regex: Match everything up to linebreak but not linebreak



  • hello. This is the line from my Python code, with the regex I must change a little bit:

    words = re.findall(r'\w+', new_filename)

    Basically, this will select the content of <title></title> tag and it will save it as an html.

    For example:

    <title>My name is Peter | Prince Justin (en)</title>

    must be save as:

    my-name-is-peter.html (so, without everything after | )

    My regex \w+ will select also the linebreak | and after it. I need to change this regex, in order to select all words before linebreak.

    I try also, this 2 regex, but are not good: \w+.*\| or \w+.*?[\s\S]\|

    Can anyone help me?



  • @Hellena-Crainicu It looks like you are asking about usage of Python’s regex machinery, and not the regex within Notepad++. Is this correct?



  • I work only with notepad++, just running the code in Python.



  • @Hellena-Crainicu But you’re asking about a regex to feed into a call to re.findall(), correct? Or are you asking how to convert lines of text that look like your <title>..<\title> example that are in a text file loaded in the np++ editor?

    If it’s the latter, I have a solution but I’m confused.



  • @Neil-Schipper I am using \w+ as you can see. But I need to stop selecting on the linebreak |, othewise I will get my-name-is-peter-prince-justin.html instead of my-name-is-peter.html



  • @Hellena-Crainicu I’m not getting the clarity I’m hoping for. Here are two very different things people do on computers:

    1. running a python program that processes an input file, and maybe changes it or produces an output file, etc.

    2. having a file loaded in an editor, and running a search and replace operation on it

    Which of these are you trying to do (that requires regex assistance as you described)?



  • it is just about the regex… maybe @guy038 will can help me. He is the master of regex.



  • For my own amusement, I solved the problem in the editor.

    I broke the problem into:

    1. consume from start line to first ‘>’
    2. capture everything up to and excluding (space followed by literal ‘|’) into group 1
    3. consume everything else up to and including EOL

    The search phrase ^.*?>(.+?)(?= \|).*?$ does this. Then replace with \1.html. Then a separate S&R can convert all spaces to ‘-’.

    But I still don’t know what you’re asking for, because you refuse to tell me!



  • Again, for my own amusement (since I’ve never used re.sub() before, only match & split):

    >>> t1 = re.sub(r"^.*?>(.+?)(?= \|).*?$", r"\1.html", "<title>My name is Peter | Prince Justin (en)</title>")
    >>> t2 = re.sub(r"\s", r"-", t1)
    >>> t2
    'My-name-is-Peter.html'
    >>>
    


  • I must split all html files, not just one. I don’t think I can use the replacement…

        new_filename = title.get_text() 
        new_filename = new_filename.lower()
        words = re.findall(r'\w+', new_filename)
        new_filename = '-'.join(words)
        new_filename = new_filename + '.html'
        print(new_filename)


  • I try now this regex: \w+.*(?= \|)

    words = re.findall(r"\w+.*(?= \|)", new_filename)

    almost works, but I get: my name is peter.html (but without little dash)



  • You guys are OFF-TOPIC.
    This is not an appropriate place to discuss Python’s regular expression engine.
    Please find a more appropriate forum for that and confine discussions here to Notepad++ related topics.
    Just because you write Python code in Notepad++ doesn’t make discussion of that code a Notepad++ topic.



  • I find the regex which I needed: \b\w+\b(?=[\w\s]+\|)

    and in Python should be:

    words = re.findall(r'\b\w+\b(?=[\w\s]+\|)', new_filename)

    thanks @Neil-Schipper You give me a good ideea ;)


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