我正在尝试获取属于活跃玩家的头像以及电子邮件以“a”开头的玩家的名称。
SELECT NAME
FROM AVATAR
INNER JOIN PLAYERAVATAR ON PLAYERAVATAR.PLAYER_ID = PLAYERAVATAR.AVATAR_ID
INNER JOIN PLAYER ON PLAYER.ACTIVE = 1;
我有这个，但它不起作用。我并不擅长这个，所以任何帮助都会受到赞赏。谢谢。
+-----------+--------------------+-----+--+---+--+
| AVATAR | | | | | |
+-----------+--------------------+-----+--+---+--+
| AVATAR_ID | NUMBER(38,0) | No | | 1 | |
| NAME | VARCHAR2(500 BYTE) | Yes | | 2 | |
| DOB | VARCHAR2(500 BYTE) | Yes | | 3 | |
| HOARD | NUMBER(38,0) | Yes | | 4 | |
| STRENGH | NUMBER | Yes | | 5 | |
| GENDER | VARCHAR2(500 BYTE) | Yes | | 6 | |
| SPECIES | VARCHAR2(500 BYTE) | Yes | | 7 | |
+-----------+--------------------+-----+--+---+--+
+-----------+--------------------+-----+--+---+--+
| PLAYER | | | | | |
+-----------+--------------------+-----+--+---+--+
| PLAYER_ID | NUMBER(38,0) | No | | 1 | |
| NAME | VARCHAR2(500 BYTE) | Yes | | 2 | |
| EMAIL | VARCHAR2(500 BYTE) | Yes | | 3 | |
| ACTIVE | NUMBER(38,0) | Yes | | 4 | |
| PASSWORD | VARCHAR2(500 BYTE) | Yes | | 5 | |
+-----------+--------------------+-----+--+---+--+
+--------------+--------------+-----+--+---+--+
| PLAYERAVATAR | | | | | |
+--------------+--------------+-----+--+---+--+
| PLAYER_ID | NUMBER(38,0) | Yes | | 1 | |
| AVATAR_ID | NUMBER(38,0) | Yes | | 2 | |
+--------------+--------------+-----+--+---+--+
答案 0 :(得分：0)
您可以这样做：
DISTINCT如果您想选择唯一头像名称，可以使用SELECT DISTINCT
AVATAR.NAME
FROM
AVATAR
INNER JOIN
PLAYERAVATAR ON PLAYERAVATAR.AVATAR_ID = AVATAR.AVATAR_ID
INNER JOIN
PLAYER ON PLAYER.PLAYER_ID = PLAYERAVATAR.PLAYER_ID
WHERE
PLAYER.ACTIVE = 1
AND
PLAYER.EMAIL LIKE 'a%' ;
，如下所示：
public class SignUpUser {
public var fullName : String?
public var id : Int?
public var city : String?
public var email : String?
public var address : String?
public var lastName : String?
public var countryCode : String?
public var firstName : String?
public var zipCode : Int?
public var contactNumber : Int?
public var sex : String?
public var dob : String?
public var signupType : String?
public var verified : String?
public var emailTokenExpiration : String?
public var updatedAt : String?
public var createdAt : String?
/**
Returns an array of models based on given dictionary.
Sample usage:
let user_list = User.modelsFromDictionaryArray(someDictionaryArrayFromJSON)
- parameter array: NSArray from JSON dictionary.
- returns: Array of User Instances.
*/
public class func modelsFromDictionaryArray(array:NSArray) -> [SignUpUser]
{
var models:[SignUpUser] = []
for item in array
{
models.append(SignUpUser(dictionary: item as! NSDictionary)!)
}
return models
}
/**
Constructs the object based on the given dictionary.
Sample usage:
let user = User(someDictionaryFromJSON)
- parameter dictionary: NSDictionary from JSON.
- returns: User Instance.
