Return a variable outside a function in Phonegap DB function

Question

I have a phonegap db functioning JS script below. I want to return the final string variable "feeds" outside the whole function. This only returns "undefined". Please help me with required changes to return the "feeds" variable.`

function getProviders() {
    var feeds = "";
    var db = window.openDatabase("db", "1.0", "desc", 1000000);

    db.transaction(function(tx) {
        var db = window.openDatabase("db", "1.0", "desc", 1000000);
        tx.executeSql("SELECT * FROM `feed_provider`", [], function(tx, results) {
            var len = results.rows.length;

            for (var i = 0; i < len; i++) {
                feeds += results.rows.item(i).id + "|" + results.rows.item(i).name + "|" + results.rows.item(i).status + "|" + results.rows.item(i).feed_url + ",";
            }
        }, sqlerror);
    }, sqlerror2);
    return feeds;
}

@techytree: Is there any reason why you are not using callbacks? — karthick
– karthick, Commented Oct 3, 2013 at 6:47

Jason P · Accepted Answer · 2013-10-02 21:46:09Z

5

I'm going to assume either db.transaction or tx.executeSql is async, in which case I would use a deferred:

function getProviders() {
    var feeds = "";
    var def = $.Deferred();
    var db = window.openDatabase("db", "1.0", "desc", 1000000);

    db.transaction(function(tx) {
        var db = window.openDatabase("db", "1.0", "desc", 1000000);
        tx.executeSql("SELECT * FROM `feed_provider`", [], function(tx, results) {
            var len = results.rows.length;

            for (var i = 0; i < len; i++) {
                feeds += results.rows.item(i).id + "|" + results.rows.item(i).name + "|" + results.rows.item(i).status + "|" + results.rows.item(i).feed_url + ",";
            }
            def.resolve(feeds);
        }, sqlerror);
    }, sqlerror2);
    return def.promise();
}

Call it like this:

getProviders().done(function(feeds) { 
    // do something with feeds
});

answered Oct 2, 2013 at 21:46

Jason P

27k3 gold badges34 silver badges45 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Teshan N. Over a year ago

I have tried this and now it returns an object which shows [object Object] when alerted. How can I use the string in 'feeds' variable through this object? Think that I want to alert the string value in feeds variable. How can I do that?

Collectives™ on Stack Overflow

Return a variable outside a function in Phonegap DB function

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related