Question

$ .each对象如何索引 - 1？

因为obj[i - 1].day无效

＆＃13;

var checkin_status = [{
    "startdate": "2015-01-08",
    "totaldays": "4",
    "roadmap": [{
        "gifttype": "stars",
        "quantity": 100,
        "day": 1
    }, {
        "gifttype": "stars",
        "quantity": 500,
        "day": 3
    }, {
        "gifttype": "stars",
        "quantity": 1000,
        "day": 10
    }, {
        "gifttype": "stars",
        "quantity": 1200,
        "day": 20
    }, {
        "gifttype": "stars",
        "quantity": 2200,
        "day": 30
    }]           
 }];
    
var checkin_info = checkin_status[0].roadmap;
    
$.each(checkin_info, function (i, obj) {
    var sum_day = obj[i - 1].day + obj.day;
    var middle_day = sum_day / 2;
    console.log(middle_day);
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

sum_day输出应为[4,13,30,50]
分为2，新输出应为[2,6.5,15,50]这是[1,3,10,20,30]

Answer 1

你根本不需要each，你将一个数组映射到另一个数组，我将它分成下面的两个操作来演示，但它不需要。这一切都可以在一个map中完成：

＆＃13;

var checkin_status = [{
    "startdate": "2015-01-08",
    "totaldays": "4",
    "roadmap": [{
        "gifttype": "stars",
        "quantity": 100,
        "day": 1
    }, {
        "gifttype": "stars",
        "quantity": 500,
        "day": 3
    }, {
        "gifttype": "stars",
        "quantity": 1000,
        "day": 10
    }, {
        "gifttype": "stars",
        "quantity": 1200,
        "day": 20
    }, {
        "gifttype": "stars",
        "quantity": 2200,
        "day": 30
    }]           
 }];

var checkin_info = checkin_status[0].roadmap;
var result1 = checkin_info.slice(1,checkin_info.length).map(function(e,i){
    return checkin_info[i].day + e.day;
});
console.log(result1); // logs [4,13,30,50]

var result2 = result1.map(function(e){
    return e/2;  
});
console.log(result2); // logs [2,6.5,15,50]

＆＃13;

这当然可以缩短为

var checkin_info = checkin_status[0].roadmap;
var result = checkin_info.slice(1,checkin_info.length).map(function(e,i){
    return (checkin_info[i].day + e.day)/2;
});

如果你真的必须拥有$.each，你可以使用一些相同的原则 - 切掉第一个元素：

$.each(checkin_info.slice(1,checkin_info.length), function (i, obj) {
    var sum_day = checkin_info[i].day + obj.day;
    var middle_day = sum_day / 2;
    console.log(middle_day);
});

Answer 2

特殊情况第一个条目，如下：

var checkin_status = [{
    "startdate": "2015-01-08",
    "totaldays": "4",
    "roadmap": [{
        "gifttype": "stars",
        "quantity": 100,
        "day": 1
    }, {
        "gifttype": "stars",
        "quantity": 500,
        "day": 3
    }, {
        "gifttype": "stars",
        "quantity": 1000,
        "day": 10
    }, {
        "gifttype": "stars",
        "quantity": 1200,
        "day": 20
    }, {
        "gifttype": "stars",
        "quantity": 2200,
        "day": 30
    }]           
 }];

var checkin_info = checkin_status[0].roadmap;

    $.each(checkin_info, function (i, obj) 
    {

        var sum_day = 0;
        if(i==0) 
        {
                  //this is the first entry, so we can't look back to the previous day.   
                  sum_day = obj.day; 
        }
        else
        {
                   //you were calling obj[i-1] here, which wasn't going to work.
                   //i presume you wanted the previous day here.
                   sum_day = checkin_info[i - 1].day + obj.day;
        }


        var middle_day = sum_day / 2;
        console.log(middle_day);
    });

Answer 3

$。每个都不利于性能，因为每个交互都会调用函数。最好在vanilla js中使用for循环。

var checkin_status = [
{"startdate":"2015-01-08",
"totaldays":"4",
"roadmap":[
            {"gifttype":"stars","quantity":100,"day":1},
            {"gifttype":"stars","quantity":500,"day":3},
            {"gifttype":"stars","quantity":1000,"day":10},
            {"gifttype":"stars","quantity":1200,"day":20},
            {"gifttype":"stars","quantity":2200,"day":30},
          ]

}];

    var checkin_info = checkin_status[0].roadmap,
        len = checkin_info.length,
        i,
        sum_day = [],
        middle_day = [],
        sum = 0;

    for(i=0; i< (len -1); i++) {
        sum = checkin_info[i].day + checkin_info[i + 1].day;
        sum_day.push(sum);
        middle_day.push(sum/2);
    }
    console.log(sum_day)
    console.log(middle_day)

jquery $ .each数组索引-1

3 个答案: