方法一:Linq ChannelList就是一个List类型的数据,IsOpen 是其元素的属性 channelCount = (from channel in DevicesManager.Instance.CurrentDevice.ChannelList where channel.IsOpen group channel by channel.ChannelID).Count(); 方法二:泛型委托Predicate<T> public delegate bool Predicate&
sql 查询某个条件下多条数据中最新的一条数据或最老的一条数据 test_user表结构如下: 需求:查询李四.王五.李二创建的最初时间或者最新时间 1:查询最初的创建时间: SELECT * FROM( SELECT * FROM test_user ) AS tu WHERE NOT EXISTS ( SELECT * FROM( SELECT * FROM test_user ) AS tu2 WHERE tu2.user_name=tu.user_name AND DATE(tu.tim
--批量对符合条件的表记录进行更新 --aa代表查询出的符合条件数据的别名 --aa后的表示需要符合的条件 --loop后开始写更新操作 begin for aa in (select a.objectid from geographic_info a, geographic_info_bf b where a.objectid = b.objectid and b.pgeocode = 'xxx') loop update geographic_info set qxcode = 'xxx',
import datetime import sys import oss2 from itertools import islice import pandas as pd import re import json from pandas.tseries.offsets import Day from multiprocessing import Process, JoinableQueue, cpu_count, Manager import time def mkbuck(bk): au
var data = [{}, {}, {}, {Id:1}] var datawilldele = [];//2,4,5 data.forEach(function (v, i,arry) { if ($.isEmptyObject(v)) { datawilldele.push(i); } }); for (var i = 0; i < datawilldele.length; i++) { data.splice(datawilldele[i]-i,1); }
表结构: 需求:找到相同的内容并删除 方法1: 通过分组找出筛选出count大于1的数据 group by可以利用聚合函数(count. sum. avg)进行分组 having 对分组的数据进行下一步筛选 通过inner join找出相同数据 select * from reviews a INNER JOIN () b on a.r_content = b.r_content and a.r_userid = b.r_userid; 加行号 select a.*, , ) as rownu
select * from (select count(name) as isone, name from tbl_org_departments group by name) t where t.isone > 1; 解释: 子句查出下列结果集作为主句的表, 当isone列为1时说明没有重复, 大于1时说明存在重复 执行子句时结果:
delete from TableName where id in( SELECT ID FROM(SELECT * FROM TableName t0WHERE(t0.Field1,t0.Field2) IN ( SELECT t1.Field1, t1.Field2 FROM TableName AS t1 GROUP BY t1.Field1, t1.Field2 HAVING count(*) > 1 )AND t0.ID NOT IN ( SELECT min(t2.ID) FROM