We Keep Coding

Javascript variable is undefined but it is defined

I've this method in my global controller object of my JavaScript application. Now I get the error, that the statement self.texts.buttons.disabledFinishedJobs is undefined. But I don't understand that because the console.log() statement outputs the expected value. What can be the reason?
toggleFinishedJobs: function() {
var self = this;
console.log(self.texts.buttons.disabledFinishedJobs[0]);
if (this.disabledFinished) {
$(".status_99").show();
this.disabledFinished = false;
$("btn_finishedJobs").text(self.texts.buttons.disabledFinishedJobs[0]);
} else {
$(".status_99").hide();
this.disabledFinished = true;
$("btn_finishedJobs").text(self.texts.buttons.disabledfinishedJobs[0]);
}
}

Try this:
$("btn_finishedJobs").text(self.texts.buttons.disabledFinishedJobs[0]);
^-Typo error
instead of
$("btn_finishedJobs").text(self.texts.buttons.disabledfinishedJobs[0]);

Looks like you have a typo at the end of your code. self.texts.buttons.disabledfinishedJobs instead of self.texts.buttons.disabledFinishedJobs.

Basic var misunderstanding

I have created this:
var where = function(){
sym.getSymbol("Man").getPosition()
}
console.log(where);
if (where()<=0){
var playMan = sym.getSymbol("Man").play();
} else {
var playMan = sym.getSymbol("Man").playReverse();
}
This is for Edge Animate hence all the syms. I am trying to access the timeline of symbol Man, then if it is at 0 play it. But it isnt working and the reason, I think, is that I have an incomplete understanding of how a var works. In my mind I am giving the variable 'where' the value of the timeline position of symbol 'Man'. In reality the console is just telling me I have a function there, not the value of the answer. I have run into this before and feel if I can crack it I will be a much better human being.
So if anyone can explain in baby-language what I am misunderstanding I would be grateful.
Thanks
S

var where = function () { ... };
and
function where() { ... }
are essentially synonymous here. So, where is a function. You are calling that function here:
if (where()<=0)
However, the function does not return anything. You need to return the value from it, not just call sym.getSymbol("Man").getPosition() inside it.
That, or don't make it a function:
var where = sym.getSymbol("Man").getPosition();
if (where <= 0) ...
The value will only be checked and assigned once in this case, instead of updated every time you call where().

Try
var where = function()
{
return sym.getSymbol("Man").getPosition();
};
Your code wasn't returning anything.

var where = function() {
return sym.getSymbol("Man").getPosition()
}
console.log(where);
if(where()<=0) {
var playMan = sym.getSymbol("Man").play();
} else {
var playMan = sym.getSymbol("Man").playReverse();
}

Why does this function return as undefined?

String.prototype.parse = function(f) {
alert(this.replace(f, ""));
};
var a = "Hello World";
parse.apply(a, ["Hello"]);
Is the code correct?

No, that’s not correct. The function is defined as String.prototype.parse, so it is not available as parse (in fact, parse is undefined).
You could run it like the following:
String.prototype.parse.apply(a, ["Hello"]);
But actually, the reason why you add the function to the prototype of String is that you extend String objects with that function. So you actually should just run the function like this:
a.parse("Hello");
edit:
Oh, and in response to your question title “Why does this function return as undefined?”: The function doesn’t return anything, because you don’t tell the function to return anything. You could for example define it like this to return the replaced string (instead of alerting it):
String.prototype.parse = function(f) {
return this.replace(f, "");
};
And then you could alert the return value of the function:
alert(a.parse("Hello"));

There is no such variable parse defined in your code sample. If you really want to apply the function later on, you should do this:
// Capture function as a local variable first
var parse = function(f) { alert(this.replace(f, "")); };
String.prototype.parse = parse;

Problem with Javascript object and accessing property which exists

I have something like this:
var test = {};
function blah() {
test[2] = 'filled';
}
blah(); // ! Hopefully confusion is now averted..
console.log(test);
//result test -> 2:"filled"
console.log(test[2]);
//result undefined
I don't understand why I'm getting 'undefined' in the second instance when according to the first instance, the property of that object clearly exists!
Does anyone have any ideas?
Thanks
OK, it seems that folk are getting confused as to what context the code exists in, for clarity sake I have now added the call to the blah(). but please refer to the comment under Jeff B's response!
Here is an example of relevant code so to say:
mydb = ..... //gets created here with relevant credentials
var test = {};
mydb.transaction(
function(transaction) {
transaction.executeSql("select * from mytable;", [], function(transaction,result) {
var r = result.rows.item(0);
test[2] = r.title;
}, errorHandler);
});
console.log(test);
//result test -> 2:"the title"
console.log(test[2]);
//result undefined
#Dancrumb
Your mention of the single-threadedness of Javascript gave me an idea, and I tried this:
window.setTimeout(function(){ alert(test[2]); },2000);
and it worked! I got the expected value to alert. Can you suggest how I can get around this without using a 'hack' like that above?

Because you aren't calling blah()?
Also, you want:
var test = [];
or:
var test = new Array();

EDIT
I ran the following code:
mydb = openDatabase('note','','Example',1024);
var test = {};
mydb.transaction(
function(transaction) {
transaction.executeSql("select * from mytable;", [], function(transaction,result) {
var r = result.rows.item(0);
test[2] = r.title;
}, errorHandler);
});
console.log(test);
console.log(test[2]);
in Safari 4.0.5
I got the following:
Object
No Properties
undefined
This is what I would expect to see. The object test does not have any properties assigned to it until the callback from mydb.transaction occurs and, since Javascript is single threaded, this cannot happen before the calls to console.log.
Since you're getting a different outcome, can you outline what browser and what version you are using?

This is pretty clearly an asynchronous issue. The simplest way of getting code to run after you set test[2], is to either put the code right there, or use another callback, and call it after you set test[2].

console.log does not print undefined?

I am new to Javascript. I am trying to understand where "this" is bound to using different examples. I am using console.log to print some values as shown below.
function FuncObject(value) {
this.answer = value;
this.get_answer = function () {
return this.answer;
}
};
var f = new FuncObject(42);
var fanswer = f.get_answer;
console.log(fanswer())
console.log prints "function" instead of "undefined". document.writeln seems to print "undefined" which is the right one because this is bound to the window object which does not have answer. Now printing function confuses me. Now I am wondering what i should be using for logging. I am unable to find an explanation for this.
thanks mohan

Just incase you didn't notice, there's a typo in your posted code of
this.get_answer = funcition ()
With that in mind, I'm not entirely sure of your experience level so let me cover all the bases.
function FuncObject(value) {
this.answer = value;
this.get_answer = function () {
return this.answer;
}
};
var f = new FuncObject(42);
var fanswer = f.get_answer;
console.log(fanswer())
You're setting fanswer = f.get_answer where f.get_answer is a function, so as such it sets fanswer to the function equivalent of this.get_answer.
If you want the return value of f.get_answer you need to call f.get_answer(), which returns 42.
With what you put, console.log(fanswer()) does print undefined as expected.
If you simply do console.log(fanswer) it records it as function, also as expected.
I'm not sure why you would receive function as you stated in your question, because I definitely do not, jsbin.