我如何过滤并返回嵌套的对象数组？

Question

有人可以帮助我吗？ 我有一个像这样的数据结构：

contracts: [ 
0: {closedRequest: [{id: ... , name: ... } , {id: ... , name: ... }, {id: ... , name: ... }]},
1: {closedRequest: [{id: ... , name: ... } , {id: ... , name: ... }, {id: ... , name: ... }]},

]

现在我将过滤此对象以删除已确定 Id 的 closedRequest 元素。 我尝试使用过滤器，但数组没有改变。

我该怎么做？

例如，我想排除所有具有 id (on closedRequest) = 3 的元素

所以，如果我有这个结构：

contracts: [ 
0: {closedRequest: [{id: 2 , name: "a"} , {id: 3 , name: "b"}, {id: 4 , name: "c" }]},
1: {closedRequest: [{id: 2 , name: "a"} , {id: 3 , name: "b"}, {id: 4 , name: "c" }]},

]

我想要：

contracts: [ 
0: {closedRequest: [{id: 2 , name: "a"}  {id: 4 , name: "c" }]},
1: {closedRequest: [{id: 2 , name: "a"}  {id: 4 , name: "c" }]},

]

我的代码示例：

client.contracts.forEach((c) =>
          c.closedRequests.filter((cr) => cr.requestID != requestID)
        );

我想要相同的对象，但已编辑关闭请求的字段。

谢谢

Answer 1

假设contracts数组的每一项都包含许多字段，你可以使用array.map将旧数组转换为新数组，使用array.filter过滤嵌套数组：

 let contracts = [ { closedRequest: [{id: 2 , name: "a"} , {id: 3 , name: "b"}, {id: 4 , name: "c" }] }, { closedRequest: [{id: 2 , name: "a"} , {id: 3 , name: "b"}, {id: 4 , name: "c" }] } ]; let filteredContracts = contracts.map(contract => ({ ...contract, closedRequest : contract.closedRequest.filter(x => x.id !== 3) })); console.log(filteredContracts);