PHP - Change Image Background
This creates an image 1500x1500 pixels. But it creates it in black... how can i make it white.
And i have to use imagecreatetruecolor() not imagecreate(). <?php $im = imagecreatetruecolor(1500,1500); // Output the image to the browser header('Content-type: image/jpg'); imagejpeg($im); imagedestroy($im); ?> Thanks! Similar TutorialsThis topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=347058.0 hi, i am trying to generate image gallery. but the problem is that when i upload the png files its thumbnail is generated with black background. JPEG an GIF thumbnail are generated successfully. i want to make background transparent. is there any solution that how i can get rid of it. i am using imagecreatetruecolor($thumbwidth, $thumbheight ); to generate thumbnail I need a button in php to change my background color every 5 presses with a random one, every five presses the color have to change once and need to stay for the next 5 presses. here is my code <head>
<title>click 5 times</title> Hi, I've making a script that's almost a todo list, and i want to have 3 colors to the posts. Post under 7 days old should be green, and thoose older then 7 days yellow and over 14 days should be red. How could i do that? Every post i timestamp when stored in my MYSQL table. Here is my code: Code: [Select] <?php $servername='localhost'; $dbusername='root'; $dbpassword=''; $dbname='store'; connecttodb($servername,$dbname,$dbusername,$dbpassword); function connecttodb($servername,$dbname,$dbuser,$dbpassword) { global $link; $link=mysql_connect ("$servername","$dbuser","$dbpassword"); if(!$link){die("Could not connect to MySQL");} mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error()); } $result = mysql_query("SELECT * FROM henvendelser ORDER by id desc ") or die(mysql_error()); echo "<table cellspacing='12px' cellpaddomg='5px' align='center'>"; echo "<tr> <th>ID</th> <th> Opprettet </th> <th>Navn</th> <th>Telefon</th> <th>Emne</th> </tr>"; while($row = mysql_fetch_array( $result )) { echo "<tr> <td>"; echo $row['id']; echo "</td> <td>"; echo date("d.m.y", strtotime($row["date"])); echo "</td> <td>"; echo "<a href=\"detaljer.php?view=$row[id]\">$row[Navn]</a>"; echo "</td> <td>"; echo $row['Telefon']; echo "</td> <td>"; echo $row['Emne']; echo "</td> </tr>"; } echo "</table>"; ?> Hi, I have looked everywhere for a way to do this and would be grateful for a pointer in the right direction. I want to change the background color of the div with the id of "loggedin", depending on the status. The status currently echos text depending on the value. Code: [Select] <div class="grid_12" id="loggedin"> <div> <ul id="loggedin"> <li><?php global $current_user; get_currentuserinfo(); echo 'Status : ' . $current_user->status . "\n";?></li> <li> <?php if ($current_user->status=="RED") echo "Red info"; ?> <?php if ($current_user->status=="GREEN") echo "Green info"; ?> <?php if ($current_user->status=="BLUE") echo "Blue info"; ?> <?php if ($current_user->status=="YELLOW") echo "Yellow info"; ?> </li> </div> </div> Thanks I have to radio buttons at the login page. Both of them are in group1. The value of the first is 'fancy', the value of the second is 'fast'. Also note that the buttons are on the login page, and i want to change the bg on the main page. Using php I have a dynamically filled select box with the names of images. How can I get an image box on my page and change upon selection? Hi all, I am dealing with an html page in which a dynamically created PNG image is shown with: <img src="createchart.php"> The problem arises when users try to download the image since the default name that appears is just "createchart.php". Is there any way to make the browser suggest something like "chart.png"? Than you guys for your help, Mam Main URL of website home page: www.mysite.com and this shows image "A" in the header. I want to have a marketing campaign which will send people to my site via this link: www.mysite.com/index.php?ref=att When people get to my site via the 2nd link, I want to show image "B" instead of image "A". I put this at top of home page... Code: [Select] require_once("includes/session.php"); if (isset($_GET['ref']) && ($_GET['ref'] == "att")) { $_SESSION['ref'] = "att"; } else { } and this code is in my header include file, a little further down on the page... Code: [Select] <?php if (isset($_SESSION['ref']) && $_SESSION['ref'] == "att") { echo "<span class='phone'><img src='/images/B.png' /></span>"; } else { echo "<span class='phone'><img src='/images/A.png' /></span>"; } ?> Does this look right? because I can't get it to work. It's showing image A regardless of which URL I use. I have to be missing something obvious because I really thought this would be super easy to do. If anyone sees any problems, let me know. But I'll keep testing and hopefully find my mistake. Thanks! Hi,
I want to show one image for the new visitors of my website and another for the returning visitors. What I want to do is that after the visitor reload the page he will see another image instead of the first one.
