两种视图模型之间的敲除传递值

问题描述

如何在两个不同的视图模型中传递元素的visible属性.假设在一个视图模型中我具有visible是false，在另一个视图模型中的click函数中，我想使该可见成为true.可以使用淘汰赛吗?

How to pass the visible property of an element, in two different view model.Suppose in 1 view model i have visible is false, in another view model in click function i want to make that visible true. Can It be possible Using Knockout.

   ViewModel1 = function() {
        var self = this;
        this.dataItem = ko.observable(false);
      };

Viewmodel 2

Viewmodel 2

   ViewModel2 = function() {
      var self = this;

      // Click Function
      this.showItem= function(){
          ViewModel1.dataItem = ko.observable(true);
      };
    };

推荐答案

您应该尝试出色的 knockout-postbox .它旨在促进各个视图模型之间的去耦通信.

You should try an excellent knockout-postbox. It is designed to facilitate decoupled communication between separate view models.

在您的情况下，您可以像这样使用它:

In your case you can use it like:

注意: syncWith 用于双向通信，如果要进行单向通信，则应尝试将 subscribeTo 与 publishOn 一起使用>方法.

Note: syncWith used for bidirectional communication, if you want unidirectional communication then you should try subscribeTo with publishOn methods.

Viewmodel 1

ViewModel1 = function() {
               var self = this;
               this.dataItem = ko.observable(false).syncWith("visible", true);
             };

Viewmodel 2

ViewModel2 = function() {
               var self = this;

               self.dataItem = ko.observable().syncWith("visible", true);

               // Click Function
               this.showItem= function(){
                    self.dataItem = ko.observable(true);
               };
             };

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