grep:匹配所有字符，直到(不包括)第一个空格 [英] grep: match all characters up to (not including) first blank space

问题描述

我有一个具有以下格式的文本文件:

I have a text file that has the following format:

characters(that I want to keep) (space) characters(that I want to remove)

例如:

foo garbagetext
hello moregarbage
keepthis removethis
(etc.)

因此，我试图在Linux中使用grep命令，以使每行中的字符最多保持不超过第一个空格.我尝试了许多尝试，例如:

So I was trying to use the grep command in Linux to keep only the characters in each line up to and not including the first blank space. I have tried numerous attempts such as:

grep '*[[:space:]]' text1.txt > text2.txt
grep '*[^\s]' text1.txt > text2.txt
grep '/^[^[[:space:]]]+/' text1.txt > text2.txt

试图将不同的例子拼凑在一起，但是我没有运气.它们都产生一个空白的text2.txt文件.我是新来的.我在做什么错了?

trying to piece together from different examples, but I have had no luck. They all produce a blank text2.txt file. I am new to this. What am I doing wrong?

*

我要保留的部分包括大写字母.因此，我想保留任何/所有字符，并且不包括每行中的空格(从空格开始删除所有内容).

The parts I want to keep include capital letters. So I want to keep any/all characters up to and not including the blank space (removing everything from the blank space onward) in each line.

**

(我要删除的)垃圾文本可以包含任何内容，包括空格，特殊字符等.因此，例如:

The garbage text (that I want to remove) can contain anything, including spaces, special characters, etc. So for example:

AA rough, cindery lava [n -S]

运行grep -o '[^ ]*' text1.txt > text2.txt后，上面的行变为:

After running grep -o '[^ ]*' text1.txt > text2.txt, the line above becomes:

AA
rough,
cindery
lava
[n
-S]

在text2.txt中. (我只想保留AA)

in text2.txt. (All I want to keep is AA)

解决方案(由Rohit Jain提供，并由beny23进一步输入):

SOLUTION (provided by Rohit Jain with further input by beny23):

 grep -o '^[^ ]*' text1.txt > text2.txt

推荐答案

您将量词*放在错误的位置.

You are putting quantifier * at the wrong place.

请尝试以下方法:-

grep '^[^\s]*' text1.txt > text2.txt

或者甚至更好:-

grep '^\S*' text1.txt > text2.txt  

\S表示匹配非空白字符.锚点^用于在行的开头进行匹配.

\S means match non-whitespace character. And anchor ^ is used to match at the beginning of the line.

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