Closures and callbacks

Question

I'm working on a database function and need to return results from fetchAll so I can use it elsewhere in my code but am not sure how to do it:

function fetchAll(sql,params,tableref){
  var fields = new Array();
  var resultout = new Array();

  for (i in tableref){     
    fields.push(i);  
  }       

  getResults(sql,params,fields,function(results){
     // I WANT TO RETURN RESULTS
     resultout.push(results);       
  });          

// TO HERE SO I CAN RETURN from Fetchall
console.log(resultout);
}

function getResults(query,params,fields,callBack){
  var result = new Array(); 
  thisDB.transaction(function (tx) {
    tx.executeSql(query,params, function(tx, rs){
       for(var i=0; i<rs.rows.length; i++) {
          var row = rs.rows.item(i);
          var rowresults = new Object();

          for (x=0;x<fields.length;x++){
            rowresults[fields[x]] = row[fields[x]];      
          }

          result.push(rowresults);
       }
       callBack(result);
    });
  }); 

return result;  
}

I think i'm missing something obvious.

Thanks

Antony

You should place the console.log call inside the callback (below the resultout.push), if I'm understanding you correctly. — mkey, Dec 20 '12 at 15:59
Can you please describe what exactly your problem is? What is this code supposed to do exactly and what does it do instead? — Philipp, Dec 20 '12 at 15:59
The short answer is *don't* have `fetchAll` return a value. Instead, have `fetchAll` accept a callback that gets called in the callback of `getResults`. You can't have a function return a value synchronously when it relies on an asynchronous method to get that value. — apsillers, Dec 20 '12 at 16:18

score 0 · Answer 1 · answered Dec 20 '12 at 16:03

0

If I'm understanding correctly your question, to see the result you need this

function fetchAll(sql,params,tableref){
    var fields = new Array();
    var resultout = new Array();

    for (i in tableref){     
        fields.push(i);  
    }       

    getResults(sql,params,fields,function(results){
        // I WANT TO RETURN RESULTS
        resultout.push(results);
        console.log(resultout);
    });
}

The callback will be executed "after" so basically in your example you'd see an empty result. This is because of the asynchronous nature of the callback.

answered Dec 20 '12 at 16:03

mkey

535
2
12

Wow - thanks for your responses. What I'm trying to do is basically get fetchAll to return an object of rows/fields from the database. If theres a better way to do it - please say. – user1018244 Dec 20 '12 at 16:26
From the top of my head, this looks OK, you just need to be aware of the asynchronous nature of your function. – mkey Dec 20 '12 at 16:43

Pau Fracés · Answer 2 · 2012-12-20T16:42:28.617

I think you are trying to move from an asynchronous request to a synchronous one

Is a good topic, you can find many posts and solutions addressing this on the web

You have several options:

Use callback instead return in your function: Your function must receive another param (the callback) and call that callback passing to it the value that you want to return

function fetchAll (sql, params, tableref, callback) {
  var fields = new Array();

  for (i in tableref) {     
    fields.push(i);  
  }       

  getResults(sql, params, fields, function (results) {
     // I WANT TO RETURN RESULTS
     callback(results);       
  });
}

Then you can log the results like this:

fetchAll(sql, params, tableref, function (results) { console.log(results); });

Find a synchronous version of getResults, often there is one function for that, maybe getResultsSync ?
With node you can use https://github.com/laverdet/node-fibers or http://github.com/maxtaco/tamejs for converting from asynchronous to synchronous style (I'm using tamejs for that purpose currently and it is great!)
And others I don't know enough to talk about

Closures and callbacks

2 Answers2