要做的是使用正则表达式来匹配以数据开头，以以后结尾且中间有数字的任何行.到目前为止，这是我构想的:

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$ 

但是，输出包括所有数据行.我想我可以通过管道输出grep -v datalater，但是我觉得一个表达式应该可以解决问题.

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.

推荐答案

使用+代替*.

+至少匹配上述一个或多个.
*匹配零个或多个.

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

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

在grep中，您需要转义+，我们可以使用\d这是一个字符类并匹配一位数字.

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$

在示例文件中，您还会有一行:

In you example file you also have a line:

datafhj893724897290384later

由于数据和数字之间存在字母，因此当前无法匹配.我们可以通过添加[^0-9]*来解决此问题，以匹配数据后直到数字的所有内容.

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.

我们的最终命令将是:

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

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