32

I have the following:

<text top="52" left="20" width="383" height="15" font="0"><b>test</b></text>

and I have the following:

fileText = re.sub("<b>(.*?)</b>", "\1", fileText, flags=re.DOTALL)

In which fileText is the string I posted above. When I print out fileText after I run the regex replacement I get back

<text top="52" left="20" width="383" height="15" font="0"></text>

instead of the expected

<text top="52" left="20" width="383" height="15" font="0">test</text>

Now I am fairly proficient at regex and I know that it should work, in fact I know that it matches properly because I can see it in the groups when I do a search and print out the groups but I am new to python and am confused as to why its not working with back references properly

csteifel
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1 Answers1

83

You need to use a raw-string here so that the backslash isn't processed as an escape character:

>>> import re
>>> fileText = '<text top="52" left="20" width="383" height="15" font="0"><b>test</b></text>'
>>> fileText = re.sub("<b>(.*?)</b>", r"\1", fileText, flags=re.DOTALL)
>>> fileText
'<text top="52" left="20" width="383" height="15" font="0">test</text>'
>>>

Notice how "\1" was changed to r"\1". Though it is a very small change (one character), it has a big effect. See below:

>>> "\1"
'\x01'
>>> r"\1"
'\\1'
>>>
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    Odd, in neither Python 2 nor 3 it returns the full string for me. I only get the get the replaced text, not the whole string. –  Dec 24 '13 at 18:31
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    Tested it in the interpreter too, there it works. It doesn't work in an executed file tho. Very odd :P –  Dec 24 '13 at 18:37
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    Alternatively, a double backslash "\\1" will do. – Klaws Jan 26 '22 at 16:59