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 follows   Property  fadeDiv  does not exist on type        I think I have the correct references for JQuery  JQueryUI and JSRender for TypeScript  but from what I ve read this is looking like a d ts issue  There are no errors in JavaScript  but I don t want to have Visual Studio saying there are errors if I can help it   Both times fadeDiv is mentioned in the JavaScript there is a red line under it and both errors say the same thing as above       lt reference path  quot    scripts typings jquery jquery d ts quot    gt       lt reference path  quot    scripts typings jqueryui jqueryui d ts quot    gt       lt reference path  quot typings jsrender jsrender d ts quot    gt   var SUCSS          document  ready function         SUCSS fadeDiv         SUCSS fadeDiv   function      var mFadeText  number    function          var mFade    quot FadeText quot         This part actually retrieves the info for the fadediv       ajax           type   quot POST quot             url   quot  js General aspx  FadeDiv1 quot           url   quot  js sucss General aspx  FadeDivList quot             data   quot   iInput   quot    JSON stringify jInput     quot   quot           contentType   quot application json  charset utf-8 quot           dataType   quot json quot           error  function  xhr  status  error                   Show the error               alert xhr responseText                      success  function  msg                mFadeText   msg d Fade                 Replace the div s content with the page method s return              if  msg d FadeType    0     FadeDivType   List                 var template     templates  quot  theTmpl quot                    var htmlOutput   template render msg d                      quot  id  lblFadeDiv  quot   html htmlOutput                             else    FadeDivType   String                    quot  id  lblFadeDiv  quot   html msg d FadeDivString                                    complete  function                  if  mFadeText    0                       quot  id  lblFadeDiv  quot   fadeIn  slow   delay 5000  fadeOut  slow                                         For those who might read this later  the SUCSS is the namespace  In typescript it appears I would have wanted to do something like this    document  ready function          SUCSS fadeDiv        module SUCSS       export function fadeDiv            So the function is made public by use of the export and I could call the SUCSS fadeDiv to run on page load by calling it with the SUCSS fadeDiv     I hope that will be helpful

User · Answer

I suggest the following change  let propertyName       as any

User · Answer

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       gt  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

User · Answer

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     lt any gt 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   lt any gt       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

User · Answer

You can assign the any type to the object  let bar  any       bar foo    quot foobar quot

User · Answer

let propertyName  data  propertyName

User · Answer

Near the top of the file  you need to write var fadeDiv       instead of fadeDiv       so that the variable is actually declared   The error  Property  fadeDiv  does not exist on type        seems to be triggering on a line you haven t posted in your example  there is no access of a fadeDiv property anywhere in that snippet

User · Answer

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

[typescript] TypeScript: Property does not exist on type '{}'

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