Python 类型错误：无法连接电子邮件中的“str”和“list”对象
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TypeError: cannot concatenate 'str' and 'list' objects in email
提问by Enzo Vidal
I am working on sending an email in python. Right now, I want to send entries from a list via email but I am encountering an error saying "TypeError: cannot concatenate 'str' and 'list' objects" and I have no idea to debug it. The following is the code that I have. I'm still new in this language (3 weeks) so I have a little backgroud.
我正在用 python 发送电子邮件。现在，我想通过电子邮件发送列表中的条目，但遇到一个错误，提示“TypeError：无法连接 'str' 和 'list' 对象”，我不知道要调试它。以下是我拥有的代码。我还是这门语言的新手（3 周），所以我有一点背景。
import smtplib
x = [2, 3, 4] #list that I want to send
to = '' #Recipient
user_name = '' #Sender username
user_pwrd = '' #Sender Password
smtpserver = smtplib.SMTP("mail.sample.com",port)
smtpserver.ehlo()
smtpserver.starttls()
smtpserver.ehlo()
smtpserver.login(user_name,user_pwrd)
#Header Part of the Email
header = 'To: '+to+'\n'+'From: '+user_name+'\n'+'Subject: \n'
print header
#Msg
msg = header + x #THIS IS THE PART THAT I WANT TO INSERT THE LIST THAT I WANT TO SEND. the type error occurs in this line
#Send Email
smtpserver.sendmail(user_name, to, msg)
print 'done!'
#Close Email Connection
smtpserver.close()
采纳答案by David Sanders
The problem is with msg = header + x. You're trying to apply the +operator to a string and a list.
问题在于msg = header + x. 您正在尝试将+运算符应用于字符串和列表。
I'm not exactly sure how you want xto be displayed but, if you want something like "[1, 2, 3]", you would need:
我不确定您希望如何x显示，但是，如果您想要“[1, 2, 3]”之类的内容，则需要：
msg = header + str(x)
Or you could do,
或者你可以这样做
msg = '{header}{lst}'.format(header=header, lst=x)
回答by Irshad Bhat
Problem is that in the code line msg = header + x, the name headeris a string and xis a list so these two cannot be concatenated using +operator. The solution is to convert xto a string. One way of doing that is to extract elementsfrom the list, convertthem to strand .join()them together. So you should replace the code line:
问题是在代码行中msg = header + x，名称header是一个字符串并且x是一个列表，因此这两个不能使用+运算符连接。解决方案是转换x为字符串。这样做的一种方式是提取元素从list，转换他们str和.join()他们在一起。所以你应该替换代码行：
msg = header + x
by:
经过：
msg = header + "".join([str(i) for i in x])
