can't make nodejs mysql query results a global variable

Question

I'm trying to use the mysql module to get some data from a mysql database and then write it to an HTML page but it seems stuck inside the query function itself. The code looks like this:

rooms = [];

var mysql = require('mysql');

var con = mysql.createConnection({
    host: "localhost",
    user: "MYUSERNAME",
    password: "MYPASSWORD",
    database: "travel"
});

con.connect(function(err) {
    if (err) throw err;
    con.query("SELECT * FROM rooms", function (err, result, fields) {
        if (err) throw err;
        var rooms = result;
        console.log(rooms[9]);
    });
});

console.log(rooms);

The first console.log outputs the results properly, but the second one returns the empty array as declared in the first line and prints first. I'm new to Javascript so I'm probably missing something very obvious. Thanks in advance.

a while ago, i got a similar issue with node.js. the problem wasn't that i couldn't fill the global variable within a function. the problem was, that the database query wasn't just finished yet. you need to make sure, you're trying to log rooms when the database call is finished! just look up a tutorial about async functions and callback functions. saves you a lot of time. — David
– David, Commented Feb 19, 2018 at 16:59
Possible duplicate of How do I return the response from an asynchronous call? — Patrick Hund
– Patrick Hund, Commented Feb 19, 2018 at 17:02
You are assuming that the data is fetched from the database synchronously, it's not, it's an async function call, see related question: stackoverflow.com/questions/14220321/… — Patrick Hund
– Patrick Hund, Commented Feb 19, 2018 at 17:03
There are also a closure problem because rooms is declare inside anonymous function and not in global — ekans
– ekans, Commented Feb 19, 2018 at 17:04

zelda11 · Accepted Answer · 2018-02-19 17:01:30Z

2

I think you are recreating another variable because adding "var " before. Have you tried without it?

If it doesn't work, here another posible solution:

global.rooms = [];
global.rooms = result;

https://nodejs.org/api/globals.html#globals_global

answered Feb 19, 2018 at 17:01

zelda11

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Comments

Vector · Accepted Answer · 2019-06-02 14:55:15Z

I don't think it matters for you anymore since it's been roughly a year since you asked this basic question, however maybe someone else that searches for this question might find this information helpful since I had the same problem.

**} else if(req.url === "/api/labeat") {
    res.writeHead(200, {"Content-Type": "application/json"});
    res.end(JSON.stringify(information));**

IMPORTANT When you try to return something to a website beside the variable declarations, when you use res.end();, make sure to turn the result or whatever kind of information you're trying to work with into a string or buffer since the res.end() expects either a string or a buffer. Also specify the Content-Type as I did (application/json).

Omprakash Sharma · Accepted Answer · 2018-02-19 17:06:52Z

0

var rooms = [];

var mysql = require('mysql');

var con = mysql.createConnection({
    host: "localhost",
    user: "MYUSERNAME",
    password: "MYPASSWORD",
    database: "travel"
});

con.connect(function(err) {
    if (err) throw err;
    con.query("SELECT * FROM rooms", function (err, result, fields) {
        if (err) throw err;
        rooms = result;
        console.log(rooms[9]);
    });
});

console.log(rooms);

you missed out to declare rooms as global variable.

answered Feb 19, 2018 at 17:06

Omprakash Sharma

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Comments

David · Accepted Answer · 2018-02-19 17:10:05Z

0

approach using a callback function

rooms = [];

var mysql = require('mysql');

var con = mysql.createConnection({
    host: "localhost",
    user: "MYUSERNAME",
    password: "MYPASSWORD",
    database: "travel"
});


function query(callback) {
  con.connect(function(err) {
      if (err) throw err;
      con.query("SELECT * FROM rooms", function (err, result, fields) {
          if (err) throw err;
          var rooms = result;
          callback(rooms);
          console.log(rooms[9]);
      });
  });
}

function log_it_out() {
  console.log(rooms);
}

query(log_it_out);

answered Feb 19, 2018 at 17:10

David

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