禁用基于其他输入的输入Knockoutjs

Question

我想设置一个敲除的observable，将禁用输入，如果两个其他输入不是都在1和30之间。现在，当我在jsFiddle中运行代码时，我的输入被禁用。不幸的是，我永远无法重新启用输入。这里是jsFiddle上的代码感谢您的任何帮助。

HTML

<!-- This is a *view* - HTML markup that defines the appearance of your 
UI --> 
<p>Alcohol Days: 
    <input data-bind="value: alcoholDays" /> 
</p> 
<p>Alcohol 5+ Days: 
    <input data-bind="value: alcoholFivePlusDays" /> 
</p> 
<p>Alcohol 4- Days: 
    <input data-bind="value: alcoholFourLessDays" /> 
</p> 
<p>Drug Days: 
    <input data-bind="value: drugDays" /> 
</p> 
<p>Both Days: 
    <input data-bind="value: bothDays, enable: enableBothDays" /> 
</p> 
<p>Enable Both Days: <strong data-bind="text: enableBothDays"></strong> 

</p> 
<p>Alcohol Days: <strong data-bind="text: alcoholDays"></strong> 

</p> 
<p>Drug Days: <strong data-bind="text: drugDays"></strong> 

</p> 
<button data-bind="click: capitalizeLastName">Go caps</button> 

的JavaScript

// This is a simple *viewmodel* - JavaScript that defines the data and behavior of your UI 
function AppViewModel() { 

    var self = this; 

    self.alcoholDays = ko.observable(""); 
    self.alcoholFivePlusDays = ko.observable(""); 
    self.alcoholFourLessDays = ko.observable(""); 
    self.drugDays = ko.observable(""); 
    self.bothDays = ko.observable(""); 

    self.enableBothDays = ko.computed(function() { 

     if ((parseInt(self.alcoholDays) > 0 && parseInt(self.alcoholDays) <= 30) && (parseInt(self.drugDays) > 0 && parseInt(self.drugDays) <= 30)) { 
      return true; 
     } else { 
      return false; 
     } 

    }, self); 

} 

// Activates knockout.js 
ko.applyBindings(new AppViewModel()); 

韦德

Answer 1

hi check this fiddle我修复了问题。

1）清除了您提琴绑定错误

2）重新计算结构观察到

self.enableBothDays = ko.computed({ 
    read: function() { 
     alert('In'); 
     var alcDays = Number(self.alcoholDays()); 
     var drgDays = Number(self.drugDays()); 
     alert(alcDays+','+drgDays); 
     var temp = false; 
    if (alcDays > 0 && alcDays <= 30 && drgDays > 0 && drgDays <= 30) { 
     temp = true; 
    } else { 
     temp = false; 
    } 
     alert(temp); 
     return temp; 
} 
}); 

3）改变启用条件

fiddle..

痕作为回答

Answer 2

使用此 启用：enableBothDays（）==真 或者 启用：enableBothDays ==真

Answer 3

固定小提琴 http://jsfiddle.net/SNv6n/20/

你分别致电self.alcoholDays代替self.alcoholDays（）。如果您将括号添加到您的计算中的这些调用中，并添加了函数“capitalizeLastName”，它就可以工作。

self.capitalizeLastName = function() { 
    alert('TODO'); 
} 
self.enableBothDays = ko.computed(function() { 

    if ((parseInt(self.alcoholDays()) > 0 && parseInt(self.alcoholDays()) <= 30) && (parseInt(self.drugDays()) > 0 && parseInt(self.drugDays()) <= 30)) { 
     return true; 
    } else { 
     return false; 
    } 

}, self);