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I wonder what version of opera you guys are running as on my opera 12.12 its the same as in IE9 white screen.
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How can catch the exceptions if my parent and child pages are from different domains.
My agenda is to:
load content using iframes from a cross domain. If the loading fails, catch the exception and retry. If the retry fails too, catch the exception and throw exception.
I know it's easy to do it if the parent and child are from same domain.
Please help!!! Thanks in advance
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Hello everyone,
I'm searching for a way to do a overlay with a loading indicator cover only 1 div on the page.
the reason i want to do this is that i'll have multiple sections in my pages that will do ajax calls to the server to get data and display in fields, most will be calculation results, so i don't want to replace the div's content, just the field value.
in my searchs, i only found solutions for overlays that cover the entire screen, that's good, but i want the rest of the page to be accessible while the ajax request goes on the background.
right now, i've the following code to create a overlay:
var lockOverlayHtml = '<div class="overlayDiv"></div>';
var lockElement = (function (container) {
var element = $(container);
var overlay = $(lockOverlayHtml);
overlay.offset(element.offset());
overlay.height(element.outerHeight(false));
overlay.width(element.outerWidth(false));
overlay.appendTo(document.body);
});
problem is, i can't think in a way to remove only this overlay when the ajax request ends, as it can be used to cover any element on the screen and i can have multiple ones at a time.
what can i use to identify this overlay and remove it?
i thought in something like generating a id using the coordinates, is it a good idea?
also, don't know if i can ask this here, but the overlayDiv class looks like this:
.overlayDiv
{
position: absolute;
opacity: 0.5;
z-index: 1000;
background-color: #606060;
}
this will work even when the browser scrolls? or i need to do something else in that case?
I'm brazilian and english (well, human languages in general) aren't my best skill, so, sorry by my english. (if you want we can speak in C# or VB.Net =p)
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I am showing a multi column table in OnClientShowing event of AutoComplete extender based on the return value (The value is concatenated with | delimeter) from the service method.
While creating dynamic table the TD element is created with value of the code and table display is name.
When user clicks on the blank space the table the OnClientShowing shows the correct value in eventArgs.get_value() but when user clicks on the text area of Table the eventArgs.get_value() returns NULL.
eventArgs.get_text() returns the the name displayed in the table but name is of no use to me. I need TD value. How can I get it.
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please help me to write ...
A) Write the code for the login.html document (Client side) which gives the following interface. When the user clicks on the button you must check if the user enters data or not, if the fields are empty you must present an alert message and do not send the data to the server side. If the user entered login and password, then your program must call the validate.php document.
B) Write the code of the server side validate.php page such that.
1) The validate.php checks if the login is valid or not. A valid login must
start with a letter and followed by any one or more of the characters:_, 0-9, A-Z, a-z
For instance, a1, ab_1, and Z_2d are valid logins while _a1, b*, and
c23_@2 are invalid logins. But, there are no restrictions on the password; the user
can use any character in the keyboard. (hint. You must use the regular expression
in php).
2) The validate.php page reads the login and passwords from a database
named"students" to verify whether the student is a valid user. The "students"
database contains two tables named "login_pssword" and “marks”. The
login_pssword table has three fields “Student_Id”,"login" and"password”.
Check if the user name and passwords exist. If it exists print the message
"Thank you for logging in". If the password or login is not in the database,
display the alert message "wrongpassword and/or login please try again".
The “marks” table has five fields: student_Id, course_name, first_mark,
second_mark, Final_mark.
3)if a valid user logged in, give him/her two choices:
1. Add a new student
2. Remove a student
Design two different forms, one for each choice above to enable the student to perform the choice he/she selected.
