pocobueno1388

Try this SELECT value1, value2, (value1 * value2) as total FROM test_oh WHERE total >= 300

$location = $_SERVER['REQUEST_URI']; header("Location: $location");

All you need to do is select the information from the database, and use a while loop to display it all. EX <?php //select info from db $query = mysql_query("SELECT * FROM table"); //display it using while loop while ($row = mysql_fetch_assoc($query)){ echo $row['col']; } ?> You just need to add the table in there. Read these tutorials http://www.tizag.com/phpT/fileopen.php http://www.tizag.com/phpT/fileread.php

Your going to need to explain a lot more.

This is a nice class for this http://htmlpurifier.org/

You need to have some sort of way to know whose profile is being viewed. You can do that by either using a session variable for the user that holds their ID or something of the sort, or you could do it through the URL, such as: www.your-site.com/profile.php?userID=5 Then you can use either one of those variables to create your query. Also, there is no need for a while loop when you only expecting one row from the db.

My guess is your image input type isn't acting as a submit button, and that $_POST value isn't going through, which would give you a blank page because of the if statement returning false. Try this and see if you get the error message. <?php if (isset($_POST['Register'])){ //check email validity function checkEmail($Email) { if(eregi("^[a-zA-Z0-9_]+@[a-zA-Z0-9\-]+\.[a-zA-Z0-9\-\.]+$]", $Email)) { return FALSE; } list($EmailName, $EmailDomain) = split("@",$Email); if(getmxrr($Domain, $MXHost)) { return TRUE; } else { if(fsockopen($Domain, 25, $errno, $errstr, 30)) { return TRUE; } else { return FALSE; } } } //check captcha input match If($_SESSION['security_code'] != $_POST['security_code'] || empty($_SESSION['security_code'])) { Header("location: Login.php?register&SecurityCodeError"); Die; } //check username $Username = mysql_real_escape_string($_POST['Username']); If(strlen($Username) > 20 OR strlen($Username)< 1 OR !(preg_match("/^\w+$/",$Username))){ Header("location: Login.php?register&UsernameError"); Die; } //check password $Password = mysql_real_escape_string($_POST['Password']); If(strlen($Password) > 20 OR strlen($Password) < 5 OR !(ctype_alnum($Password))){ Header("location: Login.php?register&PasswordError1"); Die; } //check re-entry of password $Password2 = mysql_real_escape_string($_POST['Password2']); If ($Password != $Password2) { Header("location: Login.php?register&PasswordError2"); Die; } //check string length of email $Email = mysql_real_escape_string($_POST['Email']); If(strlen($Email) > 50){ Header("location: Login.php?Register&EmailError"); Die; } //calls function to check email validity if(checkEmail($Email) == FALSE){ Header("location: Login.php?Register&EmailError"); Die; } //username exists check $CheckUsername = mysql_query("SELECT Username FROM `usertable` WHERE `Username` = '$Username'"); If (mysql_num_rows($CheckUsername)>0) { Header("location: Login.php?register&UsernameTakenError"); Die; } // email exists check $CheckEmail = mysql_query("SELECT Email FROM `usertable` WHERE `Email` = '$Email'"); If (mysql_num_rows($CheckEmail)>0) { Header("location: Login.php?register&EmailTakenError"); Die; } $IP = $_SERVER["REMOTE_ADDR"]; $Date = date("Y-m-d H:i:s",time()); $Password = md5($Password); $EmailCode = md5(uniqid(rand(), true)); $Query = "INSERT INTO `usertable` (Username,Password,Email,IP,RegisterDate,ActivateCode) Values ('$Username', '$Password', '$Email', '$Country', '$IP', '$Date', '$EmailCode')"; mysql_query($Query) or die(mysql_error()); Unset($_SESSION['security_code']); Header("location: Login.php?Register&Success"); } else { echo 'ERROR: No such variable as $_POST[register]'; } ?>

If I understand what your wanting, try this: <?php $aUsers = array( "Ädams, Egbert", "Zaun, Jillie", ); $sql = "SELECT * FROM `members`"; $show = @mysql_query($sql,$connection) or die(mysql_error()); while ($row = mysql_fetch_array($show)){ $aUsers[] = $row['name']; } //new array with names added echo '<pre>',print_r($aUsers),'</pre>'; ?>

Where are you getting the variable $_SESSION['new'] from? Also, you would need to call session_start() at the top of the page in order to maintain the session.

