当找到与sed匹配时替换整行

时间:2021-07-24 16:51:21

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'.

如果匹配模式,我需要用sed替换整行。例如,如果该行是'一二二四四'并且如果'六'那么,那么整行应该用'fault'替换。

3 个解决方案

#1


47  

You can do it with either of these:

您可以使用以下任一方法执行此操作:

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

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

它获得包含六个的完整行并用故障替换它。

Example:

例:

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

It is based on this solution to Replace whole line containing a string using Sed

它基于此解决方案使用Sed替换包含字符串的整行

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:

更一般地,您可以使用表达式sed'/match/s/.*/replacement/'文件。这将在包含匹配的行中执行sed的/ match / replacement /'表达式。在你的情况下,这将是:

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

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

如果我们有'一二*十一三四'并且我们想把'八'和'十一'作为我们的“坏”字怎么办?

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

在这种情况下,我们可以将-e用于多个条件:

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

and so on.

等等。

Or also:

或者:

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

#2


3  

This might work for you (GNU sed):

这可能适合你(GNU sed):

sed -e '/six/{c\fault' -e ';d}' file

or:

要么:

sed '/six/{c\fault'$'\n'';d}' file

#3


3  

Above answers worked fine for me, just mentioning an alternate way

上面的答案对我来说很好,只是提到另一种方式

Match single pattern and replace with a new one:

匹配单个模式并替换为新模式:

sed -i '/six/c fault' file

Match multiple pattern and replace with a new one(concatenating commands):

匹配多个模式并替换为新模式(连接命令):

sed -i -e '/one/c fault' -e '/six/c fault' file

#1


47  

You can do it with either of these:

您可以使用以下任一方法执行此操作:

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

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

它获得包含六个的完整行并用故障替换它。

Example:

例:

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

It is based on this solution to Replace whole line containing a string using Sed

它基于此解决方案使用Sed替换包含字符串的整行

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:

更一般地,您可以使用表达式sed'/match/s/.*/replacement/'文件。这将在包含匹配的行中执行sed的/ match / replacement /'表达式。在你的情况下,这将是:

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

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

如果我们有'一二*十一三四'并且我们想把'八'和'十一'作为我们的“坏”字怎么办?

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

在这种情况下,我们可以将-e用于多个条件:

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

and so on.

等等。

Or also:

或者:

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

#2


3  

This might work for you (GNU sed):

这可能适合你(GNU sed):

sed -e '/six/{c\fault' -e ';d}' file

or:

要么:

sed '/six/{c\fault'$'\n'';d}' file

#3


3  

Above answers worked fine for me, just mentioning an alternate way

上面的答案对我来说很好,只是提到另一种方式

Match single pattern and replace with a new one:

匹配单个模式并替换为新模式:

sed -i '/six/c fault' file

Match multiple pattern and replace with a new one(concatenating commands):

匹配多个模式并替换为新模式(连接命令):

sed -i -e '/one/c fault' -e '/six/c fault' file