问题
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.
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;
}
}
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.
I thought something like this.
$cursor = $collection->find(array('owner' => new MongoId($owner), '$not'=>array('_id'=>new MongoId(INSERT ID HERE))));
However, that just stops the whole thing from running.
回答1:
You need to do a $ne (not equal) to make sure the current pet you are viewing is excluded from the search by owner.
Example in the mongo
shell:
var viewingPetId = ObjectId("515535b6760fe8735f5f6899");
var ownerId = ObjectId("515535ba760fe8735f5f689a");
db.mypet.find(
{
_id: { $ne: viewingPetId },
owner: ownerId
}
)
回答2:
Use $ne
as (notice no need to use ObjectId()
, string will autocast to ObjectId
):
db.organizations.find({"_id" : {$ne:"563c50e05cdb2be30391e873"}})
来源:https://stackoverflow.com/questions/15020261/mongodb-not-id