原始SQL到laravel查询

Question

任何人都可以帮我写下面的SQL语句的数据库查询版本。我需要一些关于select语句和分区连接的帮助。

到目前为止，我已设法做到这一点。

$query = DB::table(raw('SapakInAdminOrder a'))
->select(raw('a.*'))
->leftJoin(raw('currency cu'), 'a.currency', '=', 'cu.id')
->leftJoin(raw('moodboards m'), 'a.orderMoodboardID', '=', 'm.id')
 ->join(raw('clients b'), 'a.clientID', '=', 'b.id')
 ->leftJoin(raw('moodboards mc'), 'b.moodboardID', 'mc.id')
 ->join(raw('sapakim c'), 'b.sapakID', '=', 'c.id')
 ->leftJoin(raw('sapakim sm'), 'c.managerid', '=', 'sm.id')
 ->leftJoin(raw('products p'), 'a.productKey', '=', 'p.id')
 ->where(function ($query) {
     $query->whereNull('a.isDeleted');
     $query->orWhere('a.isDeleted', '!=', 1);
 });

但我需要实现这一目标。

select * from (select ROW_NUMBER() OVER(ORDER BY case when (indesign.status=4 or indesign.statusdate is null) then getdate()+2 else indesign.statusdate end ASC) AS RowNum,a.*
FROM sapakInAdminOrder a 
left join currency cu on cu.id=a.currency
left join moodboards m on m.id=a.orderMoodboardID 
inner join Clients b on a.clientID=b.id 
left join moodboards mc on mc.id=b.moodboardID 
inner join Sapakim c on b.sapakID=c.id 
left join Sapakim sm on sm.id=c.managerid  
left join products p on p.id=a.productKey 
left join (select * from (select ROW_NUMBER() over(PARTITION BY orderID ORDER BY id DESC) r, * from orderCommunication ) f where r=1) chat on chat.orderId = a.id
left join (select id,[status],orderid,approveSMSDate,coverImage,statusDate from (SELECT  id,[status],statusDate,approveSMSDate,coverImage,orderid,ROW_NUMBER() OVER(PARTITION BY orderid ORDER BY id DESC) AS r  FROM  SapakimInAdminDesigns) f where  r=1) indesign on a.id=indesign.orderid
where (a.isDeleted is null or a.isDeleted != 1) and 
      c.inAdminManagerID=(select id from sapakim where sapakguid='test')  and 
      c.sapakguid='test' and 
      a.isFreeDesign=0 and 
      a.transactionID = -1 and 
      (a.designerPaid is null or a.designerPaid=0) and 
      (chat.sentToPrinter is null and chat.sentToManager is null and chat.sentToDesigner is null)
) bb where RowNum>=1 and RowNum<31 
ORDER BY  RowNum asc

我可以做简单的但不能真正地围绕分区连接和select语句。

我真的很感激这方面的帮助。

提前致谢。

Answer 1

我也遇到了这种类型的问题，但我得到了一个解决方案，它对我来说很好！

对这种类型的查询使用DB :: unprepared（）：

$path = base_path() . "/storage/agency.sql";
$sql = file_get_contents($path);
DB::unprepared(DB::raw($sql));

Laravel中的所有SQL命令都是默认准备的，但有时您需要在未准备模式下执行命令，因为某些数据库中的某些命令无法在准备模式下运行。

Answer 2

这是你的意思吗？

$result = DB::Select("select * from (select ROW_NUMBER() OVER(ORDER BY case when (indesign.status=4 or indesign.statusdate is null) then getdate()+2 else indesign.statusdate end ASC) AS RowNum,a.*
FROM sapakInAdminOrder a 
left join currency cu on cu.id=a.currency
left join moodboards m on m.id=a.orderMoodboardID 
inner join Clients b on a.clientID=b.id 
left join moodboards mc on mc.id=b.moodboardID 
inner join Sapakim c on b.sapakID=c.id 
left join Sapakim sm on sm.id=c.managerid  
left join products p on p.id=a.productKey 
left join (select * from (select ROW_NUMBER() over(PARTITION BY orderID ORDER BY id DESC) r, * from orderCommunication ) f where r=1) chat on chat.orderId = a.id
left join (select id,[status],orderid,approveSMSDate,coverImage,statusDate from (SELECT  id,[status],statusDate,approveSMSDate,coverImage,orderid,ROW_NUMBER() OVER(PARTITION BY orderid ORDER BY id DESC) AS r  FROM  SapakimInAdminDesigns) f where  r=1) indesign on a.id=indesign.orderid
where (a.isDeleted is null or a.isDeleted != 1) and 
      c.inAdminManagerID=(select id from sapakim where sapakguid='test')  and 
      c.sapakguid='test' and 
      a.isFreeDesign=0 and 
      a.transactionID = -1 and 
      (a.designerPaid is null or a.designerPaid=0) and 
      (chat.sentToPrinter is null and chat.sentToManager is null and chat.sentToDesigner is null)
) bb where RowNum>=1 and RowNum<31 
ORDER BY  RowNum asc"); 

Answer 3

也许您应该为这些分区查询创建数据库视图？然后您可以从数据库加入视图。

从技术上讲，框架通常不支持这些分析功能。

Answer 4

查看这个名为Orator的新Laravel工具。它允许您简单地粘贴“遗留”SQL并返回Laravel的DB Query Builder语法。

Laravel Orator News Article

Online Conversion Tool

Answer 5

假设您要在前端显示此内容，请使用与雄辩模型的关系，https://laravel.com/docs/master/eloquent-relationships#one-to-one

在模型用户中，有一个由一对一的关系调用，由方法phone（）识别，如果你去show视图我假设通过一个$ user变量做以下

<?php dd($user->phone); ?>

如果你不是这个意思，我可能会错过问题的全部内容。