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I do not know exactly but I think that your knowledgeable is enough on this job. Might I asked that you question "How old are you?"...
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Younger than my feet & older than my head.
> 40
re: console.log
What are you using for javascript development?
now you answer my question.
Where are you? Hungary/Turkey/Bulgaria...?
"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 14-Apr-13 6:29am.
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I am living in turkey.. You are older than me 1 year I'm 39 years old. Me hakan and Glad to meet you Thankyou for everything... and issue is not computer literature word.
modified 16-Apr-13 3:05am.
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Hello.
I am a bit older than 40.
the greater than symbol (>) was a clue.
Benim kod tamam mı?
Have a go with Bing translator www.bing.com/translator
Proje kodu Yöneticiler üzgün değil deneyin.
"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 15-Apr-13 16:09pm.
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Are you trying to copy the numbers so that you start another week's draw?
You can't use delete that way for arrays elements.
But your code still doesn't make any sense. There are no shortcuts to comparing *all* elements of one array using one command.
You need to compare each element of arrIns one by one, with oel[i] then copy *that element* to arr
var arr = [];
var c = 0;
for (var i=0; i < oel.length; i++) {
var e = 0;
while (e < arrIns.length) {
if (oel[i] != arrIns[e]) {
arr[c] = oel[i];
c++;
}
e++;
}
}
oel.length = 0;
"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 15-Apr-13 6:07am.
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I just got a 39, in march 02 ..
I think there is a little misunderstanding here. 1 ... 49 number of the remaining elements of the numbers array of the last 3 weeks trying to get the lucky numbers, I was try to do, just as.
if(oel[i] != arrIns[11])
{ oel[i]; };
delete oel[i];
Quote:
that is exactly the last 3 weeks I wanted to get back remaining array of numbers. The number that is 49 - 18 = 31
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[ I am mostly working from my memory, and it is getting difficult. ]
Ben çok benim hafızasından çalışıyorum ve bu zor gidiyor.
Yani, bu ekler:
document.write(arr.toString());
[ Let me know if that output is what you expected. If not you will have to try posting your current code, so that I can see it all in one go, and not have to try an put this all together in my head ]
Bu çıktı beklediğiniz ise bizi. Aksi takdirde, geçerli deftere nakil ki ben her tek seferde görmek ve bir yerine hep birlikte kafamın içinde denemek zorunda değil denemek gerekecek.
"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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document.write(arr.toString());
I have just wanted to do this.
var arrIns = [ "1","21","24","28","42","48",
"12","23","34","36","37","46",
"3","18","19","25","44","45" ];
var arrLot = [];
var arr = [];
var add = 0;
for(var i=1; i<50; i++)
{
arrLot[i] = i;
if(arrLot[i] != arrIns[0])
if(arrLot[i] != arrIns[1])
if(arrLot[i] != arrIns[2])
if(arrLot[i] != arrIns[3])
if(arrLot[i] != arrIns[4])
if(arrLot[i] != arrIns[5])
if(arrLot[i] != arrIns[6])
if(arrLot[i] != arrIns[7])
if(arrLot[i] != arrIns[8])
if(arrLot[i] != arrIns[9])
if(arrLot[i] != arrIns[10])
if(arrLot[i] != arrIns[11])
if(arrLot[i] != arrIns[12])
if(arrLot[i] != arrIns[13])
if(arrLot[i] != arrIns[14])
if(arrLot[i] != arrIns[15])
if(arrLot[i] != arrIns[16])
if(arrLot[i] != arrIns[17])
if(arrLot[i] != arrIns[18])
if(arrLot[i] != arrIns[19])
{ arr[add++] = arrLot[i];}
}
var taken = new Array(49);
for (var i=49; i>0; i--) taken[i]=false;
var lucky = [];
var z = 0;
do{
var q = Math.round(31 * Math.random());
if(!taken[arr[q]]) {
lucky[z] = arr[q];
taken[arr[q]] = true;
document.write(lucky[z] + " ");
z++;
}
}while(z<6);
modified 16-Apr-13 8:52am.
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[ But, isn't arrIns just going to keep getting longer every week? ]
Ama arrIns sadece her hafta uzun elde tutmak için gitmiyor?
[ Can you please explain what purpose is of the arrLot array, does it replace oel from last code you gave?.]
Lütfen arrLot dizi oel verdiği son kodundan yerine yok amaç ne olduğunu açıklayabilir misiniz?
[ One thing I am not understanding, and I have kept asking you information about, is the carry over from one week to another. That's what your code appears to be doing, but without a clear understanding, I cannot begin to grasp the way you intend manage this code any length of time. Please use the translator and give me an over view of the whole lotto process you are trying to achieve. And the methods you are using to develop with. ]
Bir şey ben değil anlamak, ve hakkında bilgi isteyen tutmuş, taşıma üzerinden bir hafta başka olduğunu. Bu ne yapıyor olması kodunuzu görünür, ama net bir anlayış olmadan, ben uzun süre bu kodu yönetmek istediğiniz şekilde kavramak başlar olamaz. Çevirmen ve vermek beni bir yere görüntülemek tüm Loto süreci size ulaşmak için çalışıyoruz kullanın lütfen.
Ve ile geliştirmek için kullandığınız yöntemleri.
