Laravel 查询生成器在 where 子句之后加入答案

【问题标题】：Laravel query builder join after where clauseLaravel 查询生成器在 where 子句之后加入
【发布时间】：2021-08-16 22:16:18
【问题描述】：

我正在使用 laravel 8。我有这个 mysql 命令，我想将其转换为 laravel 查询生成器样式：

select allocation.*, leav_leave_types.leave_type_code 
from (
    select * from leav_employee_annual_leave_allocations 
    where leave_year_id = $year_id and employee_id = $user_id
) as allocation
left join leav_leave_types on (leav_leave_types.id = allocation.leave_type_id)

其实我想先应用where 子句，然后再执行left join 以获得更好的性能。

如何将其转换为查询构建器样式？

【问题讨论】：

leav_employee_annual_leave_allocations 将受益于INDEX(employee_id, leave_year_id)

标签： php mysql laravel eloquent query-builder

【解决方案1】：

您的查询中当前不在documentation 中的唯一内容是使用子查询作为主表。

这可以通过将Closure 或Builder 实例传递给table() 或from() 方法来完成。

DB::table(closure, alias)
DB::table(builder, alias)
DB::query()->from(closure, alias)
DB::query()->from(builder, alias)

使用闭包：

DB::table(function ($sub) use ($user_id, $year_id) {
        $sub->from('leav_employee_annual_leave_allocations')
            ->where('leave_year', $year_id)
            ->where('employee_id', $user_id);
    }, 'allocation')
    ->select('allocation.*', 'leav_leave_types.leave_type_code')
    ->leftJoin('leav_leave_types', 'leav_leave_types.id', 'allocation.leave_type_id')
    ->get();

DB::query()
    ->select('allocation.*', 'leav_leave_types.leave_type_code')
    ->from(function ($sub) use ($user_id, $year_id) {
        $sub->from('leav_employee_annual_leave_allocations')
            ->where('leave_year', $year_id)
            ->where('employee_id', $user_id);
    }, 'allocation')
    ->leftJoin('leav_leave_types', 'leav_leave_types.id', 'allocation.leave_type_id')
    ->get();

使用 Builder 实例

$sub = DB::table('leav_employee_annual_leave_allocations') // or DB::query()->from('leav_employee_annual_leave_allocations')
    ->where('leave_year', $year_id)
    ->where('employee_id', $user_id);

DB::table($sub, 'allocation')
    ->select('allocation.*', 'leav_leave_types.leave_type_code')
    ->leftJoin('leav_leave_types', 'leav_leave_types.id', 'allocation.leave_type_id')
    ->get();

// personally my favorite way. I find it very readable.
$sub = DB::table('leav_employee_annual_leave_allocations') // or DB::query()->from('leav_employee_annual_leave_allocations')
    ->where('leave_year', $year_id)
    ->where('employee_id', $user_id);

DB::query()
    ->select('allocation.*', 'leav_leave_types.leave_type_code')
    ->from($sub, 'allocation')
    ->leftJoin('leav_leave_types', 'leav_leave_types.id', 'allocation.leave_type_id')
    ->get();

生成的 SQL 如下所示

select "allocation".*, "leav_leave_types"."leave_type_code" from (
    select * from "leav_employee_annual_leave_allocations"
    where "leave_year" = ? and "employee_id" = ?
) as "allocation"
left join "leav_leave_types" on "leav_leave_types"."id" = "allocation"."leave_type_id"

如果您希望在连接条件周围生成括号，则应使用以下符号之一。

leftJoin('leav_leave_types', ['leav_leave_types.id' => 'allocation.leave_type_id'])

leftJoin('leav_leave_types', function ($join) {
    $join->on(['leav_leave_types.id' => 'allocation.leave_type_id']);
})

leftJoin('leav_leave_types', function ($join) {
    // will generate a parenthesis if there's more than one condition
    $join->on('leav_leave_types.id', 'allocation.leave_type_id')
         ->on(...) // and condition
         ->orOn(...); // or condition
})

【讨论】：

【解决方案2】：

或者，您可以将 SQL 转为

select *, 
       ( SELECT leave_type_code
             FROM leav_leave_types
             WHERE id = allocation.leave_type_id
       ) AS leave_type_code
    FROM leav_employee_annual_leave_allocations AS allocation
    where leave_year_id = $year_id and employee_id = $user_id

（这可能更有效。）

在任何一种情况下，leav_employee_annual_leave_allocations 都会受益于INDEX(employee_id, leave_year_id)。

【讨论】：