我需要检查一个JavaScript数组,看看是否有任何重复的值。最简单的方法是什么?我只需要找到复制的值是什么——实际上我不需要它们的索引,也不需要它们复制了多少次。

我知道我可以循环遍历数组并检查所有其他值是否匹配,但似乎应该有更简单的方法。

类似的问题:

获取JavaScript数组中所有唯一的值(删除重复值)


当前回答

在这篇文章是有用的重复检查,如果你正在使用Jquery。

如何使用jquery在数组中找到重复项

var unique_values = {}; var list_of_values = []; $('input[name$="recordset"]').     each(function(item) {          if ( ! unique_values[item.value] ) {             unique_values[item.value] = true;             list_of_values.push(item.value);         } else {             // We have duplicate values!         }     });

其他回答

var array = ['a', 'b', 'c', 'a'];

function unique(array) {
    var unique_arr = [];
    array.forEach(function(i, e) {
        if (unique_arr.indexOf(i)===-1) unique_arr.push(i);
    });
    return unique_arr;
}
console.log(unique(array));

Var arr = [4,5,1,1,2,3,4,4,7,5,2,6,10,9]; Var sorted_arr = arr.sort(); Var len = arrr .length; Var结果= []; For (var I = 0;I < len;我+ +){ If (sorted_arr[i + 1] !== sorted_arr[i]) { results.push (sorted_arr[我]); } } document . write(结果);

function remove_dups(arrayName){
  var newArray = new Array();

  label:for(var i=0; i<arrayName.length; i++ ){  

     for(var j=0; j<newArray.length;j++ ){
       if(newArray[j]==arrayName[i]){
         continue label;
       }
     }

     newArray[newArray.length] = arrayName[i];

  }

  return newArray;
}

http://jsfiddle.net/vol7ron/gfJ28/

var arr  = ['hello','goodbye','foo','hello','foo','bar',1,2,3,4,5,6,7,8,9,0,1,2,3];
var hash = [];

// build hash
for (var n=arr.length; n--; ){
   if (typeof hash[arr[n]] === 'undefined') hash[arr[n]] = [];
   hash[arr[n]].push(n);
}


// work with compiled hash (not necessary)
var duplicates = [];
for (var key in hash){
    if (hash.hasOwnProperty(key) && hash[key].length > 1){
        duplicates.push(key);
    }
}    
alert(duplicates);

The result will be the hash array, which will contain both a unique set of values and the position of those values. So if there are 2 or more positions, we can determine that the value has a duplicate. Thus, every place hash[<value>].length > 1, signifies a duplicate. hash['hello'] will return [0,3] because 'hello' was found in node 0 and 3 in arr[]. Note: the length of [0,3] is what's used to determine if it was a duplicate. Using for(var key in hash){ if (hash.hasOwnProperty(key)){ alert(key); } } will alert each unique value.

我试图改善@swilliams的答案,这将返回一个没有重复的数组。

// arrays for testing
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];

// ascending order
var sorted_arr = arr.sort(function(a,b){return a-b;}); 

var arr_length = arr.length;
var results = [];
if(arr_length){
    if(arr_length == 1){
        results = arr;
    }else{
        for (var i = 0; i < arr.length - 1; i++) {
            if (sorted_arr[i + 1] != sorted_arr[i]) {
                results.push(sorted_arr[i]);
            }
            // for last element
            if (i == arr.length - 2){
                results.push(sorted_arr[i+1]);
            }
        }
    }
}

alert(results);