我有MongoDB中的下一个文档:
比赛文件:
{
"_id": ObjectId("502aa915f50138d76d11112f7"),
"contestname": "Contest1",
"description": "java programming contest",
"numteams": NumberInt(2),
"teams": [
{
"teamname": "superstars",
"userid1": "50247314f501384b011019bc",
"userid2": "50293cf9f50138446411001c",
"userid3": "50293cdff501384464110018"
},
{
"teamname": "faculty",
"userid1": "50247314f501384b0110100c",
"userid2": "50293cf9f50138446410001b",
"userid3": "50293cdff501384464000019"
}
],
"term": "Fall 2012"
}
想象一下,我有超过这个用户可以注册的文档。我想找到用户注册的所有比赛。到目前为止我有这样的事情:
$id = "50247314f501384b011019bc";
$user = array('userid1' => $id, 'userid2' => $id, 'userid3' => $id );
$team = array('teams' => $user);
$result =$this->collection->find($team);
return $result;
有人可以帮我吗?
谢谢。
------ ------解决
$team = array('$or' => array(array('teams.userid1' => $id),
array('teams.userid2' => $id),
array('teams.userid3' => $id)
));
$result =$this->collection->find($team);
答案 0 :(得分:3)
您的数据结构很难查询,因为您有一系列嵌入式文档。只需稍微更改数据,您就可以更轻松地使用它。
我已将用户ID放入数组:
{
"contestname": "Contest1",
"description": "java programming contest",
"numteams": 2,
"teams": [
{
"teamname": "superstars",
"members": [
"50247314f501384b011019bc",
"50293cf9f50138446411001c",
"50293cdff501384464110018"
]
},
{
"teamname": "faculty",
"members": [
"50247314f501384b0110100c",
"50293cf9f50138446410001b",
"50293cdff501384464000019"
]
}
],
"term": "Fall 2012"
}
然后,您可以为:
执行PHP等效find()db.contest.find(
{'teams.members':'50247314f501384b011019bc'},
{'contestname':1, 'description':1}
)
这将返回此用户输入的匹配竞赛:
{
"_id" : ObjectId("502c108dcbfbffa8b2ead5d2"),
"contestname" : "Contest1",
"description" : "java programming contest"
}
{
"_id" : ObjectId("502c10a1cbfbffa8b2ead5d4"),
"contestname" : "Contest3",
"description" : "Grovy programming contest"
}