解析文本文档的正则表达式

Question

I was trying to parse a text documents with !if and !endif in between. I want to have the text without !if, !endif and text between them.

For example:

text
!if
text1
!endif
text2

I would like to have my output = text+text2+..

I tried something like this re.findall(r'((^(!if.*!endif))+', text). But it doesnt seem to work for me.

Answer 1

Your regex would be:

^!if$.*?^!endif$\s+

This says:

^      - Match the beginning of a line (because of the re.M flag)
!if    - Match !
$      - Match the end of a line (because of the re.M flag)
.*?    - Match any number of characters (non-greedy) (includes line breaks, because of the re.S flag)
^      - Match the beginning of a line (because of the re.M flag)
!endif - Match !endif
$      - Match the end of a line (because of the re.M flag)
\s+    - Match one or more whitespace characters

So, you should be able to use it like this, which replaces all occurrences of the above regex with an empty string (nothing):

import re
s = "text\n!if\ntext1\n!endif\ntext2"
s = re.sub("^!if$.*?^!endif$\s+", "", s, flags=re.S | re.M)
print s

This will output :

text 
text2

Note that this explicitly requires !if and !endif be on separate lines. If this isn't a requirement, you can remove the $ and ^ anchors from the middle of the regex.

^!if.*?!endif$\s+

Answer 2

I can help in sed:

sed '/^if$/,/^endif$/ d'

Here is the algorithm that sed uses:

1. set the variable match=False
2. read next line
3. check if the line is equal 'if'. If so, set the variable match=True
4. if match==True, check whether current-line=='endif'. If so, set match=False and delete the current line [and jumps to 0] .
5. print the current line
6. if not EOF , jumps to 1