'Dict' Object Has No Attribute 'Id'
This Is My Code. I'm Trying to Translate Xml String into Python List to Be Shown in Html Template. Self. Task_Xml = " New Task 0 New Task 0 New Task 31 New...
This is my code. I'm trying to translate xml string into python list to be shown in html template.
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>"
58 self.xmlData = ET.fromstring(self.db.task_xml)
59
60 self.task_list = []
61 taskList = []
62 for obj in self.xmlData.iter("object"):
63 parent_task_id = obj.find("field[@name='parent_task_id']").text
64 if parent_task_id == EMPTY_UUID:
65 taskList.append({'id': obj.get("pk"),
66 'name': obj.find("field[@name='name']").text,
67 'parent_task_id': parent_task_id ,
68 })
69 # Apprend taskList:
70 for task in taskList:
71 taskViewModel = TaskViewModel(task.id, True)
72 self.task_list.append(taskViewModel)
But I'm getting the error:
'dict' object has no attribute 'id'
And it is task.id in line 71
Do you think I have a problem with this in line 65:
'id': obj.get("pk")
2 Answers
You are accessing the dictionary wrongly. You need to use subscript with string 'id' , Example -
taskViewModel = TaskViewModel(task['id'], True)
I got the same error when accessing "id" in dictionary with dot "." like JavaScript as shown below because I also use JavaScript quite often in addition to Python:
user = {"id": 1, "name": "John"}
print(user.id) # Error
So, I accessed "id" with brackets "[]" as shown below then the error was solved:
user = {"id": 1, "name": "John"}
print(user["id"]) # 1
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