找到匹配项时用sed替换整行 [英] Replace whole line when match found with sed

问题描述

如果与模式匹配，我需要用sed替换整行. 例如，如果该行是一二六三四"，而如果存在六"，则应将整个行替换为故障".

I need to replace the whole line with sed if it matches a pattern. For example if the line is 'one two six three four' and if 'six' is there, then the whole line should be replaced with 'fault'.

推荐答案

您可以使用以下任一方法进行操作:

You can do it with either of these:

sed 's/.*six.*/fault/' file     # check all lines
sed '/six/s/.*/fault/' file     # matched lines -> then remove

它将获取包含six的完整行，并将其替换为fault.

It gets the full line containing six and replaces it with fault.

示例:

$ cat file
six
asdf
one two six
one isix
boo
$ sed 's/.*six.*/fault/'  file
fault
asdf
fault
fault
boo

它基于此解决方案到通常，您可以使用表达式 sed '/match/s/.*/replacement/' file .这将在包含match的那些行中执行sed 's/match/replacement/'表达式.在您的情况下，这将是:

More generally, you can use an expression sed '/match/s/.*/replacement/' file. This will perform the sed 's/match/replacement/' expression in those lines containing match. In your case this would be:

sed '/six/s/.*/fault/' file

---

如果我们有一二六八十一三三四"并且我们想要 包括八"和十一"作为我们的坏"字?

What if we have 'one two six eight eleven three four' and we want to include 'eight' and 'eleven' as our "bad" words?

在这种情况下，我们可以将-e用于多个条件:

In this case we can use the -e for multiple conditions:

sed -e 's/.*six.*/fault/' -e 's/.*eight.*/fault/' file

以此类推.

或者:

sed '/eight/s/.*/XXXXX/; /eleven/s/.*/XXXX/' file

这篇关于找到匹配项时用sed替换整行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！