一. 1.新建表 test id varchar2(20)name varchar2(20)addr varchar2(50)score number create table test(id varchar2(20),name varchar2(20),addr varchar2(50),score number) 2.给id增加主键约束alter table test add constraint pk_id primary key(id)3.查询出分数大于60分的人员,并按分数降序进行排列

题目分析: 1.首先想到的是数组存放数据,数组肯定是String类型. 2.String类型的数组,5行6列.要把从第0行第2列到第4行第4列的数据取出转换成数值型,再统计三科总分.最后把计算出的总分转换成String类型放回数组的第5列. 3.将string类型的数组写入到文件,再从文件读出显示到控制台 遇到的困难: 1.反复的类型转换 import java.io.File; import java.io.FileOutputStream; import java.io.FileReader

3.编写BinIoDemo.java的Java应用程序,程序完成的功能是:完成1.doc文件的复制,复制以后的文件的名称为自己的学号姓名.doc. try { FileInputStream in = new FileInputStream("e:/TextRw.txt"); FileOutputStream out = new FileOutputStream("e:/2012412440张伟.txt"); int i =0; byte[] b1 = new by

1.Oracle查询数据库中所有表的记录数,但是有可能不准建议用第二种方式进行查询 select t.table_name,t.num_rows from user_tables t 2.创建oracle函数,通过函数中查询词表记录数显示当前记录数 create or replace function count_rows(table_name in varchar2, owner in varchar2 default null) return number authid current_us

--查询并打印员工的姓名和薪水 --set serveroutput on /* 1.光标的属性 %found:假设取到了记录就是true否则是false: %notfound: */ declare --定义一个光标 cursor cemp is select ename,sal from emp; --为光标定义相应的变量 pename emp.ename%type; psal emp.sal%type; begin --打开光标 open cemp; loop --取一条记录 into后面

上一篇笔记将开始定义的存储结构处理了一下,将FormItems数组中的表单项都拿到mongodb document的最外层,和以前的关系型数据类似,之不过好多列都是动态的,不固定,不过这并没有什么影响.结果就是方便我们更好的查询和统计:还有一点就是转换之后从服务器端返回客户端的对象也是如此,这样更加方便了获取每个表单项的值(例如渲染列表). 我们的好多应用场景都是分页加载,并且都需要显示总的条数.以前是弄了两个API:一个是获取查询结果:一个是获取条数.为了这样一个功能要多发送一个API觉得有点

3267. D-query Problem code: DQUERY English Vietnamese Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in

=========查询分区的大小========= select d.NAME,t.TBL_NAME,p.PART_NAME,prm.PARAM_KEY,prm.PARAM_VALUE from TBLS t left join DBS d on t.DB_ID = d.DB_ID left join PARTITIONS p on t.TBL_ID = p.TBL_ID left join PARTITION_PARAMS prm on p.PART_ID=prm.PART_ID where

1.子查询,查询出的数据随便起一个别名,然后根据分组和条件查询出的数据,作为一个具有一列的一个表,然后外面的查询查询这个数据表的这一列的总数,即可. SELECT COUNT( * ) FROM ( SELECT COUNT( * ) FROM data_report_statistics WHERE CREATE_TIME > '2018-11-06 01:00:00' GROUP BY PROVINCE ) A;

数据库记录如下: 现在要查询统计出每个'name'的'result'分别为'success'和'fail'的次数: 利用IF条件判断满足条件为1,不满足为0,再用SUM函数求和,最后通过'name'分组. SELECT NAME, , )) AS success, , )) AS fail FROM if_test GROUP BY `name` 结果如下:

传统做法可能是select count(1) 但是往往会比较慢.推荐如下做法: ) CurrentRowCount FROM sys.sysindexes WHERE id = OBJECT_ID('TableName') 查看代码

select schema_name(t.schema_id) as [ Schema ], t. name as TableName,i. rows as [RowCount] from sys.tables as t, sysindexes as i where t.object_id = i.id and i.indid <=1

package zuoye; import java.io.File; import java.io.FileInputStream; import java.io.FileOutputStream; import java.io.IOException; public class BinIoDemo { public static void main(String[] args) throws IOException { // 判断1.doc文件是否存在,不存在创建文件 File file=n

select count(0) from (select investor_uid from lzh_borrow_investor group by investor_uid) as temp

select a.name,b.rows from sysobjects a,sysindexes b where a.name = b.name order by b.rows desc

package zuoye; import java.io.FileInputStream; import java.io.FileOutputStream; public class BinIoDemo { public static void main(String[] args) { FileInputStream in; try { in = new FileInputStream("d:/zuoye/1.doc"); byte[]b=new byte[1024]; int i

select t.table_name,t.num_rows from user_tables t

方法一: select t1.num1,t2.num2,t3.num3 from (select count(*) num1 from table1) t1, (select count(*) num2 from table2) t2, (select count(*) num3 from table3) t3 方法二: select sum(t.num1),sum(t.num2),sum(t.num3) from ( as num3 from table1 union as num3 from

SELECT * from t_user ORDER BY create_time desc SELECT create_time,  COUNT(create_time) as num from t_user GROUP BY create_time ORDER BY num desc

>use information_schema; >select sum(table_rows) from tables where TABLE_SCHEMA = "test" order by table_rows asc;

SELECT OBJECT_NAME(ii.id) TableName ,rows FROM sysindexes ii INNER JOIN sysobjects oo ON ( oo.id = ii.id AND oo.xtype = 'U ') WHERE ii.indid < 2 ORDER BY rows desc ;