Question

我有一个看起来像这样的网址

URL： '/重置/密码电子邮件=：USEREMAIL＆安培; reset_token =：passwordResetToken'

但尝试使用 $ stateParams.passwordResetToken 访问返回未定义

Answer 1

请你详细说明一下;在角结肠部分不需要'=：'

url:'/reset/password?email=userEmail&reset_token=passwordResetToken'

在控制器中你可以注入 $ stateParams 然后你就不会得到未定义的

Answer 2

您可以使用$location服务，创建此服务以解析AngularJS中的网址。

这应该有效

$location.search() // will return {email: 'userEmail', reset_token: 'passwordResetToken'}

你也可以使用

$location.search('userEmail')

Answer 3

通过查看您的网址，我认为您使用的是这样的路线：

public class WebService {

    private String connectToServer(String address, String requestMethod) {
        try {
            URL url = new URL(address);
            HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setRequestMethod(requestMethod);
            return inputStreamToString(httpURLConnection.getInputStream());
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    private String inputStreamToString(InputStream inputStream) {
        StringBuilder stringBuilder = new StringBuilder();
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
            String nextLine;
            while ((nextLine = reader.readLine()) != null) {
                stringBuilder.append(nextLine);
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return stringBuilder.toString();
    }
  line 23 : public  List<FoodModel> getFoods(){ <-----------Error is here
        String response = connectToServer("http://mtg1376.gigfa.com/api/food?day=All","GET");
        if(response != null){
            List<FoodModel> foodList = new ArrayList<);
            try{
                JSONObject mainObject = new JSONObject(response);
                JSONArray foodArray = mainObject.getJSONArray("value");
                for(int i = 0 ; i< foodArray.length();i++){
                    JSONObject foodObject = foodArray.getJSONObject(i);
                    FoodModel foodModel = new FoodModel();
                    foodModel.id = foodObject.getString("id");
                    foodModel.name = foodObject.getString("name");
                    foodModel.meal = foodObject.getString("meal");
                    foodModel.photo_id = foodObject.getString("photo_id");
                    foodModel.explanation = foodObject.getString("explanation");
                    foodModel.foodtime=foodObject.getString("foodtime");
                    foodList.add(foodModel);
                }
                return foodList;
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
        return null; }


}

然后您应该使用.state('something.resetPassword', { url: "/reset/password", params: { email: null, reset_token: null }, templateUrl: 'your_template.html' })作为：

来访问参数

$stateParams

希望这会对你有帮助！

Answer 4

url：'/ reset / password？：email＆amp;：reset_token＆amp ;: role'解决问题

如何从具有'？'符号的复杂网址中提取路径参数，'＆amp;'

4 个答案: