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src *= daffod[i]
is shorthand way of using this..
src = src * daffod[i];
it's a multiply combined with an assignment
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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dusty_dex wrote: it's a multiply combined with an assignment
Not in a jQuery selector it isn't! 
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Yep, I know this now. 
Why on Earth do they have to pick operators that are in common use already?
Operator abuse it just ain't healthy. 
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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There's only 30 symbols easy to type on a normal keyboard. We have to share those between different languages. It's confusing at first but over time you'll learn to interpret the various languages without too much thinking about it.
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You can infer from this example:-
$(“a[href*=jquery.com]”)
This selector matches all <a> elements that reference the jQuery site.
That means: it selects all <a> elements that has a value 'jquery.com' in it's attribute 'href'. (* means anywhere in the string, ^ means begins with, $ means ends with)
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Documentation The lucky numbers, I want to compare the values that are encountering the same. Anyone can help on this topic?
.
.
if(arrLot[i] == arrIns[18])
if(arrLot[i] == arrIns[19]){
lucky[add++] = arrLot[i];
if(lucky[add++] = arrLot[i]) {
lucky[add++] = gb;
}
}
}
var gb = getBNS();
function getBNS()
{
var mf = 0;
return Math.floor(mf+=1);
}
do{
if(gb.mf > 1){
lucky[ct].sort(function(a,b){
return a-b});
}
document.write(lucky[ct] + " : " + gb + "<BR>");
ct++;
}while(ct<17);
mesela burada 18 adet sayı elemanlarını kendi aralarında karşılaştırdıktan sonra hangi sayının ne kadar karşılaştığını hesaplayan ve dökümanlayan bir kod yazmaya çalışıyorum. Tanımla problemi var. mesela şöyle bir çıktı veriyor:
<pre> Undefined : 1
Undefined : 1
Undefined : 1
Undefined : 1
var arrIns = [
"1","21","24","28","42","48",
"9","13","14","16","24","49",
"3","18","19","25","44","45"
];
Burada çıktı
24 : 2
1 : 1
21 : 1
olmalı..
modified 18-Apr-13 13:14pm.
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This is not a well framed question! We cannot work out what you are trying to do/ask from the post. Please elaborate and be specific.
Use the "Edit" link to edit your question and provide better information.
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Yes it is not a framed question .. Simple and very basic for programming languages. values with each other documentation as luck would have encountered a very simple problem with numbers
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[ The road to understanding comes from learning. Find a book on Javascript in your own native language and learn it. ]
Anlama yolunda öğrenme gelir. Javascript bir kitap kendi dillerinde ve bunu öğrenin.
I tried searching (google.com.tr) for books and came up with nothing, so you will just have to use this.
w3schools.com/js/ in English
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
modified 18-Apr-13 13:28pm.
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Can I recommend as well that you use something like http://jsfiddle.net/[^] because I honestly am not following what it is that you are trying to do.
as if the facebook, twitter and message boards weren't enough - blogged
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Hey, that's a nice link. It sure beats suggesting node or rhino shell. 
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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JQuery I have no any knowledge about on this language..
Hakan cursed chats dealer
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<pre>Well .. indexof arrIns[i] as a parameter can i assign..
arrIns.indexOf(newArr[j]);
for(var j=0; j<18; j++)
{
document.write( arrIns[j] + " : " + arrIns.indexOf(arrIns[j]) + "<BR>" );
}
Output is:
1 : 0
21 : 1
24 : 2
28 : 3
42 : 4
48 : 5
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Yes, that looks ok. 
Is that what you wanted?
What is the task that you need to solve, is it sort numbers into numerical order?
eg,
"1","3","9","13","14","16","18","19","21","24","25","28"...
Yes/No? Easy, but you can't use arrIns.sort without some preparation first.
Your new message yesterday uncovered more of what you try to do. It was missing in the previous thread.
Was that code I gave you what you wanted, arr[i] != arrIns[] ?
If you do not tell me some code isn't what you want, I have to assume that it is what you wanted.
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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What are the codes?
Hakan cursed chats dealer
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I was referring to this code.
Or did you mean new sorting code?
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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sort code sample pls
Hakan cursed chats dealer
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Small change to previous code, see use of var t;
function zeroAdjust(n) {
var t = "0"+n;
return t.substring(t.length-2,t.length);
}
var arrIns = [ "01","21","24","28","42","48",
"12","23","34","36","37","46",
"03","18","19","25","44","45" ];
var allLucky = arrIns.toString();
var arrLot = [];
var arr = [];
var count = 0;
var t;
for(var i=1; i<50; i++)
{
t = zeroAdjust(i);
if (allLucky.indexOf(t) == -1 ) {
arr[count] = t;
count++;
}
}
document.write("arr[]:"+ arr.toString);
document.write("<br>");
document.write("length of arr[] ="+ count +"<br>");
*NEW*
function sort(ref) {
var a = [];
var i;
function strToUcc( s ) {
if (s.length < 2) {
return String.fromCharCode(s.charCodeAt(0)-48);
} else {
return String.fromCharCode( (s.charCodeAt(1)-48)+((s.charCodeAt(0)-48)*10) );
}
}
function uccToStr( s ) {
var tmp = s.charCodeAt(0);
var lo = tmp % 10;
var hi = (tmp-lo) / 10;
return String.fromCharCode(hi+48)+String.fromCharCode(lo+48);
}
i=0;
while (i < ref.length) {
a[i] = strToUcc( ref[i] );
i++;
}
a.sort();
i=0;
while (i < ref.length) {
ref[i] = uccToStr( a[i] );
i++;
}
}
sort(arrIns);
document.write("Sorted:"+ arrIns.toString());
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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var arrIns = [
"1","21","24","28","42","48",
"9","13","14","16","24","49",
"3","18","19","25","44","45"
];
Was that I wanted to do exactly. These numbers are the numbers that are caught in the combination of more than one slip would like to add Forexmple 24 number is 2 winnig number. i want to print this numbers from inside this array..
Hakan cursed chats dealer
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Sorry I do not understand. 
try and explain by breaking up the problem, and show example of array before and after.
Maybe I can work it out.
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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Try adding this line inside the loop
for(var i=1; i<50; i++)
{
// compare t with all lucky numbers
t = zeroAdjust(i);
if (allLucky.indexOf(t) == -1 ) {
arr[count] = t; // arr[] = "nn"
count++;
}
if (allLucky.indexOf(t) != -1) document.write("Match: "+t);
} // for-loop
If you need to collect all matched lucky numbers then you should create a new array:
var matched = [];
var m = 0;
change that line to
if (allLucky.indexOf(t) != -1) {
matched[m] = t;
m++;
}
"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan
That's what machines are for.
Got a problem?
Sleep on it.
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I'm customer just now... If it does compare to 49 numbers of dimension so 1 adjust as match each one of numbers and it will no return encounter for the same numbers .. JavaScript is not skilled enough
Hakan cursed chats dealer
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