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Imoddedu

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About Imoddedu

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    Member
  • Birthday 04/12/1996

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  • Languages
    HTML, CSS, PHP, C++

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    http://starcrafthatchery.com
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  • Location
    USA
  • Interests
    Coding websites and applications. Gaming, friends and music.
  1. Of course I am going to try it, I was just curious though.EDIT: It did not change, maybe it isn't supported?
  2. How is that different from just background? I currently use background right now in the CSS and it shows up
  3. Does anybody know how I can change the background image of the body with jQuery? I've tried something like this but it doesn't seem to work: $("#webDesignLink").hover(function() { $('body').css('background', 'url(img/bgWeb.png)'); });
  4. I'm afraid I don't understand that o.O
  5. Hey there! I have a question, how can I make it so that if a user hovers over a link, then a certain div appears and stays there, and then if the user hovers onto another div, the exising div fades out and a new one fades in?
  6. Imoddedu

    Parsing XML

    Also: any reason why I only get that error with those two specific lines? ->item(0)->childNodes->item(0)->nodeValue;and this line right below it:->item(0)->childNodes->item(0)->nodeValue;
  7. Imoddedu

    Parsing XML

    Hey there! I have a problem when this parses XML...I get these errors:Notice: Trying to get property of non-object in . . .just use this as the feed: http://articlefeeds.nasdaq.com/nasdaq/symbols?symbol=GOOG <?php // Get ticker string and explode $searchstring = $_GET["ticksearch"]; $ticker = explode(":", $searchstring); // Find ticker news feed $tickerfeed = "http://articlefeeds.nasdaq.com/nasdaq/symbols?symbol=" . $ticker[0];?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"><html xmlns="http://www.w3.org/1999/x
  8. I think that might've helped, but I still don't ever get $error set to anything.Here is the whole code: // Validate form startif ($_POST) { $username = $_POST['username']; $password = $_POST['password']; $firstName = $_POST['fname']; $lastName = $_POST['lname']; $passwordb = $_POST['passwordb']; $email = $_POST['email']; $###### = $_POST['######']; $age = $_POST['age']; $country = $_POST['country']; $ip = $_POST['ip']; // validate /* CHECK FOR FILLED FIELDS */ if ( ($username || $password || $firstName || $lastName || $passwordb || $email) == "") { $error = '1'; } else { $error = '0'; }
  9. Hey! I get the error: Notice: Undefined offset: 1whenever I run this code: // Filter and Split Email $splitemail = explode('@', $email); if(checkdnsrr("$splitemail[1]")) { $email = mysql_real_escape_string($email); } else { $error = '2'; } Shouldn't there be a $splitemail[1]?
  10. Imoddedu

    PHP variables

    thanks! short and simple explanation that I understood :)which leads to another question...:how come in this code while($row = mysql_fetch_array($result)) { echo $row['title']; echo "<br />"; echo "<br />"; echo $row['response']; } I tried putting [ echo "<h4>$row['title']</h4>"; but it actually put out $row['title'] the variable name, not it's value.
  11. Imoddedu

    PHP variables

    I see URL's like this all the time:index.php?k=4(the k doesn't mean anything, just representing anything that can go there.what exactly does this mean in terms of PHP? Also, how is it used for things like looking up information in a search box and displaying the result? like ?result=blahblahblah
  12. I think there is a limit on how many of google ads you can have on a page. I'm not sure though...
  13. New problem. Now that I think I got mysql_real_escape_string working, I tried to check to see if the email is valid. I keep recieving this error if I don't put in anything for the "email" field: Notice: Undefined offset: 1 in C:\Program Files (x86)\EasyPHP-5.3.3.1\www\SearchWiDe\contact.php on line 35 here is the code, reworked: <?php$error = 0;$name = "";$email = "";$category = "";$text = "";if(isset($_POST['userName']) && isset($_POST['userEmail']) && isset($_POST['userCat']) && isset($_POST['userText'])) { $name = $_POST['userName']; $email = $_POST['userEmai
  14. That exit(mysql_error()); really helps!$error = 0;$name = "";$email = "";$category = "";$text = "";if(isset($_POST['userName']) && isset($_POST['userEmail']) && isset($_POST['userCat']) && isset($_POST['userText'])) { $name = $_POST['userName']; $email = $_POST['userEmail']; $category = $_POST['userCat']; $text = $_POST['userText']; $ip = $_SERVER['REMOTE_ADDR']; /* $name = mysql_real_escape_string($name); $email = mysql_real_escape_string($email); $text = mysql_real_escape_string($text); */ // Error Messages if($name == "") { $error = 1; } elseif($email
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