我在点击按钮时调用ajax函数它返回json数据并且我将数据传递给main.js脚本文件(控制器)它获取数据并将数据绑定到ng-grid,问题这里是我把ng-grid放在from标签中,它没有显示数据
<script type="text/javascript">
$(document).ready(function () {
$("#mybutton").click(function () {
var scope = angular.element(document.getElementById("wrap")).scope(); // to get access all the varibales defined in the contoller
scope.$apply(function () {
$.ajax({
type: "POST",
url: "Website/Nggrid.asmx/GetDataForNgGrid",
success: function (result) {
// console.log(result);
var fd = JSON.parse(result); //parsing the json string
scope.updateMessage(fd);
alert("hi");
},
error: function (xmlhttprequest, Status, thrownError) {
alert(thrownError.toString());
alert(thrownError);
}
});
});
});
});
</script>
这是我在用户点击按钮时调用的功能
<body ng-controller="MyCtrl">
<%--<form id="form1" runat="server">--%>
<div id="wrap" class="gridStyle" ng-grid="gridOptions">
</div>
<button id="mybutton">
Try it</button>
<%-- </form>--%>
</body>
这是main.js
var app = angular.module('myApp', ['ngGrid']);
app.controller('MyCtrl', function ($scope) {
$scope.myData = [];
$scope.updateMessage = function (_s) {
$scope.myData = _s;
// $scope.Enable = true;
};
$scope.gridOptions = {
data: 'myData',
columnDefs: [
{ field: 'Status', displayName: 'Status', width: "*" }
]
};
});
我的问题是,当我将ng-grid放在from标签中时它不会显示数据,请在此提出建议
<form id="form1" runat="server">
<div id="wrap" class="gridStyle" ng-grid="gridOptions">
</div>
<button id="mybutton">
Try it</button>
</form>