Question

以下是python manage shell

上的代码

>>> User.objects.filter(email__icontains="gmail.com").values_list("email", flat=True)
[u'abc@gmail.com', u'vivekbsable@gmail.com', u'vivek@gmail.com', u'xyz@gmail.com', u'vivekbsable@gmail.com']
>>> for ii in User.objects.filter(email__icontains="gmail.com"):
...      ii.email = ii.email.replace("@gmail.com", "@custom.com")
...      ii.save() 
...     
... 
>>> User.objects.filter(email__icontains="gmail.com").values_list("email", flat=True)
[]
>>> User.objects.filter(email__icontains="@custom.com").values_list("email", flat=True)
[u'vivek@custom.com', u'xyz@custom.com', u'abc@custom.com', u'vivekbsable@custom.com', u'vivekbsable@custom.com']
>>>

我想在Postgresql终端（python manage dbshell）

中编写SQL命令

如何在 SQL命令中进行上述转换？

以下是我的尝试：

[Edited1] ：

通过SQL命令获取目标电子邮件ID：

dp=# SELECT  email FROM auth_user where email LIKE '%@gmail.com';
           email           
---------------------------
 vivek@gmail.com
 xyz@gmail.com
 abc@gmail.com
 vivekbsable@gmail.com
 vivekbsable@gmail.com
(5 rows)

dp=#

Answer 1

如何在SQL命令中进行上述转换？

你可以看一下Django为此生成的查询，它可能不会像在运行中那样运行（例如Django发送的缺少参数），但是它会让你很好地了解Django如何翻译它到SQL

我们的想法是打印此值：Model.objects.filter(...).values_list(...).query

query = User.objects.filter(email__icontains="@custom.com").values_list("email", flat=True).query

# Make it print it
print query 
print(query) # Python 3 or with "from future import print_function"

Answer 2

所以你想在电子邮件中替换域，这里是test select：

select email, replace(email, '@gmail.com', '@custom.com') as new_email 
from auth_user 
where email like '%@gmail.com';

更新将是：

update auth_user 
set email = replace(email, '@gmail.com', '@custom.com') 
where email like '%@gmail.com';

Answer 3

以下是我的解决方案：

UPDATE auth_user Set email = replace(email, '@gmail.com', '@custom.com') where email LIKE '%@gmail.com';

<强>演示：

进入dbshell 1. cd / var / op / project_name python manage dbshell

dp=# SELECT  email FROM auth_user where email LIKE '%@gmail.com';
           email           
---------------------------
 vivek@gmail.com
 xyz@gmail.com
 abc@gmail.com
 vivekbsable@gmail.com
 vivekbsable@gmail.com
(5 rows)

dp=# SELECT  email FROM auth_user where email LIKE '%@custom.com';
 email 
-------
(0 rows)

dp=# UPDATE auth_user Set email = replace(email, '@gmail.com', '@custom.com') where email LIKE '%@gmail.com';
UPDATE 5

dp=# SELECT  email FROM auth_user where email LIKE '%@gmail.com';
 email 
-------
(0 rows)

dp=# SELECT  email FROM auth_user where email LIKE '%@custom.com';
           email            
----------------------------
 vivek@custom.com
 xyz@custom.com
 abc@custom.com
 vivekbsable@custom.com
 vivekbsable@custom.com
(5 rows)

dp=#

用于更新表中列值的SQL命令

3 个答案: