Javascript 使用较短的数据库查询代码将数据插入MySQL数据库表

我有一些关于将数据插入MySQL数据库的问题。事实上，我已经成功了。但我认为代码太长了，所以当服务器或其他地方出现问题时，代码看起来很糟糕，很难修改

这是我的代码，代码非常长，所以我想知道是否有一些解决方案可以缩短代码

结果变量有JSON数据，看起来我可以在db查询中使用它。我已经认为我可以使用数组或对象，我尝试了一些代码，但它们不起作用

router.post("/user", function(req, res, next) {
  let id = req.body.id;
  let githubAPI = "https://api.github.com/users/";

  let options = {
    url: githubAPI + id,
    headers: {
      "User-Agent": "request"
    }
  };
  console.log(id);
  request(options, function(error, response, data) {
    if (error) {
      throw error;
    }
    // result have JSON Data
    let result = JSON.parse(data);

    let nick = result.login;
    let id = result.id;
    let node_id = result.node_id;
    let avatar_url = result.avatar_url;
    let gravatar_id = result.gravatar_id;
    let url = result.url;
    let html_url = result.html_url;
    let followers_url = result.followers_url;
    let following_url = result.following_url;
    let gists_url = result.gists_url;
    let starred_url = result.starred_url;
    let subscriptions_url = result.subscriptions_url;
    let organizations_url = result.organizations_url;
    let repos_url = result.repos_url;
    let events_url = result.events_url;
    let received_events_url = result.received_events_url;
    let type = result.type;
    let site_admin = result.site_admin;
    let name = result.name;
    let company = result.company;
    let blog = result.blog;
    let location = result.location;
    let email = result.email;
    let hireable = result.hireable;
    let bio = result.bio;
    let public_repos = result.public_repos;
    let public_gists = result.public_gists;
    let followers = result.followers;
    let following = result.following;
    let created_at = result.created_at;
    let updated_at = result.updated_at;

    if (bio == null) {
      bio = "Developer";
    }
    db.query(
      `INSERT INTO user (login, id, node_id, avatar_url, gravatar_id, url, html_url, followers_url, following_url, gists_url, starred_url, subscriptions_url, organizations_url, repos_url, events_url, received_events_url, type, site_admin, name, company, blog, location, email, hireable, bio, public_repos, public_gists, followers, following, created_at, updated_at) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?)`,
      [
        nick,
        id,
        node_id,
        avatar_url,
        gravatar_id,
        url,
        html_url,
        followers_url,
        following_url,
        gists_url,
        starred_url,
        subscriptions_url,
        organizations_url,
        repos_url,
        events_url,
        received_events_url,
        type,
        site_admin,
        name,
        company,
        blog,
        location,
        email,
        hireable,
        bio,
        public_repos,
        public_gists,
        followers,
        following,
        created_at,
        updated_at
      ]
    );
  });

---

不要硬编码对象中的SQL和翻译，只需将字段的预期顺序保持为常量，然后将解析后的JSON中的值映射到

值的数组中即可

您还可以从相同的常量列表生成SQL和占位符：

// Keep a list for the fields where order is important
const fieldOrder = [
  'login',
  'id',
  'node_id',
  'avatar_url',
  'gravatar_id',
  'url',
  'html_url',
  'followers_url',
  'following_url',
  'gists_url',
  'starred_url',
  'subscriptions_url',
  'organizations_url',
  'repos_url',
  'events_url',
  'received_events_url',
  'type',
  'site_admin',
  'name',
  'company',
  'blog',
  'location',
  'email',
  'hireable',
  'bio',
  'public_repos',
  'public_gists',
  'followers',
  'following',
  'created_at',
  'updated_at'
];

// Parse your content in the same place    
let result = JSON.parse(data);

// Extract the value by the same key names
let values = fieldOrder.map(k => result[k]);

// Generate the statement rather than hardcoding    
let sql = `INSERT into user (${fieldOrder.join(',')}) values(${fieldOrder.map(e => '?').join(',')})`

// pass these arguments to your function    
db.query(sql, values);

这相当简单，本质上是许多ORM库在其功能实现的外观下为您所做的

请注意，其中许多部分是合理的通用性和可重用的，这是此类库实现的另一个特性

因此，您无法真正避免的一个“邪恶”是保留字段列表，因为顺序可能很重要，这是清除任何意外数据的合理方法

“便宜而肮脏”的方式可能是：

let fieldOrder = Object.keys(result);

甚至：

let [fieldOrder, values] = Object.entries(result);

但是，这并不能真正让您控制在
数据中发送的有效内容，而且还可能具有潜在的破坏性

无论如何，即使在代码中的某个地方保持一个不变的列表，只要将列出相同字段名的所有位置移动到一个列表中，就可以大大减少当前列表的数量。
谢谢您的帮助！我用这种方法解决了这个问题。我想我应该学习箭头函数。