TypeScript: Property does not exist on type '{}'

Question

I am using Visual Studio 2013 fully patched. I am trying to use JQuery, JQueryUI and JSRender. I am also trying to use TypeScript. In the ts file I\'m getting an error as follo

Answer 1

I suggest the following change

let propertyName =  {} as any;

Answer 2

myFunction(
        contextParamers : {
            param1: any,
            param2: string
            param3: string          
        }){
          contextParamers.param1 = contextParamers.param1+ 'canChange';
          //contextParamers.param4 = "CannotChange";
          var contextParamers2 : any = contextParamers;// lost the typescript on the new object of type any
          contextParamers2.param4 =  'canChange';
          return contextParamers2;
      }

Answer 3

You can assign the any type to the object:

let bar: any = {};
bar.foo = "foobar"; 

Answer 4

Access the field with array notation to avoid strict type checking on single field:

data['propertyName']; //will work even if data has not declared propertyName

---

Alternative way is (un)cast the variable for single access:

(<any>data).propertyName;//access propertyName like if data has no type

The first is shorter, the second is more explicit about type (un)casting

---

You can also totally disable type checking on all variable fields:

let untypedVariable:any= <any>{}; //disable type checking while declaring the variable
untypedVariable.propertyName = anyValue; //any field in untypedVariable is assignable and readable without type checking

Note: This would be more dangerous than avoid type checking just for a single field access, since all consecutive accesses on all fields are untyped

Answer 5

let propertyName= data['propertyName'];

Answer 6

When you write the following line of code in TypeScript:

var SUCSS = {};

The type of SUCSS is inferred from the assignment (i.e. it is an empty object type).

You then go on to add a property to this type a few lines later:

SUCSS.fadeDiv = //...

And the compiler warns you that there is no property named fadeDiv on the SUCSS object (this kind of warning often helps you to catch a typo).

You can either... fix it by specifying the type of SUCSS (although this will prevent you from assigning {}, which doesn't satisfy the type you want):

var SUCSS : {fadeDiv: () => void;};

Or by assigning the full value in the first place and let TypeScript infer the types:

var SUCSS = {
    fadeDiv: function () {
        // Simplified version
        alert('Called my func');
    }
};