將參數綁定傳遞給Laravel查詢生成器？

我目前有一個SQL查詢，這是一個夢幻般的計算器成員今天早些時候幫助我的。目前，它的工作非常好，我希望它能夠工作。將參數綁定傳遞給Laravel查詢生成器？

唯一的問題是它的RAW SQL，我非常想使用Laravel查詢生成器來工作。

很抱歉，這是一個編輯 - 這裏是原始查詢：

$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) 
     '));

這裏是我的代碼至今：

$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();

通過上面我發現了以下錯誤的代碼：

SQLSTATE [HY093]：無效的參數號（SQL：（請選擇addresses。，companies。company_name as object_name from addresses內部加入companies對addresses。 object_id = companies。 id和addresses。 object_type = 2其中addresses。 soft_deleted = 0）UNION ALL（選擇addresses。，jobs。job_title作爲object_name從addresses內對addresses加入jobs。object_id = jobs。id和addresses。object_type =？））

我已經做了搜索的價值小時但我似乎無法找到這個小問題的答案。

來源

2015-03-31 learn

瘋狂的猜測：第一個'select'會改變你的'$ bindings'。爲了排除這種情況，我會在每次select之後轉儲'$ bindings'。 – 2015-04-01 00:01:29

我設法解決它。這很簡單，只需從每個連接關閉中刪除->where，並將其鏈接到連接本身，而不是嵌套它。

前：

$companies = DB::table('addresses')->select(
    'addresses.*', 
    'companies.company_name as object_name' 
)->where('addresses.soft_deleted', '=', 0)->join('companies', function($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']); 
});

後：

$companies = DB::table('addresses')->select(
     'addresses.*', 
     'companies.company_name as object_name' 
    )->where('addresses.soft_deleted', '=', 0)->join('companies', function($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();

希望這可以幫助別人。

來源

2015-03-31 22:16:40 learn

將參數綁定傳遞給Laravel查詢生成器？

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