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sql - Laravel Query：为 Query Builder 实现 Eloquent Scope

问题描述

在 Laravel Query Builder 中，我想在 Eloquent 中实现类似 Scope 的东西。参考：Laravel 查询：添加自定义功能，如 Soft Deletes。

我有一些复杂的查询（有连接和什么没有），但我希望能够轻松应用 WHERE 条件，其工作方式如下：

原来的：

Select * from t1 join t2 ... join t3 ... etc
Where t1.c1 = x OR t3.c4 like "%like"

通缉：

Select * from t1 join t2 ... join t3 ... etc
Where (t1.c1 = x OR t3.c4 like "%like") AND (t1.isTest = false AND t3.isTest = false)

我写了以下方法：

public static function scopeNoTest($query, $tables=[false])
    {
        if (!is_array($tables)) $tables = [$tables];
            foreach ($tables as $table)
            {
                $field = ($table) ? $table . '.isTest' : 'isTest';
                $query = $query->where(function ($q) use ($query, $field)
                    {
                        $q->where($field, false)
                          ->orWhereNull($field);
                    }
                );
            }
            return $query;
    }

它像这样运行：

$select       = <parameter driven select statement>
$where[$role] = <array of different where condition based on passed in parameter?
$bindings     = <query bindings based on passed in parameters>  

$query = DB::table('Transactions AS trans')
       ->leftJoin('Buyers AS b', 'trans.ID', '=', 'b.Transactions_ID')
       ->leftJoin('Sellers AS s', 'trans.ID', '=', 's.Transactions_ID')
       ->leftJoin('Agents AS ba', 'trans.BuyersAgent_ID', '=', 'ba.ID')
       ->leftJoin('Agents AS sa', 'trans.SellersAgent_ID', '=', 'sa.ID')
       ->leftJoin('TransactionCoordinators AS btc', 'trans.BuyersTransactionCoordinators_ID', '=', 'btc.ID')
       ->leftJoin('TransactionCoordinators AS stc', 'trans.SellersTransactionCoordinators_ID', '=', 'stc.ID')
       ->leftJoin('lu_UserRoles AS lu_ur', 'trans.ClientRole', '=', 'lu_ur.Value')
       ->leftJoin('Properties AS p', 'trans.Properties_ID', '=', 'p.ID')
       ->selectRaw($select);

// ... Adds code to Only Select records with isTest NOT True
$query = Model_Parent::scopeNoTest($query, ['trans', 'ba', 'sa', ]);

$query->whereRaw($where[$role].$whereUser, $bindings)->distinct();
$transactions = $query->get();

这段代码的问题在于它没有将原始的 [passed in] 查询放在括号中——所以查询是错误的！.

代码创建的 WHERE 是：

where 
     (`trans`.`isTest` = 0 or `trans`.`isTest` is null) 
 and (`ba`.`isTest` = 0 or `ba`.`isTest` is null) 
 and (`sa`.`isTest` = 0 or `sa`.`isTest` is null) 
 and trans.BuyersTransactionCoordinators_ID = 1 OR trans.SellersTransactionCoordinators_ID = 1 
 OR  trans.CreatedByUsers_ID = 1 OR trans.OwnedByUsers_ID = 1

而且我要

where 
     (`trans`.`isTest` = 0 or `trans`.`isTest` is null) 
 and (`ba`.`isTest` = 0 or `ba`.`isTest` is null) 
 and (`sa`.`isTest` = 0 or `sa`.`isTest` is null) 
 and (trans.BuyersTransactionCoordinators_ID = 1 OR trans.SellersTransactionCoordinators_ID = 1 
      OR  trans.CreatedByUsers_ID = 1 OR trans.OwnedByUsers_ID = 1)

有没有办法做到这一点 ？？

标签： sqllaravelscopewhere-clausequery-builder

解决方案

看起来以下行导致了这种情况：

$query->whereRaw($where[$role].$whereUser, $bindings)->distinct();

我认为有两种方法可以解决这个问题：

// 1
->whereRaw('(' . $where[$role].$whereUser . ')', $bindings)->

// 2
->where(function ($query) use (...) {
    $query->whereRaw(...);
})->

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