Question

我有一个需要列表作为参数的类。问题是当我导入此函数并将其用作参数时，我得到TypeError或ValueError。

我尝试搜索此问题，但似乎找不到匹配的问题。

出现问题的代码

from payroll.payroll import get_employee_names

class ManagerForm(forms.Form):
    names = forms.ChoiceField(choices=[get_employee_names()])

导入的功能：

def get_employee_names():
    # place uploaded files into variables
    tips_file = 'media/reports/Tips_By_Employee_Report.xls'

    # get managers name
    df_employee_names = pd.read_excel(
        tips_file, sheet_name=0, header=None, skiprows=7)
    df_employee_names.rename(
        columns={0: 'Employee'}, inplace=True)
    df_employee_names['Employee'] = \
        df_employee_names['Employee'].str.lower()
    # data-frame to list
    employee_names = df_employee_names.loc[:, 'Employee'].tolist()
    return employee_names

我希望它使用函数结尾处返回的列表来创建下拉菜单。

我当前遇到的主要错误是：ValueError at /select-manager-run-payroll/ too many values to unpack (expected 2)

编辑：来发现它期望一个元组作为参数。

Answer 1

ChoiceField choices参数应接收元组列表，而不是列表的1元素列表。

重写get_employee_names以提供：

def get_employee_names():
    # ...
    # expecting that employee_names is something like ['a', 'b', 'c']
    return [(name, name) for name in employee_names]

以及您的表单：

class ManagerForm(forms.Form):
    name = forms.ChoiceField(choices=get_employee_names())

为什么列表不会在此类中作为参数接受？

1 个答案: