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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