strip tags python

Question

i want the following functionality.

input : this is test  bold text  normal text
expected output: this is test normal text

Answer 1

Sam's answer should do what's wanted fairly well as far as I can tell, but it may pay to make sure that any left over <> characters are replaced with < and > respectively to prevent misuse/invalid HTML.

This approach has the advantage that it can accept incredibly malformed HTML references/tags. BeautifulSoup also handles malformed tags fairly well but html5lib, sgmllib and htmllib can choke on invalid code, some more than others if I remember correctly.

The following code also validates & HTML references:

import re
from htmlentitydefs import name2codepoint, codepoint2name

S = '1234567890ABCDEF'
DHex = {}
for i in S:
    DHex[i.lower()] = None
    DHex[i.upper()] = None

def IsHex(S):
    if not S: return False
    for i in S: 
        if i not in DHex:
            return False
    return True

def UnEscape(S, LReEscape=None):
    # Converts HTML character references into a unicode string to allow manipulation
    #
    # If LUnEscape is provided, then the positions of the escaped characters will be 
    # added to allow turning the result back into HTML with ReEscape below, validating 
    # the references and escaping all the rest
    # 
    # This is needed to prevent browsers from stripping out e.g.   (spaces) etc
    re = LReEscape != None

    LRtn = []
    L = S.split('&')
    xx = 0
    yy = 0
    for iS in L:
        if xx:
            LSplit = iS.split(';')
            if LSplit[0].lower() in name2codepoint:
                # A character reference, e.g. '&'
                a = unichr(name2codepoint[LSplit[0].lower()])
                LRtn.append(a+';'.join(LSplit[1:]))
                if re: LReEscape.append((yy, a))

            elif LSplit[0] and LSplit[0][0] == '#' and LSplit[0][1:].isdigit():
                # A character number e.g. '4'
                a = unichr(int(LSplit[0][1:]))
                LRtn.append(a+';'.join(LSplit[1:]))
                if re: LReEscape.append((yy, a))

            elif LSplit[0] and LSplit[0][0] == '#' and LSplit[0][1:2].lower() == 'x' and IsHex(LSplit[0][2:]):
                # A hexadecimal encoded character
                a = unichr(int(LSplit[0][2:].lower(), 16)) # Hex -> base 16
                LRtn.append(a+';'.join(LSplit[1:]))
                if re: LReEscape.append((yy, a))

            else: LRtn.append('&%s' % ';'.join(LSplit))
        else: LRtn.append(iS)
        xx += 1
        yy += len(LRtn[-1])
    return ''.join(LRtn)

def ReEscape(LReEscape, S, EscFn):
    # Re-escapes the output of UnEscape to HTML, ensuring e.g.   
    # is turned back again and isn't stripped at a browser level
    L = []
    prev = 0
    for x, c in LReEscape:
        if x != prev:
            L.append(EscFn(S[prev:x]))

        o = ord(c)
        if o in codepoint2name:
            L.append('&%s;' % codepoint2name[o])
        else: L.append('&#%s;' % o)
        prev = x+len(c)
    L.append(EscFn(S[prev:]))
    return ''.join(L)

def escape(value):
    # Escape left over <>& tags
    value = value.replace('&', '&')
    value = value.replace('>', '>')
    value = value.replace('<', '<')
    return value

def strip_tags(value):
    # Strip HTML tags
    value = re.sub(r'<[^>]*?>', '', value)
    print 'No Tags:', value

    # Validate & references
    LReEscape = []
    value = UnEscape(value, LReEscape)
    value = ReEscape(LReEscape, value, EscFn=escape)
    print 'References Validated:', value
    return value

if __name__ == '__main__':
    # Outputs:
    #  No Tags: this is test  bold text  normal text >< &blah & &
    #  References Validated: this is test  bold text  normal text >< &blah & &
    strip_tags('this is test  bold text  normal text >< &blah & &')