如何使此更新API更加高效和短而无重复

如何解决如何使此更新API更加高效和短而无重复

以下是API服务中更新帐户的方法。我如何在if条件下缩短代码，而不是增加太多对象？

async updateAccount(uuid: string,body: IUpdateAccountDto) {
    const found = await this.getAccountById(uuid);
    const update = (resolve,reject) => {
      if (
        found.username ||
        found.password ||
        found.email ||
        found.phone ||
        found.product
      ) {
        resolve(
          'account updated',((found.username = body.username),(found.password = body.password),(found.email = body.email),(found.phone = body.phone),(found.product = body.product)),);
      } else {
        reject('cannot update username');
      }
    };

    const updateAccount = new Promise(update);
    return updateAccount;
  }

解决方法

您可以使用Object.values(obj)和Array.some(fn)来达到目的。 注意：使用Array.every（fn）将达到&&条件

var notValidFound = {
  username: "",password: undefined,email: "",phone: null,product: ""
};

var notValidValue = Object.values(notValidFound).some(o => o);

console.log(notValidValue);

var validFound = {
  username: "some_username",email: "[e-mail]",product: ""
};

var validValue = Object.values(validFound).some(o => o);

console.log(validValue);

您可以详细了解Object.values(obj)和Array.prototype.some(fn)