60 lines
2.2 KiB
Python
60 lines
2.2 KiB
Python
# inspired by https://gist.github.com/sente/1083506
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# The MIT License (MIT)
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# Copyright (c) 2016 Stuart Powers
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# Permission is hereby granted, free of charge, to any person obtaining a copy
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# of this software and associated documentation files (the "Software"), to deal
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# in the Software without restriction, including without limitation the rights
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# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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# copies of the Software, and to permit persons to whom the Software is
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# furnished to do so, subject to the following conditions:
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# The above copyright notice and this permission notice shall be included in all
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# copies or substantial portions of the Software.
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# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
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# SOFTWARE.
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import re
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def pretty_print(xmlstr, indent=' ', line_break='\n'):
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# python2 doesn't have nonlocal
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current = [0]
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def indent_line(line):
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addition = 0
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if re.match(r'.+</\w[^>]*>$', line):
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# single line text element, don't change indentation
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addition = 0
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elif re.match(r'^</\w', line) and current[0] > 0:
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# end of element and have padding, decrement indentation by one
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current[0] -= 1
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elif re.match(r'^<\w[^>]*[^/]>.*$', line):
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# start of element, increment indentation by one
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addition = 1
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else:
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# single line element, don't change indentation
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addition = 0
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# update and store current indentation in outer function
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current[0] += addition
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# pad the line and return
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return (indent * (current[0] - addition)) + line
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# split the document into lines, indent each line, then rejoin lines
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return line_break.join(
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map(
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indent_line,
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re.sub('(>)(<)(/*)', r'\1\n\2\3', xmlstr).split('\n')
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)
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)
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