Question

有人可以帮我吗？我花了很长时间试图弄清楚。我有以下格式：

[
  [{id:'1',venue:'foa',title:'t'},{id:'5',venue:'cs',title:'b'}, ...],
  [{id:'2',venue:'fob',title:'t'},{id:'6',venue:'cs',title:'b'}, ...],    
  [{id:'3',venue:'foc',title:'t'},{id:'7',venue:'cs',title:'b'}, ...],
  [{id:'4',venue:'fod',title:'t'},{id:'8',venue:'cs',title:'b'}, ...]
]

以这种格式（最好是._underscore）制作它的最简单方法是：

{id:'1',venue:'foa',title:'t'}
{id:'5',venue:'cs',title:'b'}
{id:'2',venue:'fob',title:'t'}
{id:'6',venue:'cs',title:'b'}
{id:'3',venue:'foc',title:'t'}
{id:'7',venue:'cs',title:'b'}
{id:'4',venue:'fod',title:'t'}
{id:'8',venue:'cs',title:'b'}

Answer 1

您的问题还不够清楚，但是如果您只想展平数组，则可以使用以下代码：

var arr = [
  [{id:'1',venue:'foa',title:'t'},{id:'5',venue:'cs',title:'b'}],
  [{id:'2',venue:'fob',title:'t'},{id:'6',venue:'cs',title:'b'}],   
  [{id:'3',venue:'foc',title:'t'},{id:'7',venue:'cs',title:'b'}],
  [{id:'4',venue:'fod',title:'t'},{id:'8',venue:'cs',title:'b'}]
];

var res = [].concat(...arr);

console.log(res);

Answer 2

我不确定您到底想要什么，但是您可以尝试使用Lodash：

var result = [];

var obj = [
  [{id:'1',venue:'foa',title:'t'},{id:'5',venue:'cs',title:'b'}],
  [{id:'2',venue:'fob',title:'t'},{id:'6',venue:'cs',title:'b'}],    
  [{id:'3',venue:'foc',title:'t'},{id:'7',venue:'cs',title:'b'}],
  [{id:'4',venue:'fod',title:'t'},{id:'8',venue:'cs',title:'b'}]
];

_.each(obj, function(i) {

  _.each(i, function(j) {
    result.push(j);
  });

});

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.js"></script>

Answer 3

您可以使用flatten()。

var arr = [[{id:'1',venue:'foa',title:'t'},{id:'5',venue:'cs',title:'b'}],[{id:'2',venue:'fob',title:'t'},{id:'6',venue:'cs',title:'b'}],    [{id:'3',venue:'foc',title:'t'},{id:'7',venue:'cs',title:'b'}],[{id:'4',venue:'fod',title:'t'},{id:'8',venue:'cs',title:'b'}]],
    result = _.flatten(arr);
console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>

如何合并包含相同对象键的多个子数组？

3 个答案: