var arr1=["a","b","b","c","c","d"];
var arr2=["b","c",];
arr1 has duplicate values which are listed in arr2. Now I want to remove the duplicate values from arr1 using arr2. Is it possible?
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stackoverflow.com/questions/13486479/javascript-array-uniqueAsif– Asif2013-09-26 06:00:49 +00:00Commented Sep 26, 2013 at 6:00
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Visit this link stackoverflow.com/questions/14930516/…user2775111– user27751112013-09-26 06:06:36 +00:00Commented Sep 26, 2013 at 6:06
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2 Answers
I would use the Array's .filter() method.
arr1 = arr1.filter(function (val) {
return arr2.indexOf(val) === -1;
});
For IE8 or earlier, this code should work:
arr1 = arr1.filter(function (val) {
var i;
for (i = 0; i < arr2.length; i += 1) {
if ( val === arr2[i] ) {
return false;
}
}
return true;
});
6 Comments
Pearl
Property or Method indexOf is undefined...
Joe Simmons
Works fine for me on
Firefox 23.0.1. What does your browser say if you do this? alert(Array.prototype.indexOf); ? If it alerts a function, my code should work. If not... that's really weird because filter and indexOf were added in the same version of JavaScript.
Deepak Ingole
The indexOf() method is not supported in Internet Explorer 8 and earlier.
joydesigner
try this if the browser is <IE 9. if(!Array.indexOf){ Array.prototype.indexOf = function(obj){ for(var i=0; i<this.length; i++){ if(this[i]==obj){ return i; } } return -1; } }
Joe Simmons
Ah... then why is
.filter() supported? Weird. I updated my answer for the asker, for IE8 and earlier. |
you can use directly to remove duplicates from array1 as
$(document).ready(function(){
var arr1=["a","b","b","c","c","d"];
var arr2=[];
$.each(arr1, function(i,el){
if($.inArray(el, arr2) === -1)
arr2.push(el);
});
alert(arr2);
});
you can observe here..
