Created
July 6, 2012 19:37
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Building a XML Tree as Easy as ABC in Python
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''' | |
**************************************************************************************** | |
Building a XML Tree as Easy as ABC in Python | |
This implementation requires lxml | |
Sample: | |
print etree.tostring(xmlbuilder("main", unique=True, | |
produtos=dict(produto=dict(codigo="UFC153", | |
descricao="Cueca Velha", | |
quantidade="2", | |
preco="10.00", | |
peso="2.00", | |
frete="0.00", | |
desconto="0.00"))), pretty_print=True) | |
Created by Igor Hercowitz | |
2012-07-04 | |
****************************************************************************************** | |
''' | |
from lxml import etree | |
def xmlbuilder(parent, unique=False, use_attributes=False, **kwargs): | |
if isinstance(parent, etree._Element): | |
node_parent = parent | |
else: | |
node_parent = etree.Element(parent) | |
for k, v in kwargs.items(): | |
node = None | |
if unique and node_parent.find(k) is not None: | |
node = node_parent.find(k) | |
if isinstance(v, dict): | |
if node is None: | |
node = etree.SubElement(node_parent, k) | |
xmlbuilder(node, use_attributes=use_attributes, **v) | |
else: | |
if use_attributes: | |
node = node_parent.set(k, v) | |
else: | |
node = etree.SubElement(node_parent, k) | |
node.text = v | |
return node_parent | |
if __name__ == "__main__": | |
root = etree.Element("main") | |
xmlbuilder(root, unique=True, use_attributes=True, | |
produtos=dict(produto=dict(codigo="UFC153", | |
descricao="Cueca Velha", | |
quantidade="2", | |
preco="10.00", | |
peso="2.00", | |
frete="0.00", | |
desconto="0.00"))) | |
xmlbuilder(root, unique=True, use_attributes=True, | |
produtos=dict(produto=dict(codigo="SONIC", | |
descricao="Porco espinho camarada", | |
quantidade="20", | |
preco="4.00", | |
peso="2.00", | |
frete="0.00", | |
desconto="0.00"))) | |
print etree.tostring(root, pretty_print=True) | |
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