HI I'm thinking to do something like this , I've seen this in a website ,I'm a php designer but i don't know who something like this can be done . it's a website that sells sofas . we've the ability to change the color and fabric and see the result right away . for example ,this is the default sofa : then I can select a part of it and I can choose the fabric and color , then it make the result : thats it , How can I do that ? What should I do ? I'm really looking to hear from you King Regards Currently, I insert a small thumbnail image at the begging of each row in a set of records from a mysql query. But that image is the same image for each record in that array. Is there a way to conditionally change which image is used, based on a value either in that same table, or a related table. See attached screenshot... and here's my current code.... Code: [Select] <?php $query = "select address from nvc"; $result = mysql_query($query); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $details=''; $details.='<li class="menu"><a href=detail.php?address='.(urlencode($row['address'])).'><img src="thumbs/house.png" /><span class="name">'.$row['city'].'</a>'; $details.='</li>'; echo($details); } ?> Thank you for any input. I have a PHP script that uploads images to a folder on my server (attachments folder). Currently the folder sits within my webroot and is publicly accessible (I have to use chmod 777 due to permissions issue). So, I created the "attachments" folder outside of my webroot (so that it is not publicly accessible), but I do not know how to set the path in the PHP code to upload it to that "attachments" folder outside of the webroot. As you see in the snippet of PHP code below, the code currently uploads the the "attachments" folder within the www (webroot) directory. How do I make it upload to the "attachments" folder OUTSIDE of the www (webroot) directory? foreach($files[$form] as $file){ $str = $file[1]; if (eval("if($str){return true;}")) { $_values[$file[0]] = $_FILES[$file[0]]["name"]; $dirs = explode("/","attachments//"); $cur_dir ="."; foreach($dirs as $dir){ $cur_dir = $cur_dir."/".$dir; if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}} $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]); @unlink($_FILES[$file[0]]["tmp_name"]); }else{ $flag=true; if ($_isdisplay) { //$ExtFltr = $file[2]; //$FileSize = $file[4]; if (!eval("if($file[2]){return true;}")){echo $file[3];} if (!eval("if($file[4]){return true;}")){echo $file[5];} $_ErrorList[] = $file[0]; } } } Hey guys! I have the following php code that grabs variables (and the browsed image) from Flash. //FLASH VARIABLES $Name = $_POST['Name']; $itemNumber = $_POST['itemNumber']; $filename = $_FILES['Filedata']['name']; $filetmpname = $_FILES['Filedata']['tmp_name']; $fileType = $_FILES["Filedata"]["type"]; $fileSizeMB = ($_FILES["Filedata"]["size"] / 1024 / 1000); list($filename, $extension) = explode('.', basename($_FILES['Filedata']['name'])); $filename = $Name; $target = $filename . $itemNumber . "." . $extension; // Place file on server, into the images folder move_uploaded_file($_FILES['Filedata']['tmp_name'], "images/".$target); This works perfect, but what I want to change is the width and height of the uploaded image. Any ideas/suggestions on how this could be done? Thanks in advance!! Cheers! i am new to web designing,
i don't know much about javascript.
<select id="plan" align="center" valign="center"> I have a page that displays various articles. I would like to show a background image for each article. Let's say I have articles for beaches. I would like to have a beach photo. Airplane article, show airplanes, etc. How can I accomplish this. This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=319094.0 I was wondering if there is any way to let visitor chose image background of div and store it in session so every page have same background? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=331191.0 |