mongodb $ not _id [英] mongodb $not _id

问题描述

我需要一种搜索​​方法，但不包括用户前面屏幕上已经存在的_id.例如，我有3个宠物资料，用户已经在浏览其中一个.

I need a way to search but not include an _id which is already on the screen in front of the user. For example, I have 3 pet profiles one which the user is already viewing.

在该页面上，我有一个名为我的家人"的标题.然后，我运行此搜索:

On that page I have a heading called My Family. I then run this search:

public function fetch_family($owner)
    {
        $collection = static::db()->mypet;
        $cursor = $collection->find(array('owner' => new MongoId($owner)));

        if ($cursor->count() > 0)
            {
                $family = array();
                // iterate through the results
                while( $cursor->hasNext() ) {   
                    $family[] = ($cursor->getNext());
                }
                return $family;
            }
    }

即使我已经展示了一只宠物，它也将我家里的所有宠物都还了.因此，我想从搜索中排除一个_id.

And it returns all the pets in my family even knowing I am already showing one. So I want to exclude that one _id from the search.

我想过这样的事情.

$cursor = $collection->find(array('owner' => new MongoId($owner), '$not'=>array('_id'=>new MongoId(INSERT ID HERE))));

但是，这只是阻止了整个过程的运行.

However, that just stops the whole thing from running.

推荐答案

您需要执行 $ne (不相等)，以确保所有者正在搜索您正在查看的当前宠物.

You need to do a $ne (not equal) to make sure the current pet you are viewing is excluded from the search by owner.

mongo外壳中的示例:

var viewingPetId = ObjectId("515535b6760fe8735f5f6899");
var ownerId = ObjectId("515535ba760fe8735f5f689a");

db.mypet.find(
    {
        _id: { $ne: viewingPetId },
        owner: ownerId
    }
)

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