使用django生成自定义xml

作为一名高管，我对Python和Django完全陌生。使用django生成自定义xml

我想从PHP转移到Python。我来到一个问题，如何生成一个自定义XML文件与所有条目形式数据库。我需要创造这样的：

<inv> 
    <invID>1</invID> 
    <group>Group</group> 
    <name>Name</name> 
    <description></description> 
</inv> 
<inv> 
    <invID>2</invID> 
    <group>Group</group> 
    <name>Name</name> 
    <description></description> 
</inv>

UPDATE

对于那些想知道这里是XML的节省最终代码有明显该杀更好的方式，但这是我想出的。

def xml(request): 
#Getting all of the items in the Database 
products = Product.objects.all() 
#Putting all of it in to Context to pass to template 
context = { 
    'products': products 
} 
#calling template with all of the information 
content = render_to_string('catalog/xml_template.xml', context) 
#Saving template tp a static folder so it can be accessible without calling view 
with open (os.path.join(BASE_DIR, 'static/test.xml'), 'w') as xmlfile: 
    xmlfile.write(content.encode('utf8')) 
#Not Sure if i actually need to call the return but i did not let me run it without content 
return render(request, 'catalog/xml_template.xml', context)

来源

2016-08-02 Mikhail Nikashov

你尝试过这么远吗？你需要找出如何建立一个字符串或写入文件或其他东西？ – sixtytrees

这是我到目前为止。我从模板创建一个自定义XML，通过视图，现在我试图将该XML保存到文件。 –

def xml(request): 
#Getting all of the items in the Database 
products = Product.objects.all() 
#Putting all of it in to Context to pass to template 
context = { 
    'products': products 
} 
#calling template with all of the information 
content = render_to_string('catalog/xml_template.xml', context) 
#Saving template tp a static folder so it can be accessible without  calling view 
with open (os.path.join(BASE_DIR, 'static/test.xml'), 'w') as xmlfile: 
xmlfile.write(content.encode('utf8')) 
#Not Sure if i actually need to call the return but i did not let me run it without content 
return render(request, 'catalog/xml_template.xml', context)

来源

2016-10-14 20:33:27

使用django生成自定义xml

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