Question

我有两个对象数组，如下所示：

    var b = [{"from":2,"to":7,"id":1},{"from":3,"to":9,"id":2},{"from":2,"to":7,"id":3}]
    var c = [{"from":3,"to":9,"id":2,"style":""},{"from":2,"to":7,"id":3,"style":"dash-line"},{"from":4,"to":2,"id":4,"style":"dash-line"},{"from":2,"to":4,"id":5,"style":""},{"from":4,"to":2,"id":6,"style":"dash-line"}];

我想要的是一个来自上面两个的对象数组，它具有唯一的＆＃34;来自＆＃34; ＆＃34;至＆＃34;和＆＃34;风格＆＃34;应该是＆＃34;＆＃34;（空白）或undefined。具有唯一ID。

即

    output = [{"from":2,"to":7,"id":0},{"from":3,"to":9,"id":1},{"from":2,"to":4,"id":6,"style":""}]

我能够得到它，如下面的代码所示，但我觉得代码可以优化或有更好的方法来做到这一点。请帮帮....谢谢。

  var b = [{"from":2,"to":7,"id":1},{"from":3,"to":9,"id":2},{"from":2,"to":7,"id":3}]
    var c = [{"from":3,"to":9,"id":2,"style":""},{"from":2,"to":7,"id":3,"style":"dash-line"},{"from":4,"to":2,"id":4,"style":"dash-line"},{"from":2,"to":4,"id":5,"style":""},{"from":4,"to":2,"id":6,"style":"dash-line"}];
    var a = b.concat(c);
findUniQue(a);

function findUniQue(a){
    var tempArr =[];
    for(var i =0;i<a.length;i++){
        if(a[i].style == undefined || a[i].style != 'dash-line' ){
            var count = 0;
            if(tempArr.length>0){
             for(var j =0;j<tempArr.length;j++){
                if((a[i].from == tempArr[j].from)&&(a[i].to == tempArr[j].to)){
                    count--;
                    break;  
                }
                else{
                    count++;
                }
                if(count == tempArr.length){
                    a[i].id = i;
                    tempArr.push(a[i]);
                }
                }
             }
             else{
                a[i].id = i;
                tempArr.push(a[i]);
             }
        }
    }
    console.dir(tempArr);
}

Answer 1

function removeduplicate(){
    var array = [{id:5},{id:8},{id:9},{id:10},{id:5},{id:8}];
    var size = array.length;
    for (var i = 0; i < size - 1; i++) {
        for (var j = i + 1; j < size; j++) {
            if (array[j].id !== array[i].id)
                continue;
            array.splice(j,1);
            j--;
            size--;
        } // for j
    } // for i
    console.log(array);
}

基于两个键，即来自javascript的from和to，从两个对象数组中找到唯一值

1 个答案: