SQL和2複雜的聯接

Question

我想優化與MySQL 5.5，關於2分離請求的SQL查詢。如果可能的話，我希望在最後有一個請求。

我有一個DocumentType表（請參閱此問題末尾的架構）。 例如，我想顯示「針對程序＃4上傳的5個文檔實例5」。

要計算總的每個程序文件，我有這個疑問：

select p.id as programId, count(dt.id) as totalDocPerProgram 
from DocumentType dt 
join DocumentTypeInProgram dtip on dtip.documentType_id=dt.id 
join Program p on dtip.program_id=p.id 
group by p.id 

要計算用戶已經多少文檔已上傳，通過程序分類，下面的查詢返回userSession以及如何列表許多文檔他以前上傳：

select p2.id as programId, wsc2.id as userSession, count(di2.id) as uploadedDocs 
from DocumentType dt2 
join DocumentInstance di2 on di2.documentType_id=dt2.id 
join WebsiteCase wsc2 on di2.websiteCase_id=wsc2.id 
left join Program p2 on p2.id=wsc2.program_id 
group by wsc2.id 

我的問題： 是否有可能有一個將返回每個程序一個單一的查詢中，totalDocPerProgram如何很多文件已經上傳了？ 是否有可能擁有類似「請爲程序＃3再上傳2個文檔」？

感謝您的幫助球員，

薩科

下面是數據庫SQL定義腳本

CREATE TABLE `DocumentType` (
    `id` bigint(20) NOT NULL AUTO_INCREMENT, 
    `description` varchar(255) DEFAULT NULL, 
    `lifetime` int(11) DEFAULT NULL, 
    `maxDocumentSize` int(11) DEFAULT NULL, 
    PRIMARY KEY (`id`) 
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8; 


CREATE TABLE `Program` (
    `id` bigint(20) NOT NULL AUTO_INCREMENT, 
    `name` varchar(255) DEFAULT NULL, 
    PRIMARY KEY (`id`), 
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8; 


CREATE TABLE `DataTypeForProgram` (
    `id` bigint(20) NOT NULL AUTO_INCREMENT, 
    `cardinality` int(11) DEFAULT NULL, 
    `dataType_id` bigint(20) DEFAULT NULL, 
    `program_id` bigint(20) DEFAULT NULL, 
    PRIMARY KEY (`id`), 
    KEY `FKA6CED6BFFCC6BB96` (`program_id`), 
    KEY `FKA6CED6BF17C2EE7E` (`dataType_id`), 
    CONSTRAINT `FKA6CED6BF17C2EE7E` FOREIGN KEY (`dataType_id`) REFERENCES `DataType` (`id`), 
    CONSTRAINT `FKA6CED6BFFCC6BB96` FOREIGN KEY (`program_id`) REFERENCES `Program` (`id`) 
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8; 

CREATE TABLE `WebsiteCase` (
    `id` bigint(20) NOT NULL AUTO_INCREMENT, 
    `creationDate` datetime DEFAULT NULL, 
    `lastUpdate` datetime DEFAULT NULL, 
    `program_id` bigint(20) NOT NULL, 
    PRIMARY KEY (`id`), 
    KEY `FK_Program` (`program_id`), 
    CONSTRAINT `FK_Program` FOREIGN KEY (`program_id`) REFERENCES `Program` (`id`) 
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8; 

CREATE TABLE `DocumentInstance` (
    `id` bigint(20) NOT NULL AUTO_INCREMENT, 
    `url` varchar(255) DEFAULT NULL, 
    `documentType_id` bigint(20) DEFAULT NULL, 
    `websiteCase_id` bigint(20) DEFAULT NULL, 
    PRIMARY KEY (`id`), 
    KEY `FK8BE8C2F05F1A4ADE` (`documentType_id`), 
    KEY `FK8BE8C2F05C57B856` (`websiteCase_id`), 
    CONSTRAINT `FK8BE8C2F05C57B856` FOREIGN KEY (`websiteCase_id`) REFERENCES `WebsiteCase` (`id`), 
    CONSTRAINT `FK8BE8C2F05F1A4ADE` FOREIGN KEY (`documentType_id`) REFERENCES `DocumentType` (`id`) 
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8; 

CREATE TABLE `DocumentTypeInProgram` (
    `id` bigint(20) NOT NULL AUTO_INCREMENT, 
    `cardinality` int(11) DEFAULT NULL, 
    `documentType_id` bigint(20) DEFAULT NULL, 
    `program_id` bigint(20) DEFAULT NULL, 
    PRIMARY KEY (`id`), 
    KEY `FK190E2A0A5F1A4ADE` (`documentType_id`), 
    KEY `FK190E2A0AFCC6BB96` (`program_id`), 
    CONSTRAINT `FK190E2A0AFCC6BB96` FOREIGN KEY (`program_id`) REFERENCES `Program` (`id`), 
    CONSTRAINT `FK190E2A0A5F1A4ADE` FOREIGN KEY (`documentType_id`) REFERENCES `DocumentType` (`id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

SET FOREIGN_KEY_CHECKS = 0; 

Answer 1

你可以做到這一點，與加入你的兩個現有查詢的查詢。像這樣：

select a.programId, 
     a.totalDocPerProgram, 
     b.upLoadedDocs, 
     (a.totalDocPerProgram - b.upLoadedDocs) as remainingDocs, 
     b.userSession 
from (
    select p.id as programId, 
      count(dt.id) as totalDocPerProgram 
    from DocumentType dt 
    join DocumentTypeInProgram dtip on dtip.documentType_id=dt.id 
    join Program p on dtip.program_id=p.id 
    group by p.id) a 
join (
    select p2.id as programId, 
      wsc2.id as userSession, 
      count(di2.id) as uploadedDocs 
    from DocumentType dt2 
    join DocumentInstance di2 on di2.documentType_id=dt2.id 
    join WebsiteCase wsc2 on di2.websiteCase_id=wsc2.id 
    left join Program p2 on p2.id=wsc2.program_id 
    group by p2.id, wsc2.id) b on a.ProgramId = b.ProgramId 

我還沒有調試這一點。但是這種事情效果很好，特別是如果你包括

where b.userSession = constant 

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