Question

我目前有一个SQL查询，这是一个出色的stackoverflow成员今天早些时候帮助我。目前它的工作方式非常好，而且我希望它能够正常工作。

唯一的问题是它的RAW SQL，我非常希望能够使用Laravel Query Builder。

抱歉这是编辑 - 这是原始的查询：

    $addresses = DB::select(
        DB::raw('
            (SELECT 
                "Company" AS object_type_name,
                companies.company_name AS object_name,
                addresses.*
            FROM 
                addresses
            INNER JOIN 
                companies 
            ON 
                addresses.object_id = companies.id
            WHERE 
                addresses.object_type = 2) 
            UNION ALL 
            (SELECT
                "Job" AS object_type_name,
                jobs.job_title AS object_name,
                addresses.*
            FROM 
                addresses
            INNER JOIN 
                jobs
            ON 
                addresses.object_id = jobs.id
            WHERE 
                addresses.object_type = 4)
        '));

这是我到目前为止的代码：

    $bindings = array(
        'soft_deleted' => 0,
        'user' => 1,
        'company' => 2,
        'candidate' => 3,
        'job' => 4,
    );

    $companies = DB::table('addresses')->select(
      'addresses.*',
      'companies.company_name as object_name'
    )->where('addresses.soft_deleted', '=', 0)->join('companies', function($join) use ($bindings){
        $join->on('addresses.object_id', '=', 'companies.id')
             ->where('addresses.object_type', '=', $bindings['company']);
    });

    $jobs = DB::table('addresses')->select(
      'addresses.*',
      'jobs.job_title as object_name'
    )->join('jobs', function($join) use ($bindings){
        $join->on('addresses.object_id', '=', 'jobs.id')
             ->where('addresses.object_type', '=', $bindings['job']);
    });

    $addresses = $companies->unionAll($jobs)->get();

使用上面的代码，我收到以下错误：

SQLSTATE [HY093]：无效的参数编号（SQL :(选择addresses。，companies。company_name为object_name来自addresses内部在companies上加入addresses。object_id = companies。id和addresses。object_type = 2其中addresses 。soft_deleted = 0）联合全部（选择addresses。，jobs。job_title为object_name来自addresses内部联接{ {1}} jobs = addresses。object_id和jobs。id =？））

我已经做了几个小时的搜索，但我似乎找不到这个小问题的答案。

Answer 1

我设法解决了这个问题。就像从每个连接闭包中删除->where并将其链接到连接本身而不是嵌套它一样简单。

在：

$companies = DB::table('addresses')->select(
  'addresses.*',
  'companies.company_name as object_name'
)->where('addresses.soft_deleted', '=', 0)->join('companies', function($join) use ($bindings){
    $join->on('addresses.object_id', '=', 'companies.id')
         ->where('addresses.object_type', '=', $bindings['company']);
});

$jobs = DB::table('addresses')->select(
  'addresses.*',
  'jobs.job_title as object_name'
)->join('jobs', function($join) use ($bindings){
    $join->on('addresses.object_id', '=', 'jobs.id')
         ->where('addresses.object_type', '=', $bindings['job']);
});

后：

    $companies = DB::table('addresses')->select(
      'addresses.*',
      'companies.company_name as object_name'
    )->where('addresses.soft_deleted', '=', 0)->join('companies', function($join){
        $join->on('addresses.object_id', '=', 'companies.id');
    })->where('addresses.object_type', '=', 2);

    $jobs = DB::table('addresses')->select(
      'addresses.*',
      'jobs.job_title as object_name'
    )->where('addresses.soft_deleted', '=', 0)->join('jobs', function($join){
        $join->on('addresses.object_id', '=', 'jobs.id');
    })->where('addresses.object_type', '=', 4);

    $addresses = $companies->unionAll($jobs)->get();

希望这有助于某人。

将参数绑定传递给Laravel查询构建器？

1 个答案: