这个错误是什么?需要超过1个值来解压缩

时间:2015-11-25 13:30:24

标签: django python-2.7 django-views

我收到错误need more than 1 value to unpack我正在使用defaultDict将我的列表设为字典。

def getTabsCols(request):
    proj_name = request.GET.get('projname')
    cursor = connection.cursor()
    proj_details = TProjects.objects.get(
        attr_project_name=proj_name, attr_project_type='Structure', attr_is_active=1)
    pid = proj_details.project_id
    query = 'call SP_Get_TABCOL_NAMES('+str(pid)+')'
    cursor.execute(query)

    proj_details = TProjects.objects.get(
        attr_project_name=proj_name, attr_project_type='Structure', attr_is_active=1)
    pid = proj_details.project_id
    query = 'call SP_Get_TABCOL_NAMES('+str(pid)+')'
    cursor.execute(query)

    result = cursor.fetchall()
    tab_list = []
    tab_col_list = []
    prd = {}
    res = []

    for row in result:
        tabs = collections.OrderedDict()
        schema_name = row[1]
        table_name = row[3]
        tab = (schema_name + '.' + table_name).encode('utf8')
        tabs[tab] = row[5].encode('utf8')
        tab_list.append(tabs)
    d = collections.defaultdict(set)

    for k, v in tab_list:
        d1[k].append(v)

    d = dict((k, tuple(v)) for k, v in d1.iteritems())

    return HttpResponse(d, content_type="text/html")

0 个答案:

没有答案