解释 2 个单独的 JSON 返回对象

Question

我在登录后使用 oauth 后端生成有效令牌，但我返回 2 个单独的 json 对象，因为当我创建令牌并从我的数据库返回用户数据时，我无法将它们与 PHP 链接，所以我得到一个像这样的字符串：

{"id":"1","email":"test@test.com","username":"user","firstname":"john","lastname":"doe"} {"access_token":"d0f8...2a","expires_in":3600,"token_type":"Bearer","scope":null}

如您所见，我得到了 2 个单独的 JSON 对象，它们也都在一行上，所以我不能用新行分割，而且我觉得分割这些数据只会弄得一团糟

在 Angular 2 中，从中获取 2 个 JSON 对象的最佳方法是什么？

这是我返回数据的方式：

login() {
        this.loading = true;
        this.authenticationService.login(this.model.username, this.model.password).subscribe(
            (data) => {
                if(data) {
                    console.log(data);
                }
                /*this.authenticationService.token = (data.token)
                if(data.token) {
                    localStorage.setItem('currentUser', data.token);
                    this.loading = false;
                    this.router.navigate(['/user']);
                }else{

                }*/
            },  //changed
            (err)=> {
                console.log(err);
                this.error = 'Username or password is incorrect';
                this.loading = false;},
            ()=>console.log("Done")
        );
    }

Answer 1

您应该将这 2 个对象包装到数组或对象中，然后从您的服务器返回，这样您将获得有效的 JSON。

返回如下内容，而不是您的有效负载：

{
  "profile": {"id":"1","email":"test@test.com","username":"user","firstname":"john","lastname":"doe"},
  "auth": {"access_token":"d0f8...2a","expires_in":3600,"token_type":"Bearer","scope":null}
}

在你的 PHP 中有这样的东西：

// your current object representation of the JSON you are sending
$profile = ...;
$auth = ...;

$wrapper = new \stdClass;
$wrapper->profile = $profile;
$wrapper->auth = $auth;

$json = json_encode($wrapper);

// and just return the $json variable instead of concatenating those 2

要解决您在 cmets 中提到的问题，您可以使用输出缓冲：

ob_start(); // start gathering the output of `echo`
$server->handleTokenRequest(OAuth2\Request::createFromGlobal‌​s())->send(); // call the method that echoes
$auth = ob_get_clean();

// now you have the JSON string in $auth
$wrapper = new \stdClass;
$wrapper->profile = $profile;
$wrapper->auth = json_decode($auth); // use `json_decode` to parse it to object

$json = json_encode($wrapper); // create the JSON again

最后一个注释 - 这绝对是正确的方法，但要回答您关于解析两个对象的原始问题 - 您可以简单地拆分 }{ 上的字符串，除非您连接更多，否则不会出现在有效 JSON 中其中。

var doubleJSON = '{"id":"1","email":"test@test.com","username":"user","firstname":"john","lastname":"doe"} {"access_token":"d0f8...2a","expires_in":3600,"token_type":"Bearer","scope":null}';

var temp = doubleJSON.split(/\}\W+\{/); // use regexp, that allows white space character between `}` and `{`
var profile = JSON.parse(temp[0] + '}'); // add the `}` character to the end, that is missing because of the split call
var auth = JSON.parse('{' + temp[1]); // add the `{` character to the front, that is missing because of the split call

console.log(profile);
console.log(auth);

Answer 2

如果你只是可以在代表两个对象的两个字符串之间放一个额外的字符（对于鸡蛋：';'），你可以通过这个字符拆分来解析它：

var str = '{"id":"1","email":"test@test.com","username":"user","firstname":"john","lastname":"doe"};{"access_token":"d0f8...2a","expires_in":3600,"token_type":"Bearer","scope":null}';

var str = '[' + str.split(';').join(',') + ']';

var objects = JSON.parse(str);
var obj1 = objects[0];
var obj2 = objects[1];
console.log("obj1: ", obj1);
console.log("obj2: ", obj2);

如果你想玩 ES6，它会更短：

var str = '{"id":"1","email":"test@test.com","username":"user","firstname":"john","lastname":"doe"};{"access_token":"d0f8...2a","expires_in":3600,"token_type":"Bearer","scope":null}';
var [obj1, obj2] = str.split(";").map(el => JSON.parse(el));
console.log("obj1: ", obj1);
console.log("obj2: ", obj2);