Question

我的数组对象类似于下面的示例。

[
 {'email':'test@gmail.com', 'name':'abc'},
 {'email':'test1@gmail.com', 'name':'bbc'},
 {'email':'test2@gmail.com', 'name':'aaa'},
 {'email':'test3@gmail.com', 'name':'cba'},
 {'email':'test3@gmail.com', 'name':'cab'},
]

所以我的新数组将具有键 A 的键值对作为键，并且值都是以 A 字母开头的对象，等等。是如果字母 A 具有2个从 a 开始的对象，那么我还想按升序排序，如下面的最终输出示例所示

我想要的最终输出是这样的。

[
  "a" : [{'email':'test@gmail.com', 'name':'aaa'},{'email':'test2@gmail.com', 'name':'abc'}],
  "b" : [{'email':'test1@gmail.com', 'name':'bbc'}],
  "c" : [{'email':'test3@gmail.com', 'name':'cab'},{'email':'test3@gmail.com', 'name':'cba'}]
]

Answer 1

您可以对数组进行排序，然后将其分组。

String sql = "SELECT a FROM XspJobRequest a " +
            "WHERE a.xspJobTypeId = :xspJobTypeId " +
            "AND a.xspJobStatusId = :xspJobStatusId " +
            "AND a.nextAttemptTs < (now() at time zone 'utc') " +
            " ORDER BY a.insertTs";
    try
    {
        Query query = entityManager.createQuery(sql);
        query.setMaxResults(limitReturnCount);
        query.setParameter("xspJobTypeId", jobType.getId());
        query.setParameter("xspJobStatusId", XspJobStatusEnum.eNew.getId());
        List<XspJobRequest> retval = query.getResultList();

        if (!alreadyActive)
        {
            et.commit();
        }
        return retval;
    }

var array = [{ email: 'test@gmail.com', name: 'abc' }, { email: 'test1@gmail.com', name: 'bbc' }, { email: 'test2@gmail.com', name: 'aaa' }, { email: 'test3@gmail.com', name: 'cba' }, { email: 'test3@gmail.com', name: 'cab' }],
    grouped = array
        .sort(({ name: a }, { name: b }) => a.localeCompare(b))
        .reduce((r, o) => {
            var group = o.name[0].toLowerCase();
            (r[group] = r[group] || []).push(o);
            return r;
        }, Object.create(null));
        
console.log(grouped);

Answer 2

您可以使用reduce方法创建对象，并可以在sort方法内部使用名称对值进行排序。

const data = [{'email':'test@gmail.com', 'name':'Abc'},{'email':'test1@gmail.com', 'name':'bbc'},{'email':'test2@gmail.com', 'name':'aaa'},{'email':'test3@gmail.com', 'name':'cba'},{'email':'test3@gmail.com', 'name':'cab'},]

const sorted = data.reduce((r, o) => {
  let key = o.name.slice(0, 1).toLowerCase();
  r[key] = (r[key] || []).concat(o);
  r[key].sort((a, b) => a.name.localeCompare(b.name));
  return r;
}, {})

console.log(sorted)

Answer 3

var array = [
 {'email':'test@gmail.com', 'name':'abc'},
 {'email':'test1@gmail.com', 'name':'bbc'},
 {'email':'test2@gmail.com', 'name':'aaa'},
 {'email':'test3@gmail.com', 'name':'cba'},
 {'email':'test3@gmail.com', 'name':'cab'},
]

function getArray(array=[]){

let newObject = {};
array.forEach(i=>{
  let key = i['name'].slice(0,1)
  if( key && newObject[key] ){
     newObject[key].push(i)
    }else{
    newObject[key] = Array(i)
    }
  }) 
  return newObject

}

console.log(getArray(array))

Answer 4

您可以使用数组精简method：

const list = [
    {'email':'test@gmail.com', 'name':'abc'},
    {'email':'test1@gmail.com', 'name':'bbc'},
    {'email':'test2@gmail.com', 'name':'aaa'},
    {'email':'test3@gmail.com', 'name':'cba'},
    {'email':'test3@gmail.com', 'name':'cab'},
]

const newList = list.reduce((acc, currVal) => {
    const firstLetter = currVal.name.charAt(0);
    if(!acc[firstLetter]){
        acc[firstLetter] = [];
    }
    acc[firstLetter].push(currVal);
    return acc
}, {})

console.log(newList)

Answer 5

循环遍历数组，基于它创建一个对象，并按填充顺序对其进行排序。

const myArr = [
 {'email':'test@gmail.com', 'name':'abc'},
 {'email':'test1@gmail.com', 'name':'bbc'},
 {'email':'test2@gmail.com', 'name':'aaa'},
 {'email':'test3@gmail.com', 'name':'cba'},
 {'email':'test3@gmail.com', 'name':'cab'}
];

const finalObj = {};
function compare(a,b) {
  if (a.name < b.name)
    return -1;
  if (a.name > b.name)
    return 1;
  return 0;
}

myArr.forEach(item => {
  const alph = item.name.substr(0, 1);
 
  if (finalObj[alph]) {
    finalObj[alph] = [...finalObj[alph], item].sort(compare);
  } else {
    finalObj[alph] = [item];
  }
});

console.log(finalObj);

Answer 6

您可以使用lodash groupBy方法获得所需的结果。

var collection =[
 {'email':'test@gmail.com', 'name':'abc'},
 {'email':'test1@gmail.com', 'name':'bbc'},
 {'email':'test2@gmail.com', 'name':'aaa'},
 {'email':'test3@gmail.com', 'name':'cba'},
 {'email':'test3@gmail.com', 'name':'cab'},
]

console.log(_.groupBy(collection, (item) => {
    return item.name[0]
}))

js：对对象数组进行排序和重新排列

6 个答案: