为什么我在使用 nodejs 的 mysql 中遇到致命错误？

Question

我正在尝试在 nodejs 中使用 mysql 模块创建数据库，但是在创建表并插入一些数据后，当我尝试 SELECT 并返回结果时，我遇到了致命错误！

    D:\Programming\Farahani App\App>node mysql.js
    Error: Cannot enqueue Query after fatal error.
        at Protocol._validateEnqueue (D:\Programming\Farahani App\App\node_modules\mysql\lib\protocol\Protocol.js:212:16)
        at Protocol._enqueue (D:\Programming\Farahani App\App\node_modules\mysql\lib\protocol\Protocol.js:138:13)
        at Connection.query (D:\Programming\Farahani App\App\node_modules\mysql\lib\Connection.js:198:25)
        at Handshake.<anonymous> (D:\Programming\Farahani App\App\mysql.js:28:13)
        at Handshake.<anonymous> (D:\Programming\Farahani App\App\node_modules\mysql\lib\Connection.js:526:10)
        at Handshake._callback (D:\Programming\Farahani App\App\node_modules\mysql\lib\Connection.js:488:16)
        at Handshake.Sequence.end (D:\Programming\Farahani App\App\node_modules\mysql\lib\protocol\sequences\Sequence.js:83:24)
        at Protocol.handleNetworkError (D:\Programming\Farahani App\App\node_modules\mysql\lib\protocol\Protocol.js:369:14)
        at Connection._handleNetworkError (D:\Programming\Farahani App\App\node_modules\mysql\lib\Connection.js:418:18)
        at Socket.emit (node:events:390:28) {
      code: 'PROTOCOL_ENQUEUE_AFTER_FATAL_ERROR',
      fatal: false
    }
    undefined

[如果相关，mysql 正在我的 xampp 中运行...]

感谢您的帮助。

我的代码：

const mysql = require('mysql');

const data = {
    host: "localhost",
    user: "user",
    password: "password",
    database: "mydb"
}
const con = mysql.createConnection(data)

    con.connect(() => {


        // create database 
        con.query('CREATE DATABASE mydb', () => {

        })
        // create table
        const table = "CREATE TABLE customers (name VARCHAR(255), address VARCHAR(255))";
        con.query(table, () => {
            })
        // add val to table
        const val1 = "INSERT INTO customers (name, address) VALUES ('Company Inc', 'Highway 37')";
        con.query(val1, () => {

        })
        // log results
        con.query("SELECT * FROM customers",  (err,result)=>{
            if(err) {console.log(err)};
            console.log(result);
        })
        con.end();
    })

Answer 1

您的代码有两个问题。主要问题是你不能在 mysql 中为每个连接执行多个查询。

您可以使用connection pools 使其更易于管理。

其次，您应该检查回调中的错误。这将有助于调试并使您的代码更加健壮。

con.query("select * from my_table", (error, results, fields) => {
  if (error) throw error;
  console.log(results);
});

Answer 2

const mysql = require('mysql');

const con  = mysql.createPool({
    host: "localhost",
    user: "user",
    password: "password",
    database: "mydb"
});

        // create database 
        con.query('CREATE DATABASE mydb', (err) => {
            if(err) {console.log(err)};
        })
        // create table
        const table = "CREATE TABLE customers (name VARCHAR(255), address VARCHAR(255))";
        con.query(table, (err) => {
            if(err) {console.log(err)};
            })
        // add val to table
        const val1 = "INSERT INTO customers (name, address) VALUES ('Company Inc', 'Highway 37')";
        con.query(val1, (err) => {
            if(err) {console.log(err)};
        })
        // log results
        con.query("SELECT * FROM customers",  (err,result)=>{
            if(err) {console.log(err)};
            console.log(result);
        })