the header code :-
<html> <head> <script type = "text/javascript">
Function validate(){
Winow.event.returnvalue = false;
If ((f1.login.value == "") || (f1.password.value ==""))
alert("you must enter your name and password");
else
window.event.returnvalue = true;
}
</script> </head> <body>
<form id ="f1" action = "validate.php" method = "post" onsubmit ="validate()">
User Name <input type ="text" name = "login"/> <br />
Password <input type ="password" name ="password"/> <br />
<input type = "submit" value ="login"/>
</form> </body> </html>
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Here is the complete code without ajax functionality. You didn't mention in the question if you wanted to display result without loading the php page.The login.html could looks like this:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<script type = "text/javascript">
function validate(){
if((f1.login.value == "") || (f1.password.value == ""))
{
alert("you must enter your name and password");
return false;
}
else
{
return true;
}
}
</script>
</head>
<body>
<form id ="f1" action = "validate.php" method="post" onsubmit ="return validate(this);">
User Name <input type ="text" name = "login"/> <br />
Password <input type ="password" name ="password"/> <br />
<input type = "submit" value ="login"/>
</form>
</body>
</html>
The validate.php could look like this:
<?php
$userName = $_REQUEST["login"];
$userPassword = $_REQUEST["password"];
if(preg_match("/^[a-zA-Z][a-zA-Z0-9_]+$/", $userName, $matches) )
{
CheckWithDatabase($userName,$userPassword);
}
function CheckWithDatabase(&$usr, &$pw)
{
mysql_connect("localhost", "userleon", "Pa55word") or die(mysql_error());
mysql_select_db("students") or die(mysql_error());
$data = mysql_query("SELECT login, password FROM login_password") or die(mysql_error());
while($info = mysql_fetch_array( $data ))
{
if( ($usr == $info['login']) && ( $pw == $info['password'] ))
{
echo "Thank you for logging in";
echo "<form><input type=\"radio\" name=\"choice\" value=\"Add a new student\">Add a new student<br/><input type=\"radio\" name=\"choice\" value=\"Remove a student\"/>Remove a student</form>";
}
else
{
echo "<a href=\"javascript:history.go(-1)\">GO BACK</a>";
echo "<script type=\"text/javascript\">alert(\"wrongpassword and/or login please try again\");</script>";
}
}
}
?>
Hope it answers your question 
follow me on twitter @leon_developer
modified 25-Nov-12 20:15pm.
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Well done, you just did this guy's homework for him. Or worse, this guy is actually a paid programmer and you just did his work for which he's getting paid.
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 I am new here.How else should I answer questions?
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Look for ones which are BLATANT homework or "send me codez im clueless", and move on to the next one.
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Being new here and with good knowledge to help people is great and we welcome you to the community.
Like mentioned already by Jamie, we don't like blatant homework answers to be answered with complete code answers like you gave here. The reason is that we want the students who come here for help to learn how to develop things properly. No one learns anything when a blatant gimme-codez homework question is answered in full as you have. The OP is simply going to copy/paste your solution as their homework solution, get the marks, and learn nothing about the process of arriving at that solution. If you are already a professional developer, would you like to hire this individual to work with/for you? I doubt it and I know I don't want to. The way to answer questions like this if you feel so compelled is to lead them down the right path to a solution while still getting them to make the decisions and write the code necessary. That way every one is a winner. Again, cheers and welcome to CP.
I wasn't, now I am, then I won't be anymore.
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Leon Munir ... thanks 4 u
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Hi!
This has been bugging me for quite a while now... I have done many google searches but nothing seems to actually help me.... When i run this script i get the following error:
"SCRIPT5007: Unable to set value of the property 'onmouseover': object is null or undefined"
I don't know how it doesn't work but if i use onmouseover="show()" and obviosuly remake my function heading it works... I can't use the onmouseover="" as apparently its not the best way of doing it... Cna nayone shine some light on where it could be going wrong? Thanks for all your help!
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<style>
#gallery
{
position:relative;
display:block;
width:800px;
height:500px;
margin:20px auto;
text-align:center;
}
#image
{
display:block;
width:800px;
height:500px;
}
#imgBar
{
position:absolute;
display:block;
width:500px;
height:80px;
z-index:2;
top:400px;
left:145px;
display:none;
}
</style>
<script>
var galleryImg = document.getElementById('image');
var bar = document.getElementById('imgBar');
galleryImg.onmouseover = function ()
{
bar.style.display = "block";
};
galleryImg.onmouseout = function ()
{
bar.style.display = "none";
};
</script>
</head>
<body>
<section id="gallery">
<img src="#" id="image"/>
<div id="imgBar">
<img src="img/gallery/thumbnail1.png" class="thumbnail"/>
<img src="img/gallery/thumbnail1.png" class="thumbnail"/>
<img src="img/gallery/thumbnail1.png" class="thumbnail"/>
<img src="img/gallery/thumbnail1.png" class="thumbnail"/>
</div>
</section>
</body>
</html>
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An HTML page is interpreted top to bottom. When your javascript runs (in the head) the HTML has not yet been parsed, and therefore does not exist.
