rog*_*awg 7 forms serialization json angularjs
据我所知,如果表单字段为空,AngularJS不会序列化表单字段.但是,我需要将这些字段放在生成的JSON中,即使它们是空的.我试图查询该JSON,如果字段描述符不存在,它将失败.有关如何让AngularJS执行此操作的任何提示？
在下面的示例中,如果您在名称字段中输入"John",而在optionalval字段中没有输入任何内容,则形成的json是{name:John}.但我希望它像{name:'John',optionalval:''}.这是"创建新"表单的情况,其中可选字段可能没有任何值.但是,无论是否有值,都需要发送字段.
<!doctype html>
<html>
<head>
<title>Serialize Test</title>
<script src="js/angular.min.js"></script>
<script>
var app = angular.module('myApp', []);
app.controller('myCtrl', function($scope, $http) {
$scope.addnew = function(){
$http.put('/newthing.json', JSON.stringify($scope.item))
.success(function(){
})
.error(function(){
});
};
});
</script>
</head>
<body ng-app="myApp" ng-controller="myCtrl">
<label for="name">Name:</label>
<input type="text" id="name" ng-model="item.name" required>
<label for="optionalval">Don't put anything here:</label>
<input type="text" id="optoinalval" ng-model="item.optionalval">
<button ng-click="addnew()">Serialize</button>
</body>
</html>一种方法是使用空字符串预初始化模型.例如,通过在控制器中设置模型值:
$scope.item = {optionalval: ''};
这是一个演示:
var app = angular.module('myApp', []);
app.controller('myCtrl', function($scope, $http) {
$scope.item = {optionalval: ''};
$scope.addnew = function() {
$http.put('/newthing.json', JSON.stringify($scope.item))
.success(function() {})
.error(function() {});
};
});<script src="https://code.angularjs.org/1.4.3/angular.js"></script>
<div ng-app="myApp" ng-controller="myCtrl">
<label for="name">Name:</label>
<input type="text" id="name" ng-model="item.name" required>
<label for="optionalval">Don't put anything here:</label>
<input type="text" id="optionalval" ng-model="item.optionalval">
<button ng-click="addnew()">Serialize</button>
<pre>{{item | json}}
</div>