Try <?php $query = "SELECT nick, (Monday + Tuesday + Wednesday + Thursday + Friday + Saturday + Sunday) AS added FROM table_name ORDER BY added DESC LIMIT 10"; $result = mysql_query($query)or die(mysql_error()); $i = 0; while ($row = mysql_fetch_assoc($result)){ echo $i .'. '.$row['nick'].'<br />'; $i++; } ?>

Here are two pages I have bookmarked http://dryicons.com/free-icons/page/1/ http://speckyboy.com/2008/01/01/top-25-free-icon-resources-for-web-designers/

Instead of this line: $id = basename($_SERVER['PHP_SELF']); I think your wanting this: $id = $_GET['id'];

The edit link needs to look something like this: www.your-site.com/script_name.php?editID=UNIQUE_ID Then, at the top of your script, your code should look something like this <?php //check if they want to edit something if (isset($_GET['editID'])){ $editID = $_GET['editID']; //select the information from the database that they want to edit $query = mysql_query("SELECT user, os, asset_name, mac, ip FROM inventory WHERE unique_field='$editID'"); $row = mysql_fetch_assoc($query); //put the information in text boxes echo "User: <input type='text' name='user' value='{$row['user']}'>"; } ?> Wherever it says UNIQUE_ID is where you want to use the field from the database table thats unique, that way you know which information to select.

can you explain this more? Like this <?php $file = $imagesFolder . $row['companyid'] . '.jpg'; echo $file; That will tell you if your variable is coming out as expected.

-I don't like the header image, it doesn't look very professional -The links under the header image don't look good and are very unorganized looking. A drop down menu seems like it would fit nicely right there. -I would personally take away the "Good Evening" message, that will also allow the middle column to align with the ones to the side of it.

Try <?php include 'config.php'; include 'opendb.php'; $pageUR1 = ereg_replace("/(.+)", "", $_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"]); $curdomain = str_replace("www.", "", $pageUR1); $query = "SELECT COUNT(*) FROM page WHERE page_url='$curdomain'"; $result = mysql_query($query)or die(mysql_error()); $count = mysql_result($result, 0); if ($count > 0) echo $curdomain; else echo "Domain not in database"; ?>

Try this: <?php $file = $imagesFolder . $row['companyid'] . '.jpg'; if (file_exists($file)){ echo "<img src='" . $imagesFolder . $row['companyid'] . ".jpg' /></p>"; } else { echo echo "<img src='icon.jpg' /></p>"; } ?>

echo the query out to see if it's coming out how it's expected to.

Your close the array twice right here: $info_box_contents[] = array('align' => 'left', 'text' => '<font color="' . $font_color . '">' . BOX_HEADING_ASEARCH . '</font>'); ); Just erase the last line -->);

It's because your setting the variable every time the page is loaded, not just when the form is submitted. <?php session_start(); if (isset($_POST['go'])){ $_SESSION['bot'][] = date('U'); } print_r($_SESSION['bot']) ?> <form method="post" action="<?php echo $_SERVER['REQUEST_URI']; ?>"> <input type="submit" value="go" name="go" /> </form>

for removing HTML tags, use the function strip_tags(). Here is an example of limiting words <?php $str = "This is an example string of words This is an example string of words This is an example string of words"; $str = explode(" ", $str); //Replace "10" with how ever many words you want to be displayed for ($i=0; $i < 10; $i++){ echo $str[$i] .' '; } ?>

I went through and commented your mistakes. If this doesn't fix it, tell us what the error is. <?php $con=mysql_connect(******) or die ('A problem has occurred'); //You should use quotes around the DB name mysql_select_db("write"); //Your were missing a single quote on the last variable $query="INSERT INTO articles (title, article, author, topic) VALUES ('$_POST[title]','$_POST[article]','$_POST[author]','$_POST[topic]')"; if (!mysql_query($query,$con)) { die('Problem Encountered'); } echo "Thank You"; //Missing a semi-colon after line mysql_close($con); ?>

On this line if ($emoticon == 1) Where is the variable $emoticon coming from?

It worked fine for me... <?php $message = "Hello "; echo emoticon($message); function emoticon($msg){ $msg = str_replace("", "<IMG src=\".http://www.website.com/members/shoutbox/images/smile.gif\">", $msg); return $msg; } ?>

Take a look at this tutorial http://www.tizag.com/phpT/fileappend.php