Question

是否可以根据值获取对象？

在下面的示例json中，键名具有 type ，因此基于类型值需要获取结果。

例如，如果type ='user'，则仅针对用户对象而非员工对象获取结果。

在这里，我很努力地同时拥有用户和员工这两个密钥，请您提出如何使用的建议

var list =[
  {"doc":{"type":"user","Title":"test1","Relations":{"users":[{"name": "user1"},{"name": "user2"},{"name": "user3"}],"employee":[{"emp": "user2"}]}}},
  {"doc":{"type":"employee","Title":"test2","Relations":{"users":[{"name": "user1"}],"employee":[{"name": "emp1"},{"name": "emp2"},{"name": "emp3"}]}}}
];

const getDetails = R.chain(R.pipe(
  R.path(['doc', 'Relations']),
  R.pick(['users', 'employee']),
  R.values
))

const result = getDetails(list)

console.log(result)

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>

当前输出：

[[{"name": "user1"}, {"name": "user2"}, {"name": "user3"}], [{"emp": "user2"}], [{"name": "user1"}], [{"name": "emp1"}, {"name": "emp2"}, {"name": "emp3"}]]

期望输出：

[[{"name": "user1"}, {"name": "user2"}, {"name": "user3"}], [{"name": "emp1"}, {"name": "emp2"}, {"name": "emp3"}]]

Answer 1

您可以使用R.cond来基于一组谓词来分支逻辑，例如您的示例中type是user还是employee。

使用这种方法，除非您希望对数组进行额外的扁平化处理，否则您需要将R.chain切换为R.map以匹配期望的列表。

const list = [
  {"doc":{"type":"user","Title":"test1","Relations":{"users":[{"name": "user1"},{"name": "user2"},{"name": "user3"}],"employee":[{"emp": "user2"}]}}},
  {"doc":{"type":"employee","Title":"test2","Relations":{"users":[{"name": "user1"}],"employee":[{"name": "emp1"},{"name": "emp2"},{"name": "emp3"}]}}}
]

const getDetails = R.map(R.pipe(
  R.prop('doc'),
  R.cond([
    [R.propEq('type', 'user'), R.path(['Relations', 'users'])],
    [R.propEq('type', 'employee'), R.path(['Relations', 'employee'])]
  ])
))

const expected = [[{"name": "user1"}, {"name": "user2"}, {"name": "user3"}], [{"name": "emp1"}, {"name": "emp2"}, {"name": "emp3"}]]

console.log(R.equals(expected, getDetails(list)))

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节点Ramda +根据值查找对象

1 个答案: