解析:更改指针的值 [英] Parse: change a pointer's value

问题描述

例如，我有两个类Player和Game，而Game有一个字段player，它是指向Player的指针.正如Parse文档所说，save是递归的，因此我假设我可以通过仅保存game来保存player，就像:

For example, I have two class Player and Game, and Game has a field player which is a pointer to Player. As Parse documentation says, save is recursive, so I assume that I can save a player by only saving a game, just like:

let player = PFObject(className: "Player")
player["name"] = ABC
let game = PFObject(className: "Game")
game["player"] = player
game["round"] = 1
game.saveInBackground()

这是真的吗?

好吧，现在假设这是正确的，那么仅保存game来更改player的name怎么办?喜欢:

Ok, now assuming that is true, then what about changing the player's name by only saving game? Like:

let player = game["player"] as! PFObject
player["name"] = BCD
game["round"] = 2
game.saveInBackground()

您能回答这两个问题吗?谢谢.

Could you please answer these two questions? Thank you.

另一个问题:如果要获取player["name"]，在查询game时是否需要使用includeKey("player")?

Another question: If I want to get the player["name"], do I need to use includeKey("player") when querying the game?

推荐答案

1. 是的，当您保存包含指向其他对象的指针的对象时，递归保存是正确的.所有指向的对象将被保存.由于game指向player，因此在保存game时也会保存player.

1. Yes, it's true about the recursive saving when you save an object that includes pointers to other objects. All of the pointed-to objects will be saved. Since game points to player, player will also be saved when game is saved.

相关对象的更改(例如玩家名称)也应基于(1)保存.

The changes in a related object, such as the player's name, should also be saved based on (1).

由于查询不会以相应的递归方式自动检索相关的对象，因此在检索游戏时需要使用includeKey来检索相关的Player对象是正确的.

Since queries do not automatically retrieve related objects in a corresponding recursive manner, you are correct about needing to use includeKey to retrieve the related Player object when retrieving a Game.

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