如何傳遞一個空的PHP變量MongoDB中查詢

我想創建一個動態的MongoDB查詢其插入它的每一部分的聚集，如果條件爲真，否則不注射的那部分。如何傳遞一個空的PHP變量MongoDB中查詢

比如我要檢查，如果時間是凌晨1點至早上8點。如果是，則定義的數組到MongoDB的查詢其他人處理，沒有在傳遞。

if ($this->Not_in_1_8 == true) { 
    $this->N_IN_1_8 = array('dont_show_between_1_n_8' => array('$ne' => true)); 
}else { 
    $this->N_IN_1_8 = null; 
} 
$MidnightCheck = $this->N_IN_1_8; 
$this->campaign = Bcamp::raw(function ($collection) use ($seat_category_list, $banner_size, $seat_filter_list, $gold_network, $MidnightCheck) { 
    return $collection->aggregate([ 
     [ 
      '$match' => [ 
       '$and' => [ 
        ["targets.cats" => [ 
         '$in' => $seat_category_list 
         ] 
        ], 
        ['banners.' . $banner_size => [ 
         '$exists' => true 
         ] 
        ], 
        ['href' => [ 
         '$nin' => $seat_filter_list 
         ] 
        ], 
        ['targets.gold_network' => [ 
         '$eq' => $gold_network 
         ] 
        ], 
        ['status' => [ 
         '$ne' => "Low_Budget" 
         ] 
        ], 
        ['daily_budget_status' => [ 
         '$ne' => "Low_Budget" 
         ] 
        ], 
        $MidnightCheck 
       ] 
      ] 
     ], 
     [ 
      '$project' => [ 
       'ab' => [ 
        '$cmp' => [ 
         '$budget', '$click_cost' 
        ] 
       ] 
      ] 
     ], 
     [ 
      '$match' => [ 
       'ab' => [ 
        '$lt' => 1 
       ] 
      ] 
     ] 

    ]); 
});

但在這個例子中它會注入到空查詢，使得它錯了，我趕上錯誤：bad query: BadValue: $or/$and/$nor entries need to be full objects

我已經改成了$this->N_IN_1_8 = '';仍然沒有成功。

我需要一箇中立的變量或條件，其中通過如果要是假的條件不會對查詢的影響。任何想法？

FYI：我使用Laravel 5.3 jenssegers/laravel-mongodb的包框架與mongodb的工作

來源

2017-01-01 Mohammad_Hosseini

Prefering保持代碼的結構是，即當預先建立的陣列並不在條件中使用代碼，你可以保留$MidnightCheck爲null與array_filter包裹陣列：

return $collection->aggregate([ 
[ 
    '$match' => [ 
     '$and' => array_filter([ 
      ["targets.cats" => ['$in' => $seat_category_list]], 
      ['banners.' . $banner_size => ['$exists' => true]], 
      ['href' => ['$nin' => $seat_filter_list]], 
      ['targets.gold_network' => ['$eq' => $gold_network]], 
      ['status' => ['$ne' => "Low_Budget"]], 
      ['daily_budget_status' => ['$ne' => "Low_Budget"]], 
      $MidnightCheck 
     ]) 
    ] 
// ...

調用array_filter沒有第二個參數會從數組中過濾掉所有假值，造成不必要的$MidnightCheck將不復存在。

我覺得可能是更清晰的將是回調至前期準備條件：

$conditions = [ 
    ["targets.cats" => ['$in' => $seat_category_list]], 
    ['banners.' . $banner_size => ['$exists' => true]], 
    ['href' => ['$nin' => $seat_filter_list]], 
    ['targets.gold_network' => ['$eq' => $gold_network]], 
    ['status' => ['$ne' => "Low_Budget"]], 
    ['daily_budget_status' => ['$ne' => "Low_Budget"]], 
]; 
if ($MidnighCheck) { 
    $conditions[] = $MidnightCheck; 
} 
return $collection->aggregate([ 
    ['$match' => [ '$and' => $conditions ],] 
])

來源

2017-01-02 22:26:17 malarzm

感謝您的教育答案 - 但是當我第二個選項去我得到：Aggregate.php線93 ： $管道是不是列表（意外指數：「$匹配」） –

謝謝你這樣做！ – malarzm

如何傳遞一個空的PHP變量MongoDB中查詢

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