Put everythng into a function
function runMeOnLoad(){
var galleryImg = document.getElementById('image');
var bar = document.getElementById('imgBar');
galleryImg.onmouseover = function ()
{
bar.style.display = "block";
};
galleryImg.onmouseout = function ()
{
bar.style.display = "none";
};
}
And execute onload
<body onload="runMeOnLoad()">
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Brilliant, thank you 
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Put your script code where the "section" tag ends (before the body tag ends) and it'll work fine.
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I have a menu bar with 3 item suppose A,B,C
if i select A means display some content
selecr B means display other content in same page
same for c
(I know masterpage tell another way)
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Hi i want to check if a string contain 17+ or 1+7 or +17 or 17- or 1-7 or -17.Im trying this but its not working.
Can someone help?
var reg=/[-+]\d{2}|\d[-+]\d|\d{2}[-+]/;
if (string.test(reg))
{
document.write("true");
}
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Your regex match not only strings you are testing but also strings containing those. Use ^ and $ to check for whole string
var reg=/^([-+]\d{2}|\d[-+]\d|\d{2}[-+])$/;
'(' and ')' are to make sure that ^ and $ are not subject of OR.
No more Mister Nice Guy... >: |
modified 24-Nov-12 3:16am.
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n.podbielski wrote: match not only not those your test strings but also containing those strings
What happend? Too much whiskey, eggnog, beer or gin?
I've rad your post over and over, but can't put any sense into it. Maybe you could try to rephrase that gibberish or sober up, whatever!
"I had the right to remain silent, but I didn't have the ability!"
Ron White, Comedian
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Liquor on 7:00 AM? I doubt it. It was just too early for world to make sense! People are not make to wake up on such cruel hours! 
No more Mister Nice Guy... >: |
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I had this form validating now it won't even run the first function. Any suggestions?
function formValidation()
{
var uid = document.registration.userid;
var uname = document.registration.username;
var uadd = document.registration.address;
var uzip = document.registration.zip;
var uemail = document.registration.email;
if(userid_validation(uid))
{
if(allLetter(uname))
{
if(alphanumeric(uadd))
{
if(allnumeric(uzip))
{
if(ValidateEmail(uemail))
{
}
}
}
}
return false;
}
function userid_validation(uid)
{
var letters = /^[A-Za-z]+$/;
var uid_len = uid.value.length;
if (uid.value.match(letters))
{
return true;
}
else
{
alert("The First Name can not be empty/must contain only letters");
uid.focus();
return false;
}
}
function allLetter(uname)
{
var letters = /^[A-Za-z]+$/;
if (uname.value.match(letters))
{
return true;
}
else
{
alert("The Last Name can not be empty/must contain only letters");
uname.focus();
return false;
}
}
function alphanumeric(uadd, num)
{
var uadd_len = uadd.value.length;
var numbers = /^[0-9]+$/;
if (uadd.value.match(numbers))
{
return true;
}
else
{
alert("Your phone number must have a 10 digits all numbers");
uadd.focus();
return false;
}
}
function allnumeric(uzip)
{
var numbers = /^[0-9]+$/;
if (uzip.value.match(numbers))
{
return true;
}
else
{
alert("You must fill in your ZIP code with five numeric characters");
uzip.focus();
return false;
}
}
function ValidateEmail(uemail)
{
var mailformat = /^\w+([\.-]?\w+)*@\w+([\.-]?\w+)*(\.\w{2,3})+$/;
if(uemail.value.match(mailformat))
{
alert("Form Succesfully Submitted");
window.location.reload();
return true;
}
else
{
alert("You have entered an invalid email address!");
uemail.focus();
return false;
}
}
chrishoy78@gmail.com
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Your first block of if statements does not seem correct. It can either return false or fall through to the remaining functions.
One of these days I'm going to think of a really clever signature.
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Thanks a ton I finally put all validated that form. Good call.
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therainking78 wrote: if(ValidateEmail(uemail))
{
I don't see a closing '}' for this '{'
Schenectady? What am I doing in Schenectady?
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