Question

假设我们有两个阵列说:

 A [] => 1 2 3 4 5

 B [] => 1 2 7 4 5

jQuery 中存在任何方法,在此情况下给出两个数组的不匹配值 :

 Result [] => 3 7

Answer 1

这里的“强力”工作演示: http://jsfidd.net/mbKfT/

good read http://api.jquery.com/jQuery.inArray/

这里使用 inArray 来检查元素是否添加到 interect 阵列中。

休息演示会起诉任何疑问 :)

<强度 > 代码

var a1 = [1,2,3,4,5];
var a2 = [1,2,7,4,5];
var intersect = [];

$.each(a1, function(i, a1val) {

    if ($.inArray(a1val, a2) === -1) {   
        intersect.push(a1val);
    }
});

$.each(a2, function(i, a1val) {

    if ($.inArray(a1val, a1) === -1) {           
        intersect.push(a1val);
    }
});
$("div").text(intersect);
alert(intersect + " -- " + matches);

Answer 2

答复:无。

解决方案 : 使用标准的 javascript 环。

var nomatches = [];
for (var i=Math.min(A.length, B.length); i-->0;) {
   if (A[i]!=B[i]) {
       nomatches.push(A[i]);
       nomatches.push(B[i]);
   }
}
// do what you want with remaining items if A.length != B.length

如果,如 Rory 所假设的那样, 您不想匹配数组, 但符合逻辑, 您可以做到这一点 :

 var nomatches = [];
var setA = {};
var setB = {};
for (var i=A.length; i-->0;) setA[A[i]]=1;
for (var i=B.length; i-->0;) setB[B[i]]=1;
for (var i=A.length; i-->0;) {
    if (!setB[A[i]]) nomatches.push(A[i]);
}
for (var i=B.length; i-->0;) {
    if (!setA[V[i]]) nomatches.push(B[i]);
}

Answer 3

var nomatch = [], Bcopy = B.slice(0);
for (var i = 0, j; i < A.length; i++) {
    j = Bcopy.indexOf(A[i]);
    if (j === -1) nomatch.push(A[i]);
    else Bcopy.splice(j, 1);
}
nomatch.push.apply(nomatch, Bcopy);

注:

This code supposes that items in A and B are unique.
indexOf for arrays must be emulated in IE8 and previous versions.

Answer 4

jQuery.inAray () 将帮助:

var a = [1,2,3,4,5], b=[1,2,7,4,5];

var ret = 
a.filter(function(el) {
  return $.inArray(el, b) === -1;
}).concat(
b.filter(function(el) {
  return $.inArray(el, a) === -1;    
})
);
console.log(ret);

http://jsfidle.net/ATAUq/" rel="not follow" >demo 。

PS: 或者你只要使用 b.indexof(el)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Answer 5

function getUnique(A, B){
  var res = [];
  $.grep(A, function(element) {
    if($.inArray(element, B) == -1) res.push(element)        
  });
  $.grep(B, function(element) {
    if($.inArray(element, A) == -1) res.push(element);    
  });
  return res;
}

使用 :

var A = [1,2,3,4,5],
    B = [1,2,3,5,7];

getUnique(A, B);

http://jsfidd.net/ZjqyU/1/" rel=“notfollow” >DEMO

Answer 6

这是现代浏览器的另一种解决办法(单线浏览器, 是! ) :

var a = [1, 2, 3, 4, 5];
var b = [1, 2, 7, 4, 5];

var result = a.concat(b).filter(function(el, i) {
    return (i < a.length ? b : a).indexOf(el) == -1;
});

DEMO: http://jsfiddle.net/6Na36/

如果您也想要保存索引检查, 您可以使用此变量 :

var a = [1, 2, 3, 4, 5];
var b = [1, 2, 7, 4, 5];

var result = a.concat(b).filter(function(el, i, c) {
    return el != c[i < a.length ? i + a.length : i - a.length];
});

<强力 > DEMO: http://jsfiddle.net/6Na36/1/

注意,两个变量都与不同大小的阵列成功合作。

友情链接