Vue3：在可组合物中使用咖喱函数不起作用

Question

我有以下可组合：

export default function useReferralCodes(affiliateUserId) {
  let createReferralCode = (affiliateUserId, payload) => {
    console.log('affiliateUserId', affiliateUserId)
    console.log('payload', payload)
  }

  let curriedReferralCode = curry(createReferralCode)
  curriedReferralCode(affiliateUserId)

  return {
    createReferralCode,
    curriedReferralCode
  }
}

现在，当我在组件中使用它时，它不会像这样工作：

  methods: {
    onSubmit (payload) {
      this.curriedReferralCode(payload) // Outputs nothing
    }
  }

但这项工作：

this.curriedReferralCode(3)(payload)

是我遗漏了什么或者我真的不完全理解柯里化吗？

谢谢！

Answer 1

从curryapi、curry来看，如果提供了所有参数，则函数要么调用该函数，要么返回一个新函数，如果提供了一些参数，则该函数只需要参数的子集。

例如在您的代码中

// createReferralCode requires 2 params
let createReferralCode = (affiliateUserId, payload) => {
  console.log('affiliateUserId', affiliateUserId)
  console.log('payload', payload)
}

// curriedReferralCode is the curried version of createReferralCode 
let curriedReferralCode = curry(createReferralCode)

curriedReferralCode(3, payload) // Equivalent to createReferralCode(3, payload)
curriedReferralCode(3) // Only 1 param provided out of 2: this creates and returns a function that requires the missing param *payload*
curriedReferralCode(3)(payload) // According to what has been said above, equivalent to createReferralCode(3, payload)

修复您的代码：

// Provide the context param when you create the function
let curriedReferralCode = curry(createReferralCode)(affiliateUserId)