SQL命令更新表中的列值

Question

以下是python manage shell代碼

>>> 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終端中編寫SQL命令（ python manage dbshell ）

如何在SQL命令中進行上述轉換？

以下是我的嘗試：

[編輯1] ：

通過SQL命令獲取目標電子郵件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命令中進行上述轉換？

您可以查看Django為此生成的查詢，它可能無法像其中那樣運行（例如Django發送的缺少參數），但是它將使您對Django如何將其轉換為SQL有了一個很好的了解

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

因此，您想替換電子郵件中的域，這是測試選擇：

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

並將更新為：

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

Answer 3

以下是我的解決方案：

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

演示：

進入dbshel​​l 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=#