處理復雜的查詢參數 Express.Js

Question

我正在使用 Express.js 制作 REST APIS 我有以下快速路線：

/api/customer

我向路由添加了多個查詢參數，如下所示：

/api/customer?name=jake
/api/customer?country=america
/api/customer?name=jake&country=america 
/api/customer?name=jake&limit=10

在我的控制器中，我使用 If 處理所有這些，並且有很多情況我覺得這種方法無法擴展，有沒有更好的方法來處理這個？

這是我的控制器的代碼，我使用 Sequelize 來查詢數據庫：

async function getAllCustomer(queryLimit, page) {  
  const customers = await Customer.findAll({
    limit: queryLimit ? parseInt(queryLimit) : null,
    offset: page ? parseInt(queryLimit) * parseInt(page) : null
  });
  
  return customers;
}

async function getCustomerByFirstName(name, queryLimit, page) {
  return await Customer.findAll({
    where: {
      firstName: name,
    }
  })
}

async function getCustomerByAddress(address) {
  return await Customer.findAll({
    where: {
      customerAddress: address
    }
  })
}

async function getCustomerByNameAddress(name, address) {
  return await Customer.findAll({
    where: {
      [Op.and]: [
        {firstName: name},
        {customerAddress: address}
      ]
    }
  })
}

async function getCustomer(req, res) {
  const page = req.query.page;
  const queryLimit = req.query.limit;
  const name = req.query.name;
  const address = req.query.address;
  
  let customers;
  /* User want to find first names */
  if (name && !address) {
    const names = name.split(",")

    customers = await getCustomerByFirstName(names, queryLimit, page)


    res.status(200).send(customers)
    return;
  }

  /* User want to find addresses */
  if (!name && address) {
    const addresses = address.split(",")
    customers = await getCustomerByAddress(addresses, queryLimit, page)
    res.status(200).send(customers)
    return;
  }

  /* User want to mix both */
  if (name && address) {
    const names = name.split(",")
    const addresses = address.split(",")
    customers = await getCustomerByNameAddress(names, addresses, queryLimit, page)    

    res.status(200).send(customers)
    return;
  }

  if (!name && !address) {
    customers = await getAllCustomer(queryLimit, page)
    res.status(200).send(customers)
    return;
  }
}

Answer 1

你可以這樣做：

async function getCustomer(req, res) {
  const page = req.query.page;
  const queryLimit = req.query.limit;
  const name = req.query.name;
  const address = req.query.address;
  
  let query = { };
  if(name) {
    query.firstName = name;
  }
  
  if(address) {
    query.address = address;
  }
  
  let customers = await getCustomers(query, queryLimit, page);
  res.status(200).send(customers)
  return;
}

async function getCustomers(query, queryLimit, page) {  
  const customers = await Customer.findAll({
    where: query,
    limit: queryLimit ? parseInt(queryLimit) : null,
    offset: page ? parseInt(queryLimit) * parseInt(page) : null
  });
  
  return customers;
}

順便說一句，在您的代碼中，函數getCustomerByFirstName 、 getCustomerByAddress和getCustomerByNameAddress期望接收name和address作為字符串參數，但您傳遞的是names和addresses數組。 這可能會導致錯誤...