Develop a HTTP interceptor in the form of a class

Question

Develop a HTTP interceptor in the form of a class

I am currently grappling with the challenge of writing an angular http interceptor in plain TypeScript. The JavaScript code that I am attempting to convert is as follows:

.config(['$httpProvider', function ($httpProvider) {

    var interceptor = ['$rootScope', '$q', 'httpBuffer', function ($rootScope, $q, httpBuffer) {
        function success(response) {
            return response;
        }

        function error(response) {
            if (response.status === 401 && !response.config.ignoreAuthModule) {
                var deferred = $q.defer();
                httpBuffer.append(response.config, deferred);
                $rootScope.$broadcast('event:auth-loginRequired', response);
                return deferred.promise;
            }
            // otherwise, default behaviour
            return $q.reject(response);
        }

        return function (promise) {
            return promise.then(success, error);
        };

    }];
    $httpProvider.responseInterceptors.push(interceptor);
}])

The initial sections are straightforward - create a class with a constructor that accepts the three dependencies $rootScope, $q and httpBuffer. Also include two private methods success and response.

class MyInterceptorClass {
    constructor(private $rootScope: ng.IRootScopeService, private $q: ng.IQService, private httpBuffer: HttpBuffer) {
    }

    private success(response: ng.IHttpPromiseCallbackArg<any>): ng.IHttpPromiseCallbackArg<any> {
        return response;
    }

    private error(response: ng.IHttpPromiseCallbackArg<any>): ng.IHttpPromiseCallbackArg<any> {
        if (response.status == 401 && !((<any>response.config).ignoreAuthModule)) {
            var deferred = this.$q.defer();
            this.httpBuffer.append(response.config, deferred);
            this.$rootScope.$broadcast('event:auth-loginRequired', response);

            return deferred.promise;
        }

        // Otherwise, default behavior
        return this.$q.reject(response);
    }
}

Additionally, ensure the registration of the interceptor is clear:

.config(['$httpProvider', ($httpProvider: ng.IHttpProvider)=> {
    $httpProvider.responseInterceptors.push(['$rootScope', '$q', 'httpBuffer', MyInterceptorClass]);
}]);

The part causing me difficulty is the final section of the original JavaScript code - specifically the return value of an anonymous function. How can I replicate this in TypeScript? It seems like this would be a nameless method in TypeScript, but unfortunately that is not supported.

javascript angularjs typescript

Answer 1

Answer №1

Here are the configuration settings:

    .config(['$httpProvider', function ($httpProvider: ng.IHttpProvider) {
        $httpProvider.interceptors.push(AuthenticationInterceptor.Factory);
    }])

Your class structure should be like this:

module Common.Security {
'use strict';

export class AuthenticationInterceptor {

    public static Factory($q: ng.IQService) {
        return new AuthenticationInterceptor($q);
    }

    constructor(private $q: ng.IQService) {
    }

    //The method name must be "response" as per http://docs.angularjs.org/api/ng/service/$http
    public response(response) {
        console.log(response);
        return response || this.$q.when(response);
    }

    public responseError(rejection) {
        console.log(rejection.status);
        if (rejection.status === 401) {
        }
        // If not 401, reject the promise
        return this.$q.reject(rejection);
    }
}

Answer 2

Here are the configuration settings:

    .config(['$httpProvider', function ($httpProvider: ng.IHttpProvider) {
        $httpProvider.interceptors.push(AuthenticationInterceptor.Factory);
    }])

Your class structure should be like this:

module Common.Security {
'use strict';

export class AuthenticationInterceptor {

    public static Factory($q: ng.IQService) {
        return new AuthenticationInterceptor($q);
    }

    constructor(private $q: ng.IQService) {
    }

    //The method name must be "response" as per http://docs.angularjs.org/api/ng/service/$http
    public response(response) {
        console.log(response);
        return response || this.$q.when(response);
    }

    public responseError(rejection) {
        console.log(rejection.status);
        if (rejection.status === 401) {
        }
        // If not 401, reject the promise
        return this.$q.reject(rejection);
    }
}

Answer 3

Answer №2

Here is an example of how to write an HTTP interceptor as a class:

module services {

export class ExampleHttpInterceptor {
    public $log:ng.ILogService;
    public $injector:ng.auto.IInjectorService;

    public responseError = (_rejection)=>{
       //handle error logic goes here
    };
    public request = (config)=>{
        config.withCredentials = true;
        return config;
    };
    public requestError = (rejection)=>{
        var $q:ng.IQService = this.$injector.get("$q");
        this.$log.log("requestError", rejection);
        return $q.reject(rejection);
    };
    static $inject = ['$injector', '$rootScope', '$q', '$window'];
    constructor($injector:ng.auto.IInjectorService, public $rootScope, public $q, public $window){
        this.$log = $injector.get('$log');
        this.$injector = $injector;
    }
  }
}

