• sql笔记


    1. 10分钟之内在超过两个医院就诊

    --明细
    create table lu_tmp as
    select t.*, row_number() over(partition by aac147,ymd order by opdate) row_num
    from shen_zx_mdps t where ym = '2018-12'
    
    --查一天内重复的
    create table lu_tmp1 as
    select aac147,opdate,hicode from lu_tmp 
    where (aac147,ymd) in (select aac147,ymd from lu_tmp group by aac147,ymd having count(distinct hicode)>2)
    group by aac147,opdate,hicode
    
    --查明细
    select a.* from lu_tmp a, (
      select a.aac147,a.opdate,a.hicode from lu_tmp1 a, 
      (
             select * 
             from lu_tmp1 t  
             where (
               select count(distinct hicode) 
               from lu_tmp1 
               where aac147=t.aac147 
               and opdate>=t.opdate 
               and opdate<t.opdate+1/144
             )>2
      ) b where a.aac147=b.aac147 and a.opdate>=b.opdate and a.opdate<b.opdate+1/144
    ) b
    where a.aac147=b.aac147 and a.opdate=b.opdate and a.hicode=b.hicode
    order by a.aac147,a.opdate,a.hicode
    
    --测试
    select aac147,opdate,hicode,hiname from lu_tmp 
    where aac147='xx' 
    and opdate >= to_date('2018/12/7 10:16:03','yyyy-mm-dd hh24:mi:ss')
    and opdate < to_date('2018/12/7 10:16:03','yyyy-mm-dd hh24:mi:ss')+1/144
    

      2. 查询连续登陆7天以上的用户

    1. 排序
    2. 日期减序号
    3. 人,日期分组统计,大于7
    4. 筛选明细
    

      3.找出同一组内值都相等,都不等,占比超过80%的组及对应的值

    --判断同一组内所有值是否都相等,输出都相等且元素个数大于3的组
    select id
    from tmp 
    group by id
    where count(distinct value)=1 and count(1)>3
    
    select id
    from (
    select id, value, count(1) counts
    from tmp t
    group by id, value
    ) t
    group by id
    having count(1)=1 and sum(counts)>3
    
    
    --找出同一组内所有值都不相等且元素个数大于3的组
    select id
    from tmp t
    group by id
    where count(1)=count(distinct value)
    
    
    --找出同一组内有超过80%数据相同的组及对应的值
    select id
    from (
    select id, value, count(1) counts
    from tmp t
    group by id, value
    ) t
    group by id
    having max(counts)/sum(counts)>0.8
    
    
    select a.id, a.values, a.counts/b.counts as rate
    from (
    select id, value, count(1) counts
    from tmp t
    group by id, value
    ) a, (
    select id, count(1) counts
    from tmp t
    group by id
    ) b
    where a.id=b.id
    and a.counts/b.counts>0.8
    

      4. 字符串拆分及入库

    --函数使用示例
    SELECT
        REGEXP_SUBSTR('17,20,23', '[^,]+', 1, LEVEL, 'i') AS STR
    FROM
        DUAL  
    CONNECT BY
        LEVEL <= 3  
    	-- LENGTH(REGEXP_REPLACE('17,20,23', '[^,]+')) + 1
    	-- LENGTH('17,20,23') - LENGTH(REGEXP_REPLACE('17,20,23', ',', ''))+1
    
    --建表
    --drop table SalesList;
    create table SalesList(
        keHu                varchar2(20),   --客户
        shangPin            varchar2(20),   --商品名称
        salesNum            number(8)       --销售数量
    );
      
    --插入数据
    declare
      --谈几个客户
      cursor lr_kh is
      select regexp_substr('张三、李四、王五、赵六','[^、]+',1, level) keHu from dual
       connect by level <= 4;
      --进点货
      cursor lr_sp is
      select regexp_substr('上衣、裤子、袜子、帽子','[^、]+',1, level) shangPin from dual
       connect by level <= 4;
    begin
      --循环插入
      for v_kh in lr_kh loop
         for v_sp in lr_sp loop
            insert into SalesList
            select v_kh.keHu, v_sp.shangPin, floor(dbms_random.value(10,50)) from dual;
         end loop;
      end loop;
      commit;
    end;
    

      

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  • 原文地址:https://www.cnblogs.com/iupoint/p/12673760.html
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