向数组 Ramda 插入一个新对象
Insert a new object to array Ramda
Ramda 有 insert function。但就我而言,我不知道如何钻取对象并将新对象插入到数组中。另外一个新对象应该在数组的最后一个索引上,我们可以通过 stuff["31"].length
获取它。我的尝试太可悲了,所以我决定不展示它们:/
数据模型:
const stuff = {
"31": [
{
"id": "11",
"title": "ramda heeeelp"
},
{
"id": "12",
"title": "ramda 123"
},
//HERE I WANT TO INSERT A NEW OBJECT - {id: "13", title: "new title"}
],
"33": [
{
"id": "3",
"title": "..."
}
],
"4321": [
{
"id": "1",
"title": "hello world"
}
]
}
这是对镜头的直接使用,append
加到最后。
我会这样做:
const addObjToGroup = (groupId, newObj, data) =>
over (lensProp (groupId), append (newObj), data)
const stuff = {"31": [{"id": "11", "title": "ramda heeeelp"}, {"id": "12", "title": "ramda 123"}], "33": [{"id": "3", "title": "..."}], "4321": [{"id": "1", "title": "hello world"}]}
console .log (
addObjToGroup ('31', {id: "13", title: "new title"}, stuff)
)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
<script>const {over, lensProp, append} = R </script>
您还可以使用 evolve 将项目附加到 groupId
中的数组:
const { evolve, append } = R
const addObjToGroup = (groupId, newObj) => evolve({ [groupId]: append(newObj) })
const stuff = {"31": [{"id": "11", "title": "ramda heeeelp"}, {"id": "12", "title": "ramda 123"}], "33": [{"id": "3", "title": "..."}], "4321": [{"id": "1", "title": "hello world"}]}
const result = addObjToGroup ('31', {id: "13", title: "new title"})(stuff)
console .log (result)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
也R.useWith
在这里有帮助!
const appendTo = R.useWith(R.over, [
R.lensProp,
R.append,
R.identity
]);
// ---
const stuff = {
"31": [
{
"id": "11",
"title": "ramda heeeelp"
},
{
"id": "12",
"title": "ramda 123"
},
//HERE I WANT TO INSERT A NEW OBJECT - {id: "13", title: "new title"}
],
"33": [
{
"id": "3",
"title": "..."
}
],
"4321": [
{
"id": "1",
"title": "hello world"
}
]
}
console.log(
appendTo('31', 'HELLO WORLD', stuff),
);
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js" integrity="sha256-xB25ljGZ7K2VXnq087unEnoVhvTosWWtqXB4tAtZmHU=" crossorigin="anonymous"></script>
Ramda 有 insert function。但就我而言,我不知道如何钻取对象并将新对象插入到数组中。另外一个新对象应该在数组的最后一个索引上,我们可以通过 stuff["31"].length
获取它。我的尝试太可悲了,所以我决定不展示它们:/
数据模型:
const stuff = {
"31": [
{
"id": "11",
"title": "ramda heeeelp"
},
{
"id": "12",
"title": "ramda 123"
},
//HERE I WANT TO INSERT A NEW OBJECT - {id: "13", title: "new title"}
],
"33": [
{
"id": "3",
"title": "..."
}
],
"4321": [
{
"id": "1",
"title": "hello world"
}
]
}
这是对镜头的直接使用,append
加到最后。
我会这样做:
const addObjToGroup = (groupId, newObj, data) =>
over (lensProp (groupId), append (newObj), data)
const stuff = {"31": [{"id": "11", "title": "ramda heeeelp"}, {"id": "12", "title": "ramda 123"}], "33": [{"id": "3", "title": "..."}], "4321": [{"id": "1", "title": "hello world"}]}
console .log (
addObjToGroup ('31', {id: "13", title: "new title"}, stuff)
)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
<script>const {over, lensProp, append} = R </script>
您还可以使用 evolve 将项目附加到 groupId
中的数组:
const { evolve, append } = R
const addObjToGroup = (groupId, newObj) => evolve({ [groupId]: append(newObj) })
const stuff = {"31": [{"id": "11", "title": "ramda heeeelp"}, {"id": "12", "title": "ramda 123"}], "33": [{"id": "3", "title": "..."}], "4321": [{"id": "1", "title": "hello world"}]}
const result = addObjToGroup ('31', {id: "13", title: "new title"})(stuff)
console .log (result)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
也R.useWith
在这里有帮助!
const appendTo = R.useWith(R.over, [
R.lensProp,
R.append,
R.identity
]);
// ---
const stuff = {
"31": [
{
"id": "11",
"title": "ramda heeeelp"
},
{
"id": "12",
"title": "ramda 123"
},
//HERE I WANT TO INSERT A NEW OBJECT - {id: "13", title: "new title"}
],
"33": [
{
"id": "3",
"title": "..."
}
],
"4321": [
{
"id": "1",
"title": "hello world"
}
]
}
console.log(
appendTo('31', 'HELLO WORLD', stuff),
);
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js" integrity="sha256-xB25ljGZ7K2VXnq087unEnoVhvTosWWtqXB4tAtZmHU=" crossorigin="anonymous"></script>