Question

我刚刚开始使用rethinkDB和thinky。我想问一下如何创建一个与现有模型具有n-n关系的模型的新实例。例如，如果我有Pizzas和Toppings的模型。我已经创建了一个Pizza实例，如何创建一个顶部实例并将其链接到已经存在的披萨？

var Pizza = thinky.createModel("Pizza", {
    name: type.string(),
    size: type.string()

});
var Topping = thinky.createModel("Topping", {
    name: type.string(),

});

Pizza.hasAndBelongsToMany(Topping, "toppings", "id", "id");
Topping.hasAndBelongsToMany(Pizza, "pizzas", "id", "id");

var pizza = new Pizza({name:"My pizza", size:"Large"});
var topping = new Topping({name:"Olive"});

如何保存顶部，以便与已经创建的披萨相关？

在docs中，对于1-n关系，他们执行以下操作：

topping.pizzas = pizza
topping.saveAll({pizza: true}).then(...);

在这种情况下，它将无法工作，因为topping.pizzas是一个对象数组而不仅仅是一个对象。

Answer 1

The answer is pretty simple I found out:

var pizza = new Pizza({name:"My pizza", size:"Large"});
var topping = new Topping({name:"Olive"});

Topping.filter({name:Olive}).run().then(function (topping) {
    if (topping.length > 0){
        if (pizza.toppings == undefined){
            pizza.toppings = []
        }
    pizza.toppings.push(topping[0]);
    pizza.saveAll().then(...)
    }
}

Pretty sure there are better ways to do it, but this works.

如何使用thinky创建一个新的n-n模型并将其链接到现有模型？

1 个答案: