mongodb $not _id [英] mongodb $not _id

问题描述

我需要一种搜索​​方式，但不包括已经在用户面前的屏幕上的 _id.例如，我有 3 个宠物资料，其中一个用户已经在查看.

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.

在那个页面上，我有一个名为我的家人"的标题.然后我运行这个搜索:

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;
            }
    }

即使知道我已经展示了一只宠物，它也会返回我家中的所有宠物.所以我想从搜索中排除那个_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.

我是这么想的.

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

但是，这只会阻止整个事情的运行.

However, that just stops the whole thing from running.

推荐答案

你需要做一个$ne(不等于)以确保您当前查看的宠物不在所有者搜索范围内.

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 shell 中的示例:

Example in the mongo shell:

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

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

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