如何防止 Knockback.js 为空关系创建视图模型?
How do you prevent Knockback.js creating view models for null relations?
如果我的 backbone 模型有关系(例如,由 backbone-relational 创建),这些关系可能可以为空,导致外键字段有时为 null
。
如果我有多个击退视图模型,并且我已经指定了工厂,以便在遵循关系时我得到具有模型所需功能的视图模型,当它遇到 null
属性时,它继续并创建一个将 null
作为 model
传递的视图模型,这可能会破坏视图模型的大部分功能。
示例:
var ChildViewModel = kb.ViewModel.extend({
constructor: function (model, options) {
// this is the problem I'm trying to avoid - creating a view model with
// no model
if (!model) {
// just report the error somehow - the jsfiddle has the
// relevant HTML element
document.getElementById("error").innerHTML = "ChildModelView initialised without a model!";
}
kb.ViewModel.prototype.constructor.apply(this, arguments);
}
});
var ParentViewModel = kb.ViewModel.extend({
constructor: function (model, options) {
// specify factories here, because this way you can easily deal with
// reverse relationships, or complicated relationship trees when you
// have a large number of different types of view model.
kb.ViewModel.prototype.constructor.call(
this,
model,
{
factories: {relation1: ChildViewModel,
relation2: ChildViewModel},
options: options
}
);
}
});
// if we assume that relation2 is a nullable relationship, backbone-relational,
// for example, would give us a model that looks like this:
var model = new Backbone.Model({
id: 1,
relation1: new Backbone.Model({id: 2}), // this works fine
relation2: null // this causes a problem
});
var view_model = new ParentViewModel(model);
和 fiddle:
我刚刚发现了我认为可能是合理的解决方案。
您的工厂不必是 ViewModel "classes",但可以是工厂函数。所以:
var nullable = function (view_model_class) {
var factory = function (object, options) {
if (object === null) return object;
return new view_model_class(object, options);
};
return factory;
};
然后当您定义工厂时:
kb.ViewModel.prototype.constructor.call(
this,
model,
{
factories: {relation1: nullable(ChildViewModel),
relation2: nullable(ChildViewModel)},
options: options
}
);
如果我的 backbone 模型有关系(例如,由 backbone-relational 创建),这些关系可能可以为空,导致外键字段有时为 null
。
如果我有多个击退视图模型,并且我已经指定了工厂,以便在遵循关系时我得到具有模型所需功能的视图模型,当它遇到 null
属性时,它继续并创建一个将 null
作为 model
传递的视图模型,这可能会破坏视图模型的大部分功能。
示例:
var ChildViewModel = kb.ViewModel.extend({
constructor: function (model, options) {
// this is the problem I'm trying to avoid - creating a view model with
// no model
if (!model) {
// just report the error somehow - the jsfiddle has the
// relevant HTML element
document.getElementById("error").innerHTML = "ChildModelView initialised without a model!";
}
kb.ViewModel.prototype.constructor.apply(this, arguments);
}
});
var ParentViewModel = kb.ViewModel.extend({
constructor: function (model, options) {
// specify factories here, because this way you can easily deal with
// reverse relationships, or complicated relationship trees when you
// have a large number of different types of view model.
kb.ViewModel.prototype.constructor.call(
this,
model,
{
factories: {relation1: ChildViewModel,
relation2: ChildViewModel},
options: options
}
);
}
});
// if we assume that relation2 is a nullable relationship, backbone-relational,
// for example, would give us a model that looks like this:
var model = new Backbone.Model({
id: 1,
relation1: new Backbone.Model({id: 2}), // this works fine
relation2: null // this causes a problem
});
var view_model = new ParentViewModel(model);
和 fiddle:
我刚刚发现了我认为可能是合理的解决方案。
您的工厂不必是 ViewModel "classes",但可以是工厂函数。所以:
var nullable = function (view_model_class) {
var factory = function (object, options) {
if (object === null) return object;
return new view_model_class(object, options);
};
return factory;
};
然后当您定义工厂时:
kb.ViewModel.prototype.constructor.call(
this,
model,
{
factories: {relation1: nullable(ChildViewModel),
relation2: nullable(ChildViewModel)},
options: options
}
);