将数组映射到对象数组

Question

我有一个带有选定项目的变化数组，其中键为id：

[1, 2, 4, 6]

和对象数组：

items = [
  {
    id: 1,
    selected: false
  },
  {
    id: 2,
    selected: false
  },
  {
    id: 3,
    selected: false
  },
  {
    id: 4,
    selected: false
  },
  {
    id: 5,
    selected: true
  },
  {
    id: 6,
    selected: false
  }
]

如何将所选数组映射到项目并将“ selected”更改为true，其他项目则将其更改为false？

Answer 1

items.forEach(item => item.selected = array.contains(item.id))

只需遍历所有项目，并根据ID是否包含在数组中来设置所选属性。

Answer 2

var items = [
  {
    id: 1,
    selected: false
  },
  {
    id: 2,
    selected: false
  },
  {
    id: 3,
    selected: false
  },
  {
    id: 4,
    selected: false
  },
  {
    id: 5,
    selected: true
  },
  {
    id: 6,
    selected: false
  }
];
var array = [1, 2, 4, 6];

items.forEach(function(item) {
  item.selected = (array.indexOf(item.id) !== -1);
})

Answer 3

您可以使用Map并对数组进行迭代，并将其全部设置为false ，然后将其选定为true 。

 var selected = [1, 2, 4, 6], items = [{ id: 1, selected: false }, { id: 2, selected: false }, { id: 3, selected: false }, { id: 4, selected: false }, { id: 5, selected: true }, { id: 6, selected: false }], map = new Map; items.forEach(a => (map.set(a.id, a), a.selected = false)); selected.forEach(a => map.get(a).selected = true); console.log(items); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

具有Set的版本。

 var selected = [1, 2, 4, 6], items = [{ id: 1, selected: false }, { id: 2, selected: false }, { id: 3, selected: false }, { id: 4, selected: false }, { id: 5, selected: true }, { id: 6, selected: false }], set = new Set(selected); items.forEach(a => a.selected = set.has(a.id)); console.log(items); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 4

var arr = [1,2,4,6];
var items = [
  {
    id: 1,
    selected: false
  },
  {
    id: 2,
    selected: false
  },
  {
    id: 3,
    selected: false
  },
  {
    id: 4,
    selected: false
  },
  {
    id: 5,
    selected: true
  },
  {
    id: 6,
    selected: false
  }
];

var notSelected = $.grep(items, function(e) { return arr.indexOf(e.id) == -1 }).map(function(e) { e.selected = false; return e;});
var selected = $.grep(items, function(e) { return arr.indexOf(e.id) != -1 }).map(function(e) { e.selected = true; return e;});

结果：

notSelected.concat(selected);

Answer 5

请尝试一下。 我认为它将起作用...

$new=array();
foreach($items as $item){
   if($item['selected']==false){
$new.=array_merge($item['id']);
    }
}
print_r($new);

并更改未选中的项目为true。

foreach($items as $item){
if($item['selected']==false){
$item['selected']=true;
 }
}