0

有谁知道我为什么会遇到以下问题：

未捕获的错误：[$injector:modulerr] http://errors.angularjs.org/1.3.6/$injector/modulerr?p0=helloKinveyApp&p1=E...ogleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.3.6%2Fangular。 min.js%3A17%3A350)angular.min.js:6 (匿名函数)angular.min.js:36 (匿名函数)angular.min.js:7 rangular.min.js:35 gangular.min.js: 38 Obangular.min.js:17 dangular.min.js:18 scangular.min.js:16 Hdangular.min.js:249（匿名函数）angular.min.js:163 aangular.min.js:32 c

var app = angular.module("helloKinveyApp", ["ngRoute", "kinvey"]);

app.run(["$kinvey", function($kinvey) {
    var promise = $kinvey.init({
       appKey: clientParams.appKey
        , appSecret: clientParams.appSecret
    });

    promise.then(function() {
        console.log("Kinvey init with success");
    }, function(errorCallback) {
        console.log("Kinvey init with error: " + JSON.stringify(errorCallback));
    });
}]);

<!DOCTYPE html>
<html data-ng-app="helloKinveyApp">
    <head>
        <title>Hello Kinvey Lesson 1</title>
    </head>
    <body>
        <div data-ng-view="" />
        <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.3.6/angular.min.js"></script>
        <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.3.6/angular-route.min.js"></script>
        <script src="https://da189i1jfloii.cloudfront.net/js/kinvey-angular-1.1.4.min.js"></script>
    </body>
</html>
4

1 回答 1

0

如果您的代码的结构类似于以下，它应该可以工作（即您应该得到一个模块加载错误）：

...
<body>
    ...
    <div data-ng-view="" />
    <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.3.6/angular.min.js">
    </script>
    <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.3.6/angular-route.min.js">
    </script>
    <script src="https://da189i1jfloii.cloudfront.net/js/kinvey-angular-1.1.4.min.js">
    </script>
    ...
    <script>
        var app = angular.module("helloKinveyApp", ["ngRoute", "kinvey"]);

        app.run(["$kinvey", function($kinvey) {
            var promise = $kinvey.init({
               appKey: clientParams.appKey,
               appSecret: clientParams.appSecret
            });

            promise.then(function() {
                console.log("Kinvey init with success");
            }, function(errorCallback) {
                console.log("Kinvey init with error: " + JSON.stringify(errorCallback));
            });
        }]);
    </script>
</body>
于 2014-12-11T09:18:48.717 回答