我目前有一个匹配结果表+----------+----------+--------+-----------+--------+-----------+-----------+-----------+ | hometeam | awayteam | p1home | p2home... | p1away | p2away... | homescore | awayscore | +----------+----------+--------+-----------+--------+-----------+-----------+-----------+我用相当笨重的东西查询SELECT (SELECT COUNT(hometeam) FROM fixture_data WHERE (p1home = '$playerID' OR p2home ...
我正在尝试优化我的代码,其中我在过去七天内生成统计数据. 目前我正在使用雄辩的和查询每天记录数据的计数,这使我在一个循环内进行7个单独的查询. 例：// loop for number of days for($i = 0; $i < $days; $i++){// some logic to set comparitive unix times$oldest = $newest - $dayDuration;// count number of objects between oldest time and newest time$counts[$i] = Object::where('objecttime','>',$oldest)->where('ob...
我有一张桌子：id | name | ----------- 1 | AAA | 2 | BBB | 3 | AAA | 4 | DDD | 5 | BBB | 6 | CCC | 7 | AAA | 8 | BBB | 9 | DDD |我想算一下AAA,BBB等多少次.通过使用count(*)我得到的孔数不是单个元素.我想要所有元素的数量.像这样：Name | Count ------------ AAA | 3 BBB | 3 CCC | 1 DDD | 2不在新表中我想要它作为输出.解决方法:使用COUNT()聚合函数来获取每个NAME的记录总数.SELECT name, COUN...
表格1"id" "name" "description" "path" "country" "status" "1" "Title 1" "Description 1" "US > Consumer > Home Applicances" "US" "0" "2" "Title 2" "Description 2" "US > Business > Legal Charges" "UK" "0"我试图用同一张桌子上的不同轮数做两次计数. 我在这里遇到了其他问题,但没有人喜欢我这样做的方式. 我目前正在使用两个sqls和更多代码：...
我有2张桌子：1.customer (contains customers) 2.customer_order (contains orders)我想知道这个：no. of customers with 0 orders no. of customers with 1 order no. of customers with 2 orders etc我有这个：SELECT COUNT(co.id) FROM customer c LEFT JOIN customer_order co ON c.id=co.id_customer GROUP BY ??? ;解决方法:这给出了每个客户的订单量SELECT co.id Customer, COUNT(*) Orders FROM customer_order co GROU...
我有一个用户的表,其中有3列,我正在尝试计数,然后按created_at日期分组. 列：’created_at’,’answers_count’,’questions_count’ 如何将这3个查询合并为一个,并将所有计数按相同的created_at日期分组？ 这是3个单独的查询：-- daily new signups SELECT date_trunc('day', users.created_at) AS signup_date, count(*) AS new_users FROM users GROUP BY signup_date ORDER BY signup_date DESC;-- new user answers_count by ...
假设我有一个人员ID(1-8)和人物角色(1-4)表：CREATE TABLE personRole ( PersonId int NOT NULL, RoleId int NOT NULL );INSERT INTO personRole VALUES (1, 1), (1, 2), (2, 1), (2, 3), (3, 3), (4, 3), (1, 4), (5, 2), (6, 1), (7, 1), (7, 4), (8, 1), (8, 2), (8, 4) ;我的目标是选择拥有3个或更多角色的人员ID,角色特别是1,2和4.这是我的第一个解决方案：SELECT PersonId FROM personRole WHERE RoleID in (1,2,4) GROUP BY...
任务：id | title | description | ---------------------------------------------------------------------1 | Task1 | Descr1 | 2 | Task2 | Descr1 | 3 | Task2 | Descr1 | 4 | Task2 | Descr1 | 5 | Task2 | Descr1 | 信息：id | task_id | message | status | ---------------------------------------------------------------------1 | 1 | Message1 | HOLD2 | 1 | ...
因此,我正在努力编写一个查询,无论我应用了什么过滤器,都会返回所有类别,但计数会根据此过滤器中返回的配方数量而变化. 如果我不对它应用任何过滤器,则此查询很有效.计数似乎是正确的,但是一旦我添加这样的东西：其中c.parent_id不为null并且r.time_cook_minutes> 60我正在过滤掉大多数类别而不是仅仅计数为零. 这是我提出的一个示例查询,它不能按我想要的方式工作：select t.id, t.name, t.parent_id, a.cntr from categories as ...
有没有更好的方法呢？SELECT(SELECT count(*) FROM `tbl` WHERE `status` = 0) as 'text1',(SELECT count(*) FROM `tbl` WHERE `status` > 0) as 'text2'text1和text2是标题.解决方法:怎么样select sum(if(status=0,1,0)) as zeros, sum(if(status>0,1,0)) as greater from tbl;可能不一定会更好,但它在你的心理武器库中是一个有用的成语！
我需要有关将转换此表的SQL的帮助：=================== | Id | FK | Status| =================== | 1 | A | 100 | | 2 | A | 101 | | 3 | B | 100 | | 4 | B | 101 | | 5 | C | 100 | | 6 | C | 101 | | 7 | A | 102 | | 8 | A | 102 | | 9 | B | 102 | | 10 | B | 102 | ===================对此：========================================== | FK | Count 100 | Count 101 | Count 10...
我有一个记录所有用户跨配置文件访问的表,我需要获取自指定的UNIX时间戳(：time)以来不同访问次数和指定配置文件(：user_id)的新访问次数.id | user_id | visitor_id | time (unix) ====+==========+============+=============1 | 1 | 5 | 13000000002 | 1 | 5 | 13000000103 | 1 | 8 | 13000000154 | 1 | 9 | 13000000205 | 1 | 9 | 1...
我有一个关于mysql(innodb)的大表,它包含产品资产(13百万行).这里是我的数据库的一个小模式：product <-many2one-- file_item --one2many--> family --many2one--> download_type* file_item *表是包含数百万行的大表.我尝试使用以下sql查询按下载类型计算产品：select t.name as type, count(p.product_id) as n from file_item p inner join family f on f.id = p.family_id inner join type t on f.id_type = t.id group b...
三个非常相关的SQL问题(我实际上使用的是mySQL)： 假设一个人员表有两列名称,国家 1)我如何向有同胞的人展示？我可以按国家显示公民人数：select country, count(*) as fellow_citizens from people group by country但是我似乎无法显示那些与其他人相关的记录> 1.以下内容无效：select name, country, count(*) as fellow_citizens from people group by country where fellow_citizens > 1;……因为我不想把人分组,所以不会...
我有这个查询插入行,使用如下子查询：INSERT INTO Lecture_presence_Student (`presence_id`, `Lecture_id`, `Student_id`, `status`) VALUES((SELECT '' as presence_id, Lecture.Lecture_id, CourseEdition_students_Student.Student_id, 'onverwerkt' FROMCourseEdition_students_Student INNER JOIN Lecture ON CourseEdition_students_Student.CourseEdition_id = Lecture.CourseEdition_id) )我没有得到它,子选择查...