#How to use update statement to associate multiple tables?
Update statement multi-table association method:
1) The simplest form
SQL code
- - It is confirmed that all customer_ids less than 1000 in the customers table are 'Beijing'
--1000以内的均是公司走向全国之前的本城市的老客户:) update customers set city_name='北京' where customer_id<1000
2) Two-table (multiple-table) association update -- only the connection in the where clause
SQL Code
--这次提取的数据都是VIP,且包括新增的,所以顺便更新客户类别 update customers a -- 使用别名 set customer_type='01' --01 为vip,00为普通 where exists (select 1 from tmp_cust_city b where b.customer_id=a.customer_id )
3) Two tables (multiple tables) related update -- the modified value is calculated from another table
SQL Code
update customers a -- 使用别名 set city_name=(select b.city_name from tmp_cust_city b where b.customer_id=a.customer_id) where exists (select 1 from tmp_cust_city b where b.customer_id=a.customer_id ) -- update 超过2个值 update customers a -- 使用别名 set (city_name,customer_type)=(select b.city_name,b.customer_type from tmp_cust_city b where b.customer_id=a.customer_id) where exists (select 1 from tmp_cust_city b where b.customer_id=a.customer_id )一、随机密码生成。编写程序在26个字母大小写和9个数字组成的列表中随机生成10个8位密码。 import random def random_password(): list1 = [] #把字母加入序列中 for i in range(65,90): list1.append(chr(i)) for i in range(97,122): list1.append(chr(i)) list2 = [1,2,3,4,5,6,7,8,9] list = list1 +list2 n = 0 while n < 10: password = [] n = n + 1 m = 0 password = password + random.sample(list, 8) #把列表转化为字符串 password_middle = [str(i) for i in password] password_end = ''.join(password_middle) print("第{}个随机生成的密码是:{}".format(n,password_end)) random_password() #random.sample(seq, k)实现从序列或集合seq中随机选取k个独立的的元素 #random.randint(a, b) #A-Z:65-90;a-z:97-122;ASCII码48~57为0到9十个阿拉伯数字
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