1. 统计数据文件.暂时文件.日志文件大小 select sum(bytes)/1024/1024/1024 as GB from dba_data_files; select sum(bytes)/1024/1024/1024 as GB from dba_temp_files; select sum(bytes)/1024/1024/1024 as GB from v$log; 2. 统计Oracle数据库文件占用空间 Total Size of the database= A+B+C se
mysql count group by统计条数方法 mysql 分组之后如何统计记录条数? gourp by 之后的 count,把group by查询结果当成一个表再count一次select count(*) as count from(SELECT count(*) FROM 表名 WHERE 条件 GROUP BY id ) a; 实战例子:select count(*) as total from (select count(*) from users group by user_i
Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively. Substrings that occur multiple times are counted the numbe
[抄题]: Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively. Substrings that occur multiple times are counted the
统计某个用户下面的对象个数 包括表,视图,同义词,函数,存储过程,包,触发器,索引,约束,序列. 1. sql语句 SELECT (SELECT COUNT(*) FROM USER_TABLES) AS TABLES, (SELECT COUNT(*) FROM USER_VIEWS)AS VIEWS, (SELECT COUNT(*) FROM USER_SYNONYMS)AS SYNONYMS, (SELECT COUNT(*) FROM USER_OBJECTS WHERE OBJECT
Given a binary tree, count the number of uni-value subtrees. A Uni-value subtree means all nodes of the subtree have the same value. Example : Input: root = [5,1,5,5,5,null,5] 5 / \ 1 5 / \ \ 5 5 5 Output: 4 题意: 给定一棵二叉树,求所有节点都同值的二叉树. 思路: dfs 代码: clas
删除重复数据 准备重复数据 create table mystudent ( stuno number , stuname varchar2(20), stuage number ); insert into mystudent values(1,'zs',23); insert into mystudent values(1,'zs',23); insert into mystudent values(2,'ls',24); insert into mystudent values(2,'ls
SQL语句参考以下: select owner,segment_name,segment_type,tablespace_name,sum(bytes)/1024/1024/1024 from dba_segments where tablespace_name='TBS_LDHB' group by owner,segment_name,segment_type,tablespace_name 以上
用distinct关键字只能过滤查询字段中所有记录相同的(记录集相同),而如果要指定一个字段却没有效果,另外distinct关键字会排序,效率很低 . select distinct name from t1 能消除重复记录,但只能取一个字段,现在要同时取id,name这2个字段的值. select distinct id,name from t1 可以取多个字段,但只能消除这2个字段值全部相同的记录 所以用distinct达不到想要的效果,用group by 可以解决这个问题. 例如要显示
--统计数量 select count(*) from table; --统计某一列的数量(去空) select count(col) from table; --统计某一列的值大于或小于另一个值的数量(去不去空没试过) then 'name' end) as new_name from table;