我有来自grep的输出,我正在尝试清理,看起来像:
<words>Http://www.path.com/words</words>
我试过用…
sed 's/<.*>//'
…删除标记,但这会破坏整个行。我不知道为什么会这样,因为每个“”关闭。
最简单的方法是什么?
谢谢!
最佳答案
请尝试以下sed表达式:
sed 's/<.*>\(.*\)<\/.*>/\1/'
快速分解表达式:
<.*> - Match the first tag
\(.*\) - Match and save the text between the tags
<\/.*> - Match the end tag making sure to escape the / character
\1 - Output the result of the first saved match
- (the text that is matched between \( and \))
有关反向引用的详细信息
评论中提出了一个问题,为了完整起见,可能应该加以解决。
\(
和\)
是sed的后参考标记。它们保存一部分匹配的表达式以供以后使用。例如,如果我们有一个输入字符串:
这里面有(帕伦斯)。另外,我们可以用类似于thisparens的parens
使用反向引用。
我们发展出一个表达式:
sed s/.*(\(.*\)).*\1\\(.*\)\1.*/\1 \2/
这给了我们:
parens like this
那是怎么回事?让我们把这个表达式分解一下。
表达式分解:
sed s/ - This is the opening tag to a sed expression.
.* - Match any character to start (as well as nothing).
( - Match a literal left parenthesis character.
\(.*\) - Match any character and save as a back-reference. In this case it will match anything between the first open and last close parenthesis in the expression.
) - Match a literal right parenthesis character.
.* - Same as above.
\1 - Match the first saved back-reference. In the case of our sample this is filled in with `parens`
\(.*\) - Same as above.
\1 - Same as above.
/ - End of the match expression. Signals transition to the output expression.
\1 \2 - Print our two back-references.
/ - End of output expression.
如我们所见,从括号(
(
和)
)之间获取的后引用被替换回匹配表达式中,以便能够匹配字符串parens
。