JS Question

Sandeep Mewara

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.

Sandeep Mewara Microsoft ASP.NET MVP 2012 & 2013 [My Blog]: Sandeep Mewara's Tech Journal! [My Latest Article]: HTML5 Quick Start Web Application

dusty_dex

[ 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. :sigh: 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.

Hakan Bulut

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

Dennis E White

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

dusty_dex

to find element number in array, use

arrIns.indexOf("24");

=2

or

var something = "44";
arrIns.indexOf( something ); // not a string

=15

Your explanations are not good. Code is worse still. Do you also want to sort array? :laugh: :laugh: :laugh: ;P :laugh: :laugh: :laugh: You need to understand the difference between "1" and 1, sorting "strings" is alphabetical A-Z a-z

"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.

dusty_dex

Hey, that's a nice link. It sure beats suggesting node or rhino shell. :-D

"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.

Hakan Bulut

<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]) + "
" );
}
Output is:
1 : 0
21 : 1
24 : 2
28 : 3
42 : 4
48 : 5

dusty_dex

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.

Hakan Bulut

What are the codes?

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dusty_dex

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.

Hakan Bulut

sort code sample pls :)

Hakan cursed chats dealer

Hakan Bulut

JQuery I have no any knowledge about on this language..

Hakan cursed chats dealer

dusty_dex

Small change to previous code, see use of var t;

// **NEW** add leading zero to numbers < 10
function zeroAdjust(n) {
var t = "0"+n;
return t.substring(t.length-2,t.length);
}

// these also need leading zero if < 10
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++)
{
// compare t with all lucky numbers
t = zeroAdjust(i);
if (allLucky.indexOf(t) == -1 ) {
arr[count] = t; // arr[] = "nn"
count++;
}
} // for-loop

document.write("arr[]:"+ arr.toString);
document.write("
");
document.write("length of arr[] ="+ count +"
");

*NEW*

function sort(ref) {
var a = [];
var i;

//== conversion function (only to sort) ========================================
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) );
}
}

//== conversion function as "nn" string
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);

}
//=================================================================================

// char-code from "nn"
i=0;
while (i < ref.length) {
a[i] = strToUcc( ref[i] );
i++;
}

a.sort(); // SORT

// back to "nn"
i=0;
while (i < ref.length) {
ref[i] = uccToStr( a[i] );
i++;
}

} // fn-sort

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.

Hakan Bulut

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 :)

dusty_dex

Sorry I do not understand. :confused: 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.

dusty_dex

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.

Hakan Bulut

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 :)

dusty_dex

If you are referring to C++ pointers then you are not going to get very far. If I understood problem better. I could work it out. You are not a very good communicator, it is not a problem with javascript. The problem is English to Turkish translation. But you are not giving enough feedback.

"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.

Hakan Bulut

I know this is possible with Reverse() but IE9.0 is not supported I have to thanks for everything

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dusty_dex

If you are referring to array.reverse, then it is possible, even with IE 9.0 and as far back as IE 3.0. The browser doesn't affect Javascript features, only the revision (version) supported. Javascript features are not controlled by Microsoft, only the implementation. ECMA controls Javascript features. Most of the time I have to figure out (guess) what you want. And I don't always get a reply. a simple :thumbsup: or Yes, ok. or :thumbsdown: or No, not ok. I think your problem is storing numbers in strings, and applying C++ use (ie pointers) to them. Obviously the lack of computer books in your own language means that you could be learning from bad translation where the meaning of concepts is lost. Maybe you should learn English. Then the barriers to understanding won't slow you down.

"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.