数组 forEach 在最后一次迭代中覆盖对象的值

Question

我正在遍历文件名数组、拆分名称并将数据存储在一个对象中。用于测试目的的两个文件名是相同的，除了周“数字”应该创建两个单独的周。问题是第一个条目被最后一次迭代覆盖了，所以我最终只得到了第 2 周的条目。

代码：

const planList = [
    'military_greekHero_achilles_week_1.htm',
    'military_greekHero_achilles_week_2.htm'
];

var _completePlan = {};

planList.forEach(_plan => {

// Pull data from the file name formated: target_series_title_overview/week_weekNum.htm
    let _planPieces = _plan.split('.')[0].split('_'),// Drop the .htm
        _planTarget = _planPieces[0],
        _planSeries = _planPieces[1],
        _planTitle = _planPieces[2],
        _planOverview = _planPieces[3],
        _planWeek = _planPieces[4];

    _planOverview = _planOverview == 'overview' ? true : false;
    
// Start Building Plan Object
    _completePlan[_planTitle] = {
        info: {},
        weeks: {}
    }

// _planWeek logs 1 and 2 while iterating but entry for .weeks.1 is overwritten with .weeks.2
    _completePlan[_planTitle].weeks[_planWeek] = {
        sn: { inactive: true },
        mo: { inactive: true },
        tu: { inactive: true },
        we: { inactive: true },
        th: { inactive: true },
        fr: { inactive: true },
        st: { inactive: true }
    }
});

console.log(_completePlan);
});

我觉得我错过了一些简单的东西......有什么想法吗？

Answer 1

您只需要在尝试创建对象之前检查对象是否已经存在（从而覆盖之前的对象）：

if (!_completePlan.hasOwnProperty(_planTitle)) {
    _completePlan[_planTitle] = {
      info: {},
      weeks: {}
    }
  }

我还添加了一个有助于减少一些代码的小重组语句：

let [_planTarget, _planSeries, _planTitle, _planO, _planWeek] = _plan.split('.')[0].split('_'), // Drop the .htm
_planOverview = _planO === 'overview' ? true : false;

const planList = [
  'military_greekHero_achilles_week_1.htm',
  'military_greekHero_achilles_week_2.htm'
];

var _completePlan = {};

planList.forEach(_plan => {

  // Pull data from the file name formated: target_series_title_overview/week_weekNum.htm
  let [_planTarget, _planSeries, _planTitle, _planO, _planWeek] = _plan.split('.')[0].split('_'), // Drop the .htm
    _planOverview = _planO === 'overview' ? true : false;

  // Start Building Plan Object
  if (!_completePlan.hasOwnProperty(_planTitle)) {
    _completePlan[_planTitle] = {
      info: {}, weeks: {}
    }
  }

  _completePlan[_planTitle].weeks[_planWeek] = {
    sn: { inactive: true},
    mo: { inactive: true},
    tu: { inactive: true},
    we: { inactive: true},
    th: { inactive: true},
    fr: { inactive: true},
    st: { inactive: true}
  }
});

console.log(_completePlan);

Answer 2

每次迭代都会重置整个_completePlan[_planTitle]。因此，weeks 对象中的1 或2 对象本身并没有被“覆盖”，而是它们的父级的父级 对象被重置为{info: {}, weeks: {}}。

因此，如果 weeks 对象存在，则需要将其设置为自身，并且仅当它不为空白对象时。

你可以这样做：

const planList = [
    'military_greekHero_achilles_week_1.htm',
    'military_greekHero_achilles_week_2.htm'
];

var _completePlan = {};


planList.forEach(_plan => {

        // Pull data from the file name formated: target_series_title_overview/week_weekNum.htm
            let _planPieces = _plan.split('.')[0].split('_'),// Drop the .htm
                _planTarget = _planPieces[0],
                _planSeries = _planPieces[1],
                _planTitle = _planPieces[2],
                _planOverview = _planPieces[3],
                _planWeek = _planPieces[4];

            _planOverview = _planOverview == 'overview' ? true : false;
            
        // Start Building Plan Object
            

        // _planWeek logs 1 and 2 while iterating but entry for .weeks.1 is overwritten with .weeks.2
            _completePlan[_planTitle] = _completePlan[_planTitle] || {};
            _completePlan[_planTitle].info = _completePlan[_planTitle].info || {};
            _completePlan[_planTitle].weeks = _completePlan[_planTitle].weeks || {};


            _completePlan[_planTitle].weeks[_planWeek] = {
                sn: { inactive: true },
                mo: { inactive: true },
                tu: { inactive: true },
                we: { inactive: true },
                th: { inactive: true },
                fr: { inactive: true },
                st: { inactive: true }
    }
});
console.log(_completePlan);