How can I remove duplicate Objects from array of Objects in javascript?

Javascript selecting an array?

• I am creating a game that uses a function that checks whether an integer exists within an array. If it does, then the function will perform A otherwise it performs B. For example this is what I have. if(isitthere(image)){ //image is an integer. So image may ==1 I have several arrays (like this). array0= new Array(1,2); array1= new Array(3,4); array2= new Array(5,6); I am using a random number generator to select a number between 0-3. This is the function I am using to check whether image (integer) exists within the array. function isitthere(obj){ var num=Math.floor(Math.random()*3) var j=false; for(var hits=0;hits<array.length;hits++){ if(array[hits]==obj){ j=true; break; } } return j; } However, notice that I haven't stated the number after array in the FOR or IF loop. In other words, I have only said array and not array1 or array2 or array3. This is because I want to concatenate the word array with the random number created. So in effect, the function selects 1 of the possible 3 arrays. However this does not work! If I literally write array0 or array1 or array2, it works fine. But as soon as I write array+num it does not work and it always performs the else and skips the IF. Sorry if i havent explained myself well enough :/ but any solutions on how to add the random number to the word array so that the function selects a random array would be great!!! thank you

• Answer:

  You don't want to do that. Trust me. You want to create another array and put your three arrays in it. So it would work like this: var array0 = [1, 2]; // never use the "new Array" syntax var array1 = [3, 4]; var array2 = [5, 6]; var bigArray = [array0, array1, array2]; Then, in your function, you can do something like: var num = Math.floor(Math.random()*3); var arrayToUse = bigArray[num]; And then use arrayToUse in your loop. Edit: You can email me through my Yahoo!Answers profile. Just click on my name, and there's an Email link on my profile page.