To register the interceptor in your Angular module:

 var app =angular.module('myApp',[]);
 app.config(['$httpProvider', (_$httpProvider:ng.IHttpProvider)=>{
    _$httpProvider.interceptors.push('ExampleHttpInterceptor');

}]);

Answer 4

Here is an example of how to write an HTTP interceptor as a class:

module services {

export class ExampleHttpInterceptor {
    public $log:ng.ILogService;
    public $injector:ng.auto.IInjectorService;

    public responseError = (_rejection)=>{
       //handle error logic goes here
    };
    public request = (config)=>{
        config.withCredentials = true;
        return config;
    };
    public requestError = (rejection)=>{
        var $q:ng.IQService = this.$injector.get("$q");
        this.$log.log("requestError", rejection);
        return $q.reject(rejection);
    };
    static $inject = ['$injector', '$rootScope', '$q', '$window'];
    constructor($injector:ng.auto.IInjectorService, public $rootScope, public $q, public $window){
        this.$log = $injector.get('$log');
        this.$injector = $injector;
    }
  }
}

To register the interceptor in your Angular module:

 var app =angular.module('myApp',[]);
 app.config(['$httpProvider', (_$httpProvider:ng.IHttpProvider)=>{
    _$httpProvider.interceptors.push('ExampleHttpInterceptor');

}]);

Answer 5

Answer №3

It is important to implement the following in your code: function handleResponse = (response) => {}

function handleResponseError = (rejection) => { }

Failure to do so may result in 'this' being undefined. To gain a better understanding of why this is necessary, watch this video: https://www.example.com/video123

Answer 6

It is important to implement the following in your code: function handleResponse = (response) => {}

function handleResponseError = (rejection) => { }

Failure to do so may result in 'this' being undefined. To gain a better understanding of why this is necessary, watch this video: https://www.example.com/video123

Answer 7

Answer №4

Below are the configuration settings:

    .config(['$httpProvider', function ($httpProvider: ng.IHttpProvider) {
        $httpProvider.interceptors.push(AuthenticationInterceptor.Factory);
    }])

Your actual class structure should be like this:

module Common {
'use strict';

export class AuthenticationInterceptor {

    private static _instance: AuthenticationInterceptor;

    public static Factory($q: ng.IQService) {
        AuthenticationInterceptor._instance = new AuthenticationInterceptor($q);

        return AuthenticationInterceptor._instance;
    }

    constructor(private $q: ng.IQService) {
    }

    //The method name must be exactly "response" - http://docs.angularjs.org/api/ng/service/$http
    public response(response) {
        var self = Common.AuthenticationInterceptor._instance;

        console.log(response);
        return response || self.$q.when(response);
    }

    public responseError(rejection) {
        var self = Common.AuthenticationInterceptor._instance;

        console.log(rejection.status);
        if (rejection.status === 401) {
        }

        // Default behavior
        return self.$q.reject(rejection);
    }
}

It is important to retrieve the class instance with the full namespace because Angular may have 'this' undefined during callbacks.

Answer 8

Below are the configuration settings:

    .config(['$httpProvider', function ($httpProvider: ng.IHttpProvider) {
        $httpProvider.interceptors.push(AuthenticationInterceptor.Factory);
    }])

Your actual class structure should be like this:

module Common {
'use strict';

export class AuthenticationInterceptor {

    private static _instance: AuthenticationInterceptor;

    public static Factory($q: ng.IQService) {
        AuthenticationInterceptor._instance = new AuthenticationInterceptor($q);

        return AuthenticationInterceptor._instance;
    }

    constructor(private $q: ng.IQService) {
    }

    //The method name must be exactly "response" - http://docs.angularjs.org/api/ng/service/$http
    public response(response) {
        var self = Common.AuthenticationInterceptor._instance;

        console.log(response);
        return response || self.$q.when(response);
    }

    public responseError(rejection) {
        var self = Common.AuthenticationInterceptor._instance;

        console.log(rejection.status);
        if (rejection.status === 401) {
        }

        // Default behavior
        return self.$q.reject(rejection);
    }
}

It is important to retrieve the class instance with the full namespace because Angular may have 'this' undefined during callbacks.

Answer 9

Answer №5

Use the following code to register it by name:

$httpProvider.interceptors.push('myInterceptorClass');

Next, ensure that your class is also registered as a service:

yourApp.service('myInterceptorClass', MyInterceptorClass)

Answer 10

Use the following code to register it by name:

$httpProvider.interceptors.push('myInterceptorClass');

Next, ensure that your class is also registered as a service:

yourApp.service('myInterceptorClass', MyInterceptorClass)

Develop a HTTP interceptor in the form of a class

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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