Pin*_*aas 20 angularjs ng-grid
尝试根据同一行中的另一个值从gridcollection显示columnvalue.用户可以选择/更改包含带有值的网格的模态中的值.当模态关闭时,值将被传回.那时我想为'也称为'设置一个值:
HTML:
Also known as: <input type="text" `ng-model="displayValue(displayNameData[0].show,displayNameData[0].value)">`
Run Code Online (Sandbox Code Playgroud)
我在范围上创建了一个函数,仅在'show'值为true时才选择值:
$scope.displayValue = function (show, val) {
if (show) {
return val;
}
else {
return '';
}
}
Run Code Online (Sandbox Code Playgroud)
但是,当我关闭模态时,我收到一个错误:
Error: [ngModel:nonassign] Expression 'displayValue(displayNameData[0].show,displayNameData[0].value)' is non-assignable.
Run Code Online (Sandbox Code Playgroud)
plnkr参考:http://plnkr.co/edit/UoQHYwAxwdvX0qx7JFVW?p = preview
kar*_*una 16
正如HackedByChinese所提到的,你不能将ng-model绑定到一个函数,所以试试这样:
<input type="text" ng-if="displayNameData[0].show"
ng-model="displayNameData[0].value">
Run Code Online (Sandbox Code Playgroud)
或者,如果您希望此控件可见,您可以创建指令,添加函数$parsers
将根据以下内容设置空值show
:
angular.module('yourModule').directive('bindIf', function() {
return {
restrict: 'A',
require: 'ngModel',
link: function(scope, element, attrs, ngModel) {
function parser(value) {
var show = scope.$eval(attrs.bindIf);
return show ? value: '';
}
ngModel.$parsers.push(parser);
}
};
});
Run Code Online (Sandbox Code Playgroud)
HTML:
<input type="text" bind-if="displayNameData[0].show"
ng-model="displayNameData[0].value">
Run Code Online (Sandbox Code Playgroud)
绑定到getter/setter
有时将ngModel绑定到getter/setter函数会很有帮助.getter/setter是一个函数,它在使用零参数调用时返回模型的表示,并在使用参数调用时设置模型的内部状态.对于内部表示与模型公开给视图的内部表示不同的模型,有时使用它是有用的.
的index.html
<div ng-controller="ExampleController">
<form name="userForm">
<label>Name:
<input type="text" name="userName"
ng-model="user.name"
ng-model-options="{ getterSetter: true }" />
</label>
</form>
<pre>user.name = <span ng-bind="user.name()"></span></pre>
</div>
Run Code Online (Sandbox Code Playgroud)
app.js
angular.module('getterSetterExample', [])
.controller('ExampleController', ['$scope', function($scope) {
var _name = 'Brian';
$scope.user = {
name: function(newName) {
// Note that newName can be undefined for two reasons:
// 1. Because it is called as a getter and thus called with no arguments
// 2. Because the property should actually be set to undefined. This happens e.g. if the
// input is invalid
return arguments.length ? (_name = newName) : _name;
}
};
}]);
Run Code Online (Sandbox Code Playgroud)