*/
init?() {}
required public init?(dictionary: NSDictionary) {
fullName = dictionary["fullName"] as? String
id = dictionary["id"] as? Int
city = dictionary["city"] as? String
email = dictionary["email"] as? String
address = dictionary["address"] as? String
lastName = dictionary["lastName"] as? String
countryCode = dictionary["countryCode"] as? String
firstName = dictionary["firstName"] as? String
zipCode = dictionary["zipCode"] as? Int
contactNumber = dictionary["contactNumber"] as? Int
sex = dictionary["sex"] as? String
dob = dictionary["dob"] as? String
signupType = dictionary["signupType"] as? String
verified = dictionary["verified"] as? String
emailTokenExpiration = dictionary["emailTokenExpiration"] as? String
updatedAt = dictionary["updatedAt"] as? String
createdAt = dictionary["createdAt"] as? String
}
/**
Returns the dictionary representation for the current instance.
- returns: NSDictionary.
*/
public func dictionaryRepresentation() -> NSDictionary {
let dictionary = NSMutableDictionary()
dictionary.setValue(self.fullName, forKey: "fullName")
dictionary.setValue(self.id, forKey: "id")
dictionary.setValue(self.city, forKey: "city")
dictionary.setValue(self.email, forKey: "email")
dictionary.setValue(self.address, forKey: "address")
dictionary.setValue(self.lastName, forKey: "lastName")
dictionary.setValue(self.countryCode, forKey: "countryCode")
dictionary.setValue(self.firstName, forKey: "firstName")
dictionary.setValue(self.zipCode, forKey: "zipCode")
dictionary.setValue(self.contactNumber, forKey: "contactNumber")
dictionary.setValue(self.sex, forKey: "sex")
dictionary.setValue(self.dob, forKey: "dob")
dictionary.setValue(self.signupType, forKey: "signupType")
dictionary.setValue(self.verified, forKey: "verified")
dictionary.setValue(self.emailTokenExpiration, forKey: "emailTokenExpiration")
dictionary.setValue(self.updatedAt, forKey: "updatedAt")
dictionary.setValue(self.createdAt, forKey: "createdAt")
return dictionary
}
}
答案 1 :(得分：0)
您有错误，因为列name位于多个表中。
您应该使用别名。
试试这个：
SELECT
P.NAME
FROM
AVATAR A
INNER JOIN
PLAYERAVATAR PL ON PL.PLAYER_ID = A.PLAYER_ID
INNER JOIN
PLAYER P ON P.ACTIVE = 1 AND P.PLAYER_ID = A.PLAYER_ID
WHERE
SUBSTR(P.EMAIL, 1, 1) = 'a'
答案 2 :(得分：0)
你可以使用
select a.NAME from AVATAR a
join PLAYERAVATAR pa on a.AVATAR_ID =pa.AVATAR_ID
join PLAYER p on pa.PLAYER_ID =p.PLAYER_ID
where p.ACTIVE = 1 and p.EMAIL like 'a%'
如果您需要不同的头像名称，可以使用
select distinct(a.NAME)
答案 3 :(得分：0)
从您请求的评论部分中的简短讨论中可以清楚地看出，您的数据模型实际上并不符合要求。你说一个化身只属于一个玩家。因此，将player_id存储在头像记录中并删除桥接表。
+-----------+--------------------+-----+--+---+--+ | PLAYER | | | | | | +-----------+--------------------+-----+--+---+--+ | PLAYER_ID | NUMBER(38,0) | No | | 1 | | | NAME | VARCHAR2(500 BYTE) | Yes | | 2 | | | ... | | | | | | +-----------+--------------------+-----+--+---+--+ +-----------+--------------------+-----+--+---+--+ | AVATAR | | | | | | +-----------+--------------------+-----+--+---+--+ | AVATAR_ID | NUMBER(38,0) | No | | 1 | | | NAME | VARCHAR2(500 BYTE) | Yes | | 2 | | | PLAYER_ID | NUMBER(38,0) | No | | 3 | | | ... | | | | | | +-----------+--------------------+-----+--+---+--+
查询变为：
select name
from avatar
where player_id in
(
select player_id
from player
where active = 1
and email like 'a%'
);