업데이트 문을 사용하여 여러 테이블을 연결하는 방법은 무엇입니까?

--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

--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
)

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 = &#39;&#39;.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