如何在 nodejs 中的单个文件中提供 mysql 数据库连接

我需要为模块提供 mysql 连接.我有一个这样的代码.

var express = require('express'),app = express(),server = require('http').createServer(app);var mysql = require('mysql');var 连接 = mysql.createConnection({主机:'127.0.0.1'，用户:'root'，密码 : ''，数据库:'聊天'});connection.connect(function(err) {如果(错误){console.error('错误连接:' + err.stack);返回;}});app.get('/save', function(req,res){var post = {from:'me', to:'you', msg:'hi'};var query = connection.query('INSERT INTO messages SET ?', post, function(err, result) {如果(错误)抛出错误；});});服务器.听(3000)；

但是我们如何为所有模块提供一次 mysql 连接.

您可以创建一个 db 包装器然后需要它.node 的 require 每次都返回相同的模块实例，因此您可以执行连接并返回处理程序.来自 Node.js 文档:

对 require('foo') 的每次调用都将返回完全相同的对象，如果它解析为相同的文件.

你可以创建db.js:

var mysql = require('mysql');var 连接 = mysql.createConnection({主机:'127.0.0.1'，用户:'root'，密码 : ''，数据库:'聊天'});connection.connect(function(err) {如果(错误)抛出错误；});module.exports = 连接；

然后在您的 app.js 中，您只需需要它.

var express = require('express');var app = express();var db = require('./db');app.get('/save',function(req,res){var post = {from:'me', to:'you', msg:'hi'};db.query('INSERT INTO messages SET ?', post, function(err, result) {如果(错误)抛出错误；});});服务器.听(3000)；

这种方法允许您抽象任何连接细节，在整个应用程序中包装您想要公开和 require db 的任何其他内容，同时由于 node require 的工作方式而保持与数据库的一个连接:)

I need to provide the mysql connection for modules. I have a code like this.

var express = require('express'),
app = express(),
server = require('http').createServer(app);

var mysql      = require('mysql');
var connection = mysql.createConnection({
    host     : '127.0.0.1',
    user     : 'root',
    password : '',
    database    : 'chat'
});

connection.connect(function(err) {
    if (err) {
        console.error('error connecting: ' + err.stack);
        return;
    }
});

app.get('/save', function(req,res){
    var post  = {from:'me', to:'you', msg:'hi'};
    var query = connection.query('INSERT INTO messages SET ?', post, function(err, result) {
        if (err) throw err;
    });
});

server.listen(3000);

But how we provide one time mysql connection for all the modules.

You could create a db wrapper then require it. node's require returns the same instance of a module every time, so you can perform your connection and return a handler. From the Node.js docs:

every call to require('foo') will get exactly the same object returned, if it would resolve to the same file.

You could create db.js:

var mysql = require('mysql');
var connection = mysql.createConnection({
    host     : '127.0.0.1',
    user     : 'root',
    password : '',
    database : 'chat'
});

connection.connect(function(err) {
    if (err) throw err;
});

module.exports = connection;

Then in your app.js, you would simply require it.

var express = require('express');
var app = express();
var db = require('./db');

app.get('/save',function(req,res){
    var post  = {from:'me', to:'you', msg:'hi'};
    db.query('INSERT INTO messages SET ?', post, function(err, result) {
      if (err) throw err;
    });
});

server.listen(3000);

This approach allows you to abstract any connection details, wrap anything else you want to expose and require db throughout your application while maintaining one connection to your db thanks to how node require works :)