Php 使用sprip从OneToOne关系获取JSON
我将从我的数据库中获得类似以下结构的JSON:Php 使用sprip从OneToOne关系获取JSON,php,json,left-join,propel,one-to-one,Php,Json,Left Join,Propel,One To One,我将从我的数据库中获得类似以下结构的JSON: { "Users": [ { "Id": 1, "Name": "a", "Family": "b", "RegisterId": 1, "AccessType": 1, "Username": "abc" }, { "Id": 2, "Name": "x", "Family": "
{
"Users": [
{
"Id": 1,
"Name": "a",
"Family": "b",
"RegisterId": 1,
"AccessType": 1,
"Username": "abc"
},
{
"Id": 2,
"Name": "x",
"Family": "y",
"RegisterId": 2,
"AccessType": 1,
"Username": "xyz"
}
]
}
Id、Name、Family、RegisterId、AccessType是User\u TBL列,Username是Register\u TBL列
我可以使用以下查询获取此JSON:
SELECT u.id,u.name,u.family,u.access_type,u.register_id,r.username
FROM `user` as u LEFT JOIN `register` as r
ON u.register_id = r.id
我试图使用“推进”在行下使用,但它只显示所有用户列
$userList = UserQuery::create()-> joinWith('User.Register')-> find();
你有什么建议 我使用了下面的代码,它解决了我的问题
$userList = UserQuery::create()
-> leftJoinRegister()
-> select(array('ID','NAME','FAMILY','AccessType'))
-> withColumn('Register.Username' , 'Username')
->find();