2016-06-27 27 views
0

來論證,而必須是一個列表或元組列表中右鍵,有啥錯在這裏 我想發送到多個地址中to_email「到」的說法配置send_mail

from django.core.mail import send_mail 
from django.conf import settings 


subject = 'Where is the fault' 
from_email = settings.EMAIL_HOST_USER 
to_email = [from_email , '[email protected]'] 
contact_message = "%s: %s via %s" % (
    form_full_name , 
    form_message , 
    form_email) 

send_mail(
    'subject', 
    'contact_message.', 
    'from_email', 
    'to_email', 
    fail_silently=False, 
) 

這說明什麼手段。我已經在我的settings.py文件中完成了設置,如下所示。

EMAIL_USE_TLS = True 
EMAIL_HOST = 'smtp.gmail.com' 
EMAIL_HOST_USER = '[email protected]' 
EMAIL_HOST_PASSWORD = 'password' 
EMAIL_PORT = 587 
+3

您正在向''to_email'',一個字符串,而不是'to_email'發送一個引用列表的標識符。你用'send_mail'的所有參數犯這個錯誤;刪除引號並重試。 – jonrsharpe

+0

謝謝,我明白了 – patrick

回答

1

在你send_mail部分,你需要參考列表,而不是你指的是字符串,沒有定義...

改成這樣,它應該工作:

from django.core.mail import send_mail 
from django.conf import settings 


subject = 'Where is the fault' 
from_email = settings.EMAIL_HOST_USER 
to_email = [from_email , '[email protected]'] 
contact_message = "%s: %s via %s" % (
    form_full_name , 
    form_message , 
    form_email) 

send_mail(
    subject, 
    contact_message., 
    from_email, 
    to_email, 
    fail_silently=False, 
)