我遇到以下问题。将JSON解析为Angular 2对象
我有一个非常大的JSON字符串,它具有来自对象的所有变量。
对象:
export class User {
ssn:string;
userId:string;
firstName:string;
lastName:string;
middleName:string;
office:string;
role:string;
lockCode:string;
command:string;
street:string;
city:string;
position:string;
zip:string;
phone:string;
dsn:string;
fax:string;
email:string;
pwEffectiveDate:any;
pwVaildationDate:any;
fromDate:any;
toDate:any;
systemAccess:string;
dmType:string;
accessInfoEffectiveDate:any;
accessInfoEffectiveTo:any;
availableOffices: string[];
availbleRole:string[];
}
JSON:
@Injectable()
export class SearchService {
getData() :any[] { return [
{"snn": "26999935-7", "userId": "EVD404", "firstName": "Chaney", "lastName": "Vernon", "middleName": "X", "office": "ADURT", "role": "GC", "lockCode": "Q", "command": "5th Grp", "street": "953-1348 Faucibus Rd.", "city": "Bienne-lez-Happart", "position": "Developer", "zip": "76222", "phone": "233-969-1834", "dsn": "359-887-4719", "fax": "157-376-6377", "email": "[email protected]", "pwEffectiveDate": "13/03/17", "pwVaildationDate": "27/01/18", "fromDate": "10/11/17", "toDate": "21/12/17", "systemAccess": "GC", "dmType": "XJ", "accessInfoEffectiveDate": "26/12/2016", "accessInfoEffectiveTo": "06/06/2016", "availableOffices": "UUU", "availbleRole": "GC"},
{"snn": "43250813-7", "userId": "NSB626", "firstName": "Addison", "lastName": "Vernon", "middleName": "X", "office": "AUTRO", "role": "GC", "lockCode": "O", "command": "11th ACR", "street": "Ap #904-5416 Semper, Road", "city": "s Herenelderen", "position": "Developer", "zip": "26457", "phone": "890-600-3144", "dsn": "679-122-1054", "fax": "913-500-7495", "email": "[email protected]", "pwEffectiveDate": "11/06/17", "pwVaildationDate": "01/03/17", "fromDate": "05/08/17", "toDate": "29/09/16", "systemAccess": "LIMIT", "dmType": "NB", "accessInfoEffectiveDate": "19/04/2017", "accessInfoEffectiveTo": "13/04/2016", "availableOffices": "LLL", "availbleRole": "USER"},
然后,我希望能够调用像下面的方法时,我通过我的服务到组件:
getUserByLastName(lastName):User[]{
let temp: User[]=[];
for(let d of this.data) {
if(d.lastName == lastName){
temp.push(d);
}
}
return temp;
}
我已尝试JSON.parse,但没有奏效。我尝试了其他一些东西,但没有一个似乎坚持。
--------------------------------- Update 1 ----------- -----------------
它引起了我的注意,我应该使用Observable。以下是我在试图实现这一点,但它是目前没有工作:
getUserBySSN():Observable<User[]> {
return this._http.get(this._url)
.map((response: Response) => response.json())
.do(data => console.log("User data" + JSON.stringify(data)))
.catch(this.handleError);
}
private handleError(error: Response) {
console.log(error);
return Observable.throw(error.json().error || 'Internal Server error');
}
我创建了一个JSON文件和设置可变url
作为其路径。不过我正在给以下错误:
The type argument for type parameter 'T' cannot be inferred from the
usage. Consider specifying the type arguments explicitly. Type
argument candidate 'Response' is not a valid type argument because it
is not a supertype of candidate 'Response'. Types of property 'type'
are incompatible. Type 'string' is not assignable to type 'ResponseType'
有人建议我用.map((response: Response) => <User[]> response.json())
但我不会允许将其转换。
经过进一步的研究,我发现这是最好的方法,并试图让它起作用,所以稍后我可以使用它来对数据库进行实际的HTTP调用。
你可以给使用JSON.parse的一个例子,它不工作? – snorkpete
你为什么不在服务中使用Observable? – Aravind
如何以及从哪里获取服务中的数据? – OmarIlias