在nodejs中的query()函数中传递多个查询

Question

In this only first query will run how I pass 2 query simultaneously .

我想同时运行insert或update查询。

db.js

require('dotenv').config()
const mysql = require('mysql');


const db_config = {
    connectionLimit: 10,
    host: "localhost",
    user: "root",
    password: "",
    database: "vechain"
};

const pool = mysql.createPool(db_config);
exports.pool = pool;
exports.query = function(query) {
    try {
        return new Promise((resolve, reject) => {
            pool.query(query, function(err, result, fields) {
                if (err) reject(err);
                resolve(result);
            });
        })
    } catch (err) {
        console.log('in db_sql function error');
        console.log(err);
        res.status(500).send({ success: false, msg: 'Error', data: '', errors: err });
    }
}

**app.js**

const db = require('./db'); //here I inclue database details

//这是我的query = await db.query(‘插入测试(名称)选择"kk“(如果不存在)(从name="kmkm”的测试中选择*)；’更新测试集名称= "88“，其中id =”463“)；

Answer 1

在文档中，您可以传递两个查询。

我在配置中添加了一个选项multipleStatements: true,。

此外，我还使用...args向query函数传递多个查询。

从文档中，请注意，用于流多语句查询的接口是实验性的

示例：

require("dotenv").config();
const mysql = require("mysql");

const db_config = {
  connectionLimit: 10,
  host: "localhost",
  user: "root",
  password: "",
  database: "vechain",
  multipleStatements: true,
};

const pool = mysql.createPool(db_config);
exports.pool = pool;
exports.query = function (...args) {
  try {
    const query = args.join(";");
    console.log({ query });
    return new Promise((resolve, reject) => {
      pool.query(query, function (err, result, fields) {
        if (err) reject(err);
        resolve(result);
      });
    });
  } catch (err) {
    console.log("in db_sql function error");
    console.log(err);
    res
      .status(500)
      .send({ success: false, msg: "Error", data: "", errors: err });
  }
};

你app.js

const db = require("./db"); //here I include database details

const query1 = "SELECT 1";
const query2 = "SELECT 2";

db.query(query1, query2).then(res => {
  console.log(res);
});

Answer 2

试试看；

var check = db.query(query1,query2);