当我试图更新django中的表单时，会出现这种类型的问题

Question

我将此类型值存储在DB中：

输入：

{
    "PatientProfile__is_recruiter": "1", 
    "PatientProfile__partner": "FMCS", 
    "PatientProfile__health_insurance_provider": "MILITARY/VA", 
    "PatientProfile__has_medical_home": "0", 
    "PatientProfile__medical_history_heart_disease": "0", 
    "PatientProfile__medical_history_hypertension": "0", 
    "data_model_name": [
        "PatientProfile"
    ]
}

当我尝试更新和更新之后，我发现相同的结果如下：

{
    "PatientProfile__is_recruiter": "1", 
    "PatientProfile__partner": "FMCS", 
    "PatientProfile__health_insurance_provider": "MILITARY/VA", 
    "PatientProfile__has_medical_home": "0", 
    "PatientProfile__medical_history_heart_disease": "0", 
    "PatientProfile__medical_history_hypertension": "0", 
    "data_model_name": [
        "PatientProfile"
    ]
}

如果我没有更新此代码并将其提取到db并尝试执行，请执行。我没有任何错误。当我试图在更新后执行此代码时。下面是定义错误：

回溯(最近一次调用)：

  File "/usr/local/lib/python2.6/dist-packages/django/core/handlers/base.py", line 111, in get_response
    response = callback(request, *callback_args, **callback_kwargs)

  File "/usr/local/lib/python2.6/dist-packages/django/contrib/auth/decorators.py", line 23, in _wrapped_view
    return view_func(request, *args, **kwargs)

  File "/home/ubuntu/django-apps/project_name/../project_name/apps/accounts/decorators.py", line 44, in inner_decorator
    return func(request, *args, **kwargs)

  File "/home/ubuntu/django-apps/project_name/../project_name/apps/reports/views.py", line 97, in hiv_report_new
    return form.get_itable(pk)

  File "/home/ubuntu/django-apps/project_name/../project_name/apps/reports/forms.py", line 454, in get_itable
    custom_data =  ast.literal_eval(report_qs[0]['query'])

  File "/usr/lib/python2.6/ast.py", line 49, in literal_eval
    node_or_string = parse(node_or_string, mode='eval')

  File "/usr/lib/python2.6/ast.py", line 37, in parse
    return compile(expr, filename, mode, PyCF_ONLY_AST)

  File "<unknown>", line 1

    {


^

SyntaxError: invalid syntax

Answer 1

请使用json将dict或list存储在db中

例如：

存贮

obj = json.dumps("{
    'PatientProfile__is_recruiter': '1', 
    'PatientProfile__partner': 'FMCS', 
    'PatientProfile__health_insurance_provider': 'MILITARY/VA', 
    'PatientProfile__has_medical_home': '0', 
    'PatientProfile__medical_history_heart_disease': '0', 
    'PatientProfile__medical_history_hypertension': '0', 
    'data_model_name': [
        'PatientProfile'
    ]
}")

并存储json obj i.

在检索使用时

json.loads

因此，您将得到原来保存在db..。

:)

Answer 2

在存储时使用json.dumps：

obj = json.dumps("{
'PatientProfile__is_recruiter': '1', 
'PatientProfile__partner': 'FMCS', 
'PatientProfile__health_insurance_provider': 'MILITARY/VA', 
'PatientProfile__has_medical_home': '0', 
'PatientProfile__medical_history_heart_disease': '0', 
'PatientProfile__medical_history_hypertension': '0', 
'data_model_name': ['PatientProfile']
}")

检索时，您有两个选项，即；

示例：

>>>simplejson.loads('{"x":"y"}') 
{'x': 'y'}


>>> json.loads('{"x":"y"}') 
{u'x': u'y'} 

也就是说，如果字符串是ASCII (否则返回unicode对象)，simplejson返回字节字符串，而json总是返回unicode对象。