javascript - 使用node sqlite3异步获取所有数据

Question

我需要使用 db.all() 异步获取所有数据。我不想在回调中运行其他功能/代码。我想“导出”结果。可以吗？

* 我是 SQLite 的新手 😬 *

const sqlite3 = require('sqlite3').verbose()
const db = new sqlite3.Database('./db/subscribers.db')

const createSQLStatement = `CREATE TABLE IF NOT EXISTS subscriptions (id INTEGER PRIMARY KEY AUTOINCREMENT, subscriber TEXT)`
db.run(createSQLStatement)

exports.getAllSubscribers = async () => {
  const $sql = `SELECT * FROM subscriptions`
  return await db.all($sql, (err, rows) => {
    if (err) {
      return
    }

    // I wish to export "rows" which is an array
    return rows
  })
}

在另一个文件中

const { getAllSubscribers } = require('./foo')

getAllSubscribers().then(subs => {

  // subs is not what I expect it to be.
  // I'm getting:
  // Database { open: true, filename: './db/subscribers.db', mode: 65542 }
  // But I do need subs here to loop over and not inside of getAllSubscribers()

})
.catch(e => console.log(e))

我也看过 this doc但似乎没有什么能给我预期的结果。

Answer 1

当您使用then 和catch 时，您不会使用await。

所以你需要改进你的第一个脚本:

const sqlite3 = require('sqlite3').verbose()
const db = new sqlite3.Database('./db/subscribers.db')

const createSQLStatement = `CREATE TABLE IF NOT EXISTS subscriptions (id INTEGER PRIMARY KEY AUTOINCREMENT, subscriber TEXT)`
db.run(createSQLStatement)

exports.getAllSubscribers = () => {
  const $sql = `SELECT * FROM subscriptions`
  return db.all($sql, []);
}

在另一个文件中你可以像这样使用await:

const { getAllSubscribers } = require('./foo')
const getSubscribers = async () => {
    const subscribers = await getAllSubscribers();
};

您可以在此处找到更多示例:https://www.scriptol.com/sql/sqlite-async-await.php