如何删除X个字符后的所有单词 [英] How To Delete All Words After X Characters

问题描述

我阅读了这篇文章: sed删除行中剩余的字符除了前5个，它可以帮助我删除x之后的所有字符.但是，我很难找到如何删除x个字符后的所有单词.

I read this post: sed delete remaining characters in line except first 5 which helps me to delete all characters after x. However, I'm having a hard time finding how to delete all words after x characters.

我从以下代码开始:

echo "StackOverflow Users Are Brilliant And Hard Working" | sed 's/.//30g'
#character 30 ---------------------^

我的尝试:

echo "StackOverflow Users Are Brilliant And Hard Working" | sed 's/ .* //30g'
#character 30 ---------------------^

在这些输出中，我截断了最后一个单词或对单词进行计数.相反，我需要删除30个字符之后的单词.我在各种行/单词长度上运行此操作，因此这就是为什么我不能仅将其设置为单词的末尾.

In these outputs, I either cut off the last word or counting words. Instead, I need to remove the words after 30 characters. I am running this on various lines/word lengths so that's why I can't just set it to the end of the word.

所需的输出:

StackOverflow Users Are Brilliant

如果您知道如何将x个字符后的单词数加在一起，将非常感谢您的帮助.

If you know how to put together counting the words after x characters, your help would be greatly appreciated.

请注意:如前所述，请勿将代码更改为33或34个字符.问题的关键是要删除30个字符后的所有单词.

Please note: As stated earlier, do not change the code to 33 or 34 characters. The point of the question is to remove all WORDS after 30 characters.

推荐答案

此 awk 可以完成

$ awk 'BEGIN{FS=OFS="" }  length>30{i=30; while($i~/\w/) i++; NF=i-1; }1' file
StackOverflow Users Are Brilliant
This line has 22 chars

设置FS=OFS=""，以便将每个字符视为一个字段

Setting FS=OFS="" so that each char is considered as a field

如果length>30，则i=30; while($i~/\w/) i++;，即继续递增i，直到我们降到非数字字符为止；循环结束后，设置所需的NF.

If length>30 then i=30; while($i~/\w/) i++; i.e keep incrementing i until we land at a non-alnum character; Once loop ends set the desired NF.

带有length<=30的行将原样打印.

使用 grep

$ grep -oE "^.{1,29}\w*" file
StackOverflow Users Are Brilliant
This line has 22 chars

^.{1,29}\w*:1到29，因为如果30th char是非数字的，则不应考虑它.

^.{1,29}\w* : 1 to 29 because if 30th char is non-alnum then it shouldn't be considered.

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