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JavaScript jQuery Basics (2014) Creating a Simple Drawing Application Perform: Part 3

Rishit Shah
Rishit Shah
4,975 Points
Posted by Rishit Shah
Rishit Shah
4,975 Points

Unexpected end of input

I am getting this error when i try to run my code "Unexpected end of input". here is my code 

var color = $('.selected').css("background-color");

//when clicking on classes
$('.controls li').click(function(){
  //deselect sibling elements
    $(this).siblings().removeClass('selected');
  //select clicked one
  $(this).addClass('selected');
  //cache current color
  color = $('this').css("background-color");
});
// whe new color is clicked
$('#revealColorSelect').click(function(){

  changeColor();
  $('#colorSelect').toggle();
//})
  //show or hide select button

//when color sliders change
  //update the new color span
function changeColor(){
  var r =$("#red").val();
  var g =$("#green").val();
  var b =$("#blue").val();
//  
  $('#newColor').css( "background-color" , "rgb("+ r +  ","  + g + "," + b +")");

 }


$("input[type=range]").on("input", changeColor);

  $("#addNewColor").click(function() {
var $newColor = $("<li></li>");
  $newColor.css("background-color", $("#newColor").css("background-color") );
$(".controls ul").append($newColor);
  $newColor.click();

});
//On mouse events on the canvas
  //Draw lines
LaVaughn Haynes
LaVaughn Haynes
12,397 Points
LaVaughn Haynes
LaVaughn Haynes
12,397 Points

I edited your question to display the code correctly. Don't forget to add javascript after the 3 ticks

```javascript

1 Answer

LaVaughn Haynes
LaVaughn Haynes
12,397 Points
LaVaughn Haynes
LaVaughn Haynes
12,397 Points

At a glance it looks like you commented out the end of your click method. Also add a semicolon.

$('#revealColorSelect').click(function(){

  changeColor();
  $('#colorSelect').toggle();
//})
Rishit Shah
Rishit Shah
4,975 Points
Rishit Shah
Rishit Shah
4,975 Points

Thank you. That solved my problem