Question

由于不喜欢这种方法，我正在寻找有关如何改进此方法的建议。

我有两个带有Json的对象数组。

第一个列表具有日期范围内的所有日期，第二个列表来自SQL查询，其中可能缺少几个月。

期望的结果必须是一个包含所有月份的清单，并带有事宜计数，但是当我将它们加载到下面的函数中时，我缺少了清单中不存在的月份的问题计数。

我认为我只是在该函数中缺少了一些代码，但是我不确定是什么，因此来到这里。

我确实在JSFiddle中有一个解决方法，但是我希望使用更清晰的代码。

let x = [{labelDate: "Jun 2013"},
{labelDate: "Jul 2013"},
{labelDate: "Aug 2013"},
{labelDate: "Sep 2013"},
{labelDate: "Oct 2013"},
{labelDate: "Nov 2013"},
{labelDate: "Dec 2013"},
{labelDate: "Jan 2014"},
{labelDate: "Feb 2014"},
{labelDate: "Mar 2014"},
{labelDate: "Apr 2014"},
{labelDate: "May 2014"}];


let y = [{labelDate: "Jun 2013", matterCount: "1"},
{labelDate: "Jul 2013", matterCount: "2"},
{labelDate: "Aug 2013", matterCount: "2"},
{labelDate: "Sep 2013", matterCount: "10"},
{labelDate: "Oct 2013", matterCount: "1"},
{labelDate: "Nov 2013", matterCount: "1"},
{labelDate: "Feb 2014", matterCount: "1"},
{labelDate: "Apr 2014", matterCount: "17"},
{labelDate: "May 2014", matterCount: "21"}];

const merge = (x, y) =>
            x.map(xItem => ({
              ...y.find(yItem => yItem.labelDate === xItem.labelDate && Item),
              ...xItem
            }));

Actual outcome 
merge = [{labelDate: "Jun 2013", matterCount: "1"},
{labelDate: "Jul 2013", matterCount: "2"},
{labelDate: "Aug 2013", matterCount: "2"},
{labelDate: "Sep 2013", matterCount: "10"},
{labelDate: "Oct 2013", matterCount: "1"},
{labelDate: "Nov 2013", matterCount: "1"},
{labelDate: "Dec 2013"},
{labelDate: "Jan 2014"},
{labelDate: "Feb 2014", matterCount: "1"},
{labelDate: "Mar 2014"},
{labelDate: "Apr 2014", matterCount: "17"},
{labelDate: "May 2014", matterCount: "21"}];

desired outcome 
merge  = [{labelDate: "Jun 2013", matterCount: "1"},
{labelDate: "Jul 2013", matterCount: "2"},
{labelDate: "Aug 2013", matterCount: "2"},
{labelDate: "Sep 2013", matterCount: "10"},
{labelDate: "Oct 2013", matterCount: "1"},
{labelDate: "Nov 2013", matterCount: "1"},
{labelDate: "Dec 2013", matterCount: "0"},
{labelDate: "Jan 2014", matterCount: "0"},
{labelDate: "Feb 2014", matterCount: "1"},
{labelDate: "Mar 2014", matterCount: "0"},
{labelDate: "Apr 2014", matterCount: "17"},
{labelDate: "May 2014", matterCount: "21"}];

https://jsfiddle.net/cp39aoyf/2/

Answer 1

||设置默认值如何？

const merge = (x, y) =>
        x.map(xItem => ({
          ...y.find(yItem => yItem.labelDate === xItem.labelDate && yItem) || {...xItem, matterCount: "0"}
        }));

Demo JSFiddle.

另一种方法是简单地添加默认值，该默认值将被yItem覆盖：

const merge = (x, y) =>
    x.map(xItem => ({
      ...xItem,
      ...{matterCount: "0"},
      ...y.find(yItem => yItem.labelDate === xItem.labelDate && yItem)
    }));

Fiddle for this one.

在唯一ID上合并两个数组

1 个答案: