Node js array value accessing issue

Question

I have array and i want to store the query result in array below is the code

var button_settings=[];
con.query("SELECT * from ht_button", function (err, result, fields){
   if (err) throw err;
   button_settings.push(result);
});
console.log(button_settings);

it shows [] i want the stored result.

Answer 1

You are querying the database which is asynchronous in nature thus the console.log does not wait for the result from con.query and executes immediately after that. Thus, you get [] printed. But if you have records in result after the query is completed then you will get the array printed in console if you put the console.log(button_settings); inside the con.query function like this:

Change code to

var button_settings=[];
con.query("SELECT * from ht_button", function (err, result, fields){
   if (err) throw err;
   button_settings.push(result);
   console.log(button_settings);  //console after the query completes
});

So, to make the flow work like a synchronous where you will use the value of button_settings, create a private function like this,

var button_settings=[];
con.query("SELECT * from ht_button", function (err, result, fields){
   if (err) throw err;
   button_settings.push(result);
   _processButtonSettings();
});

//put all your code related to button_settings here
function _processButtonSettings(){
  console.log(button_settings);
  //and more code...
}

Answer 2

connection.query is async, which mean when your call your console.log(button_settings), button_settings has not been set yet.

var button_settings=[];
con.query("SELECT * from ht_button", function (err, result, fields){
   if (err) throw err;
   else setValue(result);
});
function setValue(value) {
  button_settings = value;
  console.log(button_settings);
}

Answer 3

Try to implement Promise here

console.log

fired before the query function callback

Answer 4

The correct way to solve this issue is to write console.log inside the callback. That's what is highly recommended.

Now coming to the part what you want i.e. writing console.log after con.query in the same block and also you don't want console.log in seperate function (this is what I understand from the question and your comments in other answers)

To achieve this you'll have to wrap con.query in a promise and use await to wait till the promise resolves.

var button_settings=[];
await new Promise(function(resolve, reject){
    con.query("SELECT * from ht_button", function (err, result, fields){
        if (err) reject(err)
        button_settings.push(result);
        resolve()
    });
}).catch(function(err){/* handle err */})
console.log(button_settings);

How the only problem is that you can only use await inside an async function. So let's wrap the whole in code an async function

Final code:

async function doStuff(){
    var button_settings=[];
    await new Promise(function(resolve, reject){
        con.query("SELECT * from ht_button", function (err, result, fields){
            if (err) reject(err)
            button_settings.push(result);
            resolve()
        });
    }).catch(function(err){/* handle err */})
    console.log(button_settings);
}
doStuff();

I have not tested this code with actual mysql thing... But I here's a equivalent test :

async function doStuff(){
  var button_settings=[];
  await new Promise(function(resolve, reject){
    setTimeout(function(){
      button_settings = ["foobar"];
      resolve();
    },1000)
  });
  console.log(button_settings);
}
doStuff();

Answer 5

   var res1=[];
   var res2=[];
   var res3=[];
   async function doStuff(){

    await new Promise(function(resolve, reject){
        con.query("SELECT * FROM vm_admin", function (err, result, fields) {
            if (err) throw err;
            resolve();
            res1.push(result);
        });
    });     

    await new Promise(function(resolve, reject){
        con.query("SELECT * FROM vm_banks", function (err, result, fields) {
            if (err) throw err;
            resolve();
            res2.push(result);
        });
    });     

    console.log(res1);  
    console.log('----------------');
    console.log(res2);  

}
doStuff();

I have checked above code and its working fine , thanks all of you for your quick support.