javascript - 对具有多个条件的对象进行排序

Question

我目前正在尝试对我的 reducer 中返回的对象数组进行排序。我想根据返回的 name 对应用程序进行排序，并且当前仍有一些名称为 null 我想列出 null 名称位于末尾。目前，当我尝试添加附加条件时，出现错误 Cannot read property 'toLowerCase' of null 我期望 null 项目只需将它们移动到结尾。

// Actions
export const FETCH_MYAPPS_PENDING = 'widgets/apps/FETCH_PENDING';
export const FETCH_MYAPPS_FULFILLED = 'widgets/apps/FETCH_FULFILLED';

// Reducer
const appsSort = (a, b) => {
  if (a.name != null && b.name != null) {
   if (a.name.toLowerCase() > b.name.toLowerCase()) return 1;
   if (a.name.toLowerCase() < b.name.toLowerCase()) return -1;
  }
  return 0;
};

export default function reducer(state = { data: [], pending: false }, action) {
switch (action.type) {
case FETCH_MYAPPS_FULFILLED:
  return {
    data: action.apps.sort(appsSort),
    pending: false,
    retrievedAt: Date.now(),
  };
case FETCH_MYAPPS_PENDING:
  return {
    ...state,
    pending: true,
  };
default:
  return state;
 }
}

Answer 1

我的尝试。顶部有点多余，但在检查名称之前进行空检查是有意义的。

    const appsSort = (a, b) => {
      //null checks
      if (a.name === null && b.name === null) return 0;
      if (a.name === null && b.name !== null) return 1;
      if (b.name === null && a.name !== null) return -1;
      //check for lowercase
      if (a.name.toLowerCase() > b.name.toLowerCase()) return 1;
      if (a.name.toLowerCase() < b.name.toLowerCase()) return -1;
      return 0;
    };
    
    var obj = [
      {name: 'apples', id: 1},
      {name: null, id: 3},
      {name: 'Banana', id: 2},
      {name: null, id: 4}
    ]
    
    obj.sort(appsSort);
    
    console.log(obj)