reactjs - 是的 - 如何有条件地验证至少设置了 n 个值之一？

Question

如果其他 3 个字段中有 1 个不为空，如何使用 Yup 验证使该字段成为必填字段？我尝试了以下模式，但遇到了循环依赖问题。谢谢!

const schema = yup.object().shape({
  a: yup.string().when(['b', 'c', 'd'], {
    is: (b, c, d) => !b && !c && !d,
    then: yup.string().required()
  }),
  b: yup.string().when(['a', 'c', 'd'], {
    is: (a, c, d) => !a && !c && !d,
    then: yup.string().required()
  }),
  c: yup.string().when(['a', 'b', 'd'], {
    is: (a, b, d) => !a && !b && !d,
    then: yup.string().required()
  }),
  d: yup.string().when(['a', 'b', 'c'], {
    is: (a, b, c) => !a && !b && !c,
    then: yup.string().required()
  })
}, [['a', 'b', 'c'], ['b', 'c', 'd'], ['a','c', 'd'], ['a','b','d']]) 

Answer 1

希望下面的代码能解决你的问题

/* 
   All you need to do is list all the pair-wise (2-tuples) combinations. 
   As you have 4 fields, you expect to have 6 pairs in your array. 
   i.e. ['a', 'b'],['a', 'c'],['a', 'd'], ['b', 'c'], ['b', 'd'],['c', 'd'])
*/

const schema = yup.object().shape({
  a: yup.string().when(['b', 'c', 'd'], {
    is: (b, c, d) => b || c || d,
    then: Yup.string(),
    otherwise: Yup.string().required('Required')
  }),
  b: yup.string().when(['a', 'c', 'd'], {
    is: (a, c, d) => a || c || d,
    then: Yup.string(),
    otherwise: Yup.string().required('Required')
  }),
  c: yup.string().when(['a', 'b', 'd'], {
    is: (a, b, d) => a || b || d,
    then: Yup.string(),
    otherwise: Yup.string().required('Required')
  }),
  d: yup.string().when(['a', 'b', 'c'], {
    is: (a, b, c) =>  a || b || c,
    then: Yup.string(),
    otherwise: Yup.string().required('Required')
  })
}, [['a', 'b'],['a', 'c'],['a', 'd'], ['b', 'c'], ['b', 'd'],['c', 'd'] ])

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