"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 16-Apr-13 13:54pm.
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Yes.. You are working really hard issue on!.. This is really hard to comes true in JavaScript language.
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[ I was thinking back to my first reply to this thread of yours: ]
Iyi, ben bu konuyu senin ilk zaman cevap geri düşünüyordum:
Reading-a-text-file-in-JS
[ and thought that maybe you intend to use external file with previous week's data, using string data. Yes?]
ve düşündüğünüz hafta önceki verileri, dış dosya kullanılacak dize verileri kullanarak belki düşündüm.
Evet?
[ Anyhow, keep me informed of any problems. ]
Her neyse, bana herhangi bir sorunları haberdar tutmak.
"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 originally asked you (in your earlier posts) ]
Başlangıçta (önceki mesaj) sordum
Q1. nasıl ne zaman rasgele şanslı numaraları öncekilerin yerine üretilen karar veriyorsunuz?
S2. Yeni piyango numarası + şanslı numaraları (otomatik olarak) geçerli tarihi & zamanında üretilmektedir?
[ The problem as I see it, is that you need some *automatic* method to keep track of the current week's lucky numbers, and discard the oldest. Week-2
I did it that way with my BASIC program 20+ years ago so that I wouldn't need to keep adjusting the code for extra 6 numbers every week. ]
Sorun görüyorum, bazı ihtiyacınız olduğunu * otomatik * yöntem geçerli haftaki şanslı numaralar izlemenize ve en eski atmak için. Hafta-2
Ben fazladan 6 sayı kodunu ayarlama tutmak gerek olmaz bu şekilde temel programım ile 20 + yıl önce her hafta yaptım.
[ is the translator good? ]
Çevirmen iyi mi?
"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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You were right .. has no need to use new delete method to do this.
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[ Overall, how is it looking? OK? ]
Genel olarak, ne arıyor? Tamam?
"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 have left the arrLot array alone for now. But this is my guess at what your code needs to be.
Basically I convert arrIns array to one big string, then compare sub-strings with i. If the value of i isn't found then arr[ ] = i
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++)
{
if (allLucky.indexOf(zeroAdjust(i)) == -1 ) {
arr[count] = i;
count++;
}
}
document.write("length of arr[] ="+ count +"<br>");
I have provided this just to help you move forward (if my guess was correct)
here are some more bits of code for you to look at.
csv = comma,separated,values which is what you get with an array.toString() call.
var arrIns = [ "01","21","24","28","42","48",
"12","23","34","36","37","46",
"03","18","19","25","44","45" ];
var arr_to_csv = arrIns.toString();
document.write(arr_to_csv);
var arr_from_csv = [];
var tmp = arr_to_csv.split(",");
for (var i=0; i < tmp.length; i++) {
arr_from_csv[i] = parseInt( tmp[i]);
document.write(arr_from_csv[i]);
}
The following code has arrIns as a N-dimensional array (3x6) of numbers, but it would complicate the code for comparisons, like the allLucky string compare above.
var arrIns = [ [ 1,21,24,28,42,48],
[12,23,34,36,37,46],
[ 3,18,19,25,44,45]
];
document.write("lucky numbers, one week at a time<br>");
for (var week=1; week < 4; week++) {
document.write("week no: "+ week +"<br>");
for (var i=0; i < 6; i++) {
document.write(arrIns[week-1][i]);
}
}
"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 16-Apr-13 17:35pm.
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Check the code project website for code changes.
"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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Ok... Good coding and good idea.. I will try these codes right now.
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JavaScript controls with date and time (format: dd / mm / yyyy hh: mm: ss) Thank you, and provide the connection site
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Would you care to re-phrase that as a question?
It currently makes no sense at all.
Make it work. Then do it better - Andrei Straut
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Can any one tell me how to hid the form details when I submit form to server.
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The easy way is to use the POST method instead of GET for your form.
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Submit as a POST request such that data is sent in form headers. If you do a GET, you would have to use querystring.
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I see that IE(10), after changing an OPTION's .text property from within the OnChange event handler, the SELECT element becomes unclickable (like disabled) for the user.
HTML fragment to reproduce the behavior:
<form><select size=3 id=sel>
<option>one
<option>two
<option>three
</select>
<textarea name=debugga cols=40 rows=10>
</textarea></form>
<script>
document.forms[0].sel.onchange=function ()
{ var opt = this.options[this.selectedIndex];
this.form.debugga.value += "[click:"+opt.text+"] ";
opt.text += '#';
}
</script>
When using onclick instead of onchange, the problem does not appear.
When opening an alert box after the .text change (uncomment line 'now it works'), the problem does not occur either.
Has anyone an idea how to do a workaround or if Microsoft has already commented on this bug? I cannot find a single article about this issue. Thanks in advance!
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Personally, I'd try fixing the HTML and trying again. (You haven't closed any of the 3 <option> tags)
<form>
<select size=3 id=sel>
<option>one</option>
<option>two</option>
<option>three</option>
</select>
<textarea name=debugga cols=40 rows=10></textarea>
</form>
Make it work. Then do it better - Andrei Straut
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you should test if your suggestion works before giving such an trivial answer. no -- it doesn't make any difference, since the closing tag for the option element is optional by specification.
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