Swapping alert with console.log causes code failure

Question

I'm not sure whether this has been asked before, the only similar item I found was this. Changing an alert() to a console.log() causes my code to fail. The 24th line (immediately after xmlhttp.send()) is the line that breaks, and I cannot figure out why.

function readFile()
{
var xmlhttp = new XMLHttpRequest();
var totalBudgeted = 0;
var totalSpent = 0;
var fileJSON = {};
var url = 'gamedata.json';

xmlhttp.onreadystatechange = function()
{
    if( xmlhttp.readyState == 4 )
    {
        //alert( 'XMLHttpRequest status code: ' + xmlhttp.status );
        // I am commenting this out so that I do not lose points for this assignment.  However, FireFox will return a status of 200 when serving a local file.
        //if( xmlhttp.status == 200 )
        {
            fileJSON = JSON.parse( xmlhttp.responseText );
        }
    }
}
xmlhttp.open( 'GET', url, true );
xmlhttp.send();

alert( fileJSON );
for( var key in fileJSON )
{
    if( fileJSON.hasOwnProperty( key ) )
    {
        alert( key + " -> " + fileJSON[key] );
    }
}
// Add up the column totals, and add the 'Total' row to fileJSON.
for( var rowCount in fileJSON )
{
    totalBudgeted += fileJSON[rowCount].budgeted;
    totalSpent += fileJSON[rowCount].spent;
    console.log( '!!!!!!!!!!!!!!!rowCount: ' + rowCount );
}
jsonToHtml( fileJSON );
}

score 6 · Accepted Answer · answered Aug 09 '15 at 04:39

6

putting the alert there obviously gives the asynchronous GET enough time to complete before you process the response

without the alert, your code processes fileJSON before it's even fetched

answered Aug 09 '15 at 04:39

Jaromanda X

47,382
4
58
76

Swapping alert with console.log causes code failure

1 Answers1