将xml字符串转换为html，我收到错误消息 [英] Translate xml string to html, I received error messages

问题描述

这是错误消息

 
 code>'NoneType'对象没有属性'text'
  

这是我的代码

  self.task_xml =<？xml version =1.0encoding =utf-8？> 
 < django-objects version =1.0> 
< object model =task.taskpk =31> 
< field name =nametype =CharField >新任务< / field> 
< field name =parent_task_idtype =IntegerField> 0< / field> 
< / object> 
& task.taskpk =32> 
< field name =nametype =CharField>新任务< / field> 
< field name =parent_task_id =IntegerField> 0< / field> 
< / object> 
< object model =task.taskpk =33> 
< field name = nametype =CharField> New Task< / field> 
< field name =parent_task_idtype =IntegerField> 31< / field> 
< / object> 
< ob ject model =task.taskpk =34> 
< field name =nametype =CharField>新任务< / field> 
< field name =parent_task_idtype =IntegerField> 31< / field> 
< / object> 
< / django-objects>
 
 self.xmlData = ET.fromstring（self.task_xml）
 
 self.xmlList = [] 
对于self.xmlData.iter（object）中的obj：
 self.xmlList.append（{'id'：obj.find（object [@ name ='pk']）。text，
'name'：obj.find（field [@ name ='name']）。text，
'parent_task_id'：obj.find（field [@ name ='parent_task_id']）。文本，
} 
  

我只是想从：

 < object model =task.taskpk =33> 
  

我确定这是不正确的。

  id'：obj.find（object [@ name ='pk']）。text 
  

你可以帮助我知道正确的语法。

解决方案

假设这是 lxml.etree 或 xml.etree.ElementTree ，你做错了，你是尝试找到带有对象的对象节点的节点，并且具有名称属性为 pk ，这是不正确的。

您要执行以下操作 -

  self.xmlList = [] 
对于self.xmlData.iter（object）中的obj：
 self.xmlList.append（{'id'：obj.get（'pk'），
'name'：obj.find（field [@ name ='name']）。text，
'parent_task_id'：obj.find（field [@ name ='parent_task_id']）。text，
} 
  

obj.get（'pk'）获取 pk 属性的值从对象节点。

Hi I received error messages with this.

this is the error message

'NoneType' object has no attribute 'text'

this is my code

self.task_xml = "<?xml version="1.0" encoding="utf-8"?>
    <django-objects version="1.0">
<object model="task.task" pk="31">
<field name="name" type="CharField">New Task</field>
<field name="parent_task_id" type="IntegerField">0</field>
</object>
<object model="task.task" pk="32">
<field name="name" type="CharField">New Task</field>
<field name="parent_task_id" type="IntegerField">0</field>
</object>
<object model="task.task" pk="33">
<field name="name" type="CharField">New Task</field>
<field name="parent_task_id" type="IntegerField">31</field>
</object>
<object model="task.task" pk="34">
<field name="name" type="CharField">New Task</field>
<field name="parent_task_id" type="IntegerField">31</field>
</object>
</django-objects>"

    self.xmlData = ET.fromstring(self.task_xml)

    self.xmlList = []
    for obj in self.xmlData.iter("object"):
        self.xmlList.append({'id': obj.find("object[@name='pk']").text,
                    'name': obj.find("field[@name='name']").text,
                    'parent_task_id': obj.find("field[@name='parent_task_id']").text,
                    }

I just wanted to copy the id from:

<object model="task.task" pk="33">

and I am sure this isn't correct.

'id': obj.find("object[@name='pk']").text

can you please help me know the correct syntax.

解决方案

Assuming this is either lxml.etree or xml.etree.ElementTree, you are doing it wrongly, you are trying to find a node with the object node with tag object and that has name attribute as pk , which is not correct.

You want to do the following -

self.xmlList = []
for obj in self.xmlData.iter("object"):
    self.xmlList.append({'id': obj.get('pk'),
                'name': obj.find("field[@name='name']").text,
                'parent_task_id': obj.find("field[@name='parent_task_id']").text,
                }

obj.get('pk') gets the value of the pk attribute from the object node.

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