PHP - How To Add A Background Image To This Php Table Code?
Well here is the original code in my script file, and it has a background color Green now:
[/php]$article_content = "<center><table style='color: #FFFF31; background: Green; width: 400px; border: 1px solid #9BDDFF;'><tr><td><i><b><u>Welcome to Mysidia City's fishing pool</b></u></i><br><br><br> Before you actually begin your fishing career, I'd strongly recommend you to doublecheck if you have a fishing rod already since it is apparently required for every fisherman here. <br><br><br><a href='fishing.php?act=1'>Enter</a> the fishing pool now if you have a fishing rod already or <br>Visit this <a href='fishingquiz.php'>Page</a> to take the fishing quiz and earn yourself an old rod to begin with. <td></tr></table><br><center><img src='picuploads/other/lakeofrage.png'></center>";[/php] Now I wish to replace the background color with a background image, what should I do? The image url is provided below: http://www.pokemansion.net/picuploads/other/seadrabg.png Similar TutorialsThis topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=347058.0 Hi, I have This Function For Two backgrounds row of table : Code: [Select] function background() { $bg = ''; // odd $bg2 = 'F3F3F3'; // even if ( $i%2 == 0 ) $bgs = "$bg"; else $bgs = "$bg2"; return $bgs; } PHP CODE IS : Code: [Select] echo "<TABLE>"; $i = 0; while ($f = $db->fetcharray( $r )) // for fetch { background(); // function echo "<tr> <td bgcolor=\"{$bgs}\">$id</td> <td bgcolor=\"{$bgs}\">$name</td> <td bgcolor=\"{$bgs}\">$date</td> </tr> "; $i++; } echo "</table>"; But This Not Worked. What's Problem ? Thanks 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>"; ?> This topic has been moved to Other. http://www.phpfreaks.com/forums/index.php?topic=354424.0 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 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! Hello
I am working on a site and the theme is a responsive theme but the header background is not resizing on mobile and tablets below is a link to the site
http://unitednews.sr/category/news/each category / page has a unique header , this is what I have in css for category news body.category-news #header-wrap { background: url( http://unitednews.sr/wp-content/uploads/2014/10/Header_UnitedNews.jpg); background-repeat:no-repeat; }Adding background size: cover does not work any idea ? Edited by crawling, 13 November 2014 - 11:47 AM. I am using the following css
.content { min-height: 850px; color: #CCC; padding-right: 320px; background-image:url(../images/wheelie.png); background-repeat: no-repeat; background-position: right bottom; }however the background image only shows in the desktop version of safari, I have checked firefox and chrome and mobile safari. I can't see why. This topic has been moved to CSS Help. http://www.phpfreaks.com/forums/index.php?topic=355565.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=331191.0 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 CSS Help. http://www.phpfreaks.com/forums/index.php?topic=349151.0 I am trying to add a background image via a custom style, but it breaks my site. I am a php novice, can someone point out my error please? The echo is being used other places on this page, so I know this code is ok: <?php echo $event->data->thumbnails['medium']; ?> but I am missing something here, this keeps breaking the page: <?php echo '<article data-style="'.$label_style.'" style="background-image:url(\<?php echo $event->data->thumbnails['medium']; ?>\")" class="mec-event-article mec-clear '.$this->get_event_classes($event).'"' . $colorful_bg_color . '>'; ?> Thanks for any help, as I learn! I am trying to copy an image into another image. Code: [Select] $im = imagecreatefrompng("/oldimg.png"); //this is a semi-transparent image where I want the new image to be copied imagealphablending($im, 1); //this seems to remove the black background header('Content-type: image/png'); //outputs old image with a transparent background (YAY!) imagepng($im); imagecopy($im, $newimage, 32, 0, 0, 0, 32, 32); //I want to replace a part of the old image with a new semi-transparent image header('Content-type: image/png'); //outputs image with a black background (noooooooooo!) imagepng($im); Can anyone help me? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=317523.0 I want to get image background removal with core php like(remove.bg website without using any API)
i want to do it with code any guidance how can i do this with programmatically in php. I want to place Custom background image for embedded media player Can anyone help me? Hi
I require a PHP code to get output table.
My Table
Item Qty Date aa-1 2 2014-10-01 aa-2 5 2014-10-01 aa-3 1 2014-10-01 ab-1 2 2014-10-01 ab-2 1 2014-10-01 bb-1 4 2014-10-01 bb-2 3 2014-10-01 bb-3 2 2014-10-01 aa-1 1 2014-10-02 aa-2 2 2014-10-02 aa-3 5 2014-10-02 ab-1 6 2014-10-02 ab-2 1 2014-10-02 bb-1 9 2014-10-02 bb-2 0 2014-10-02 bb-3 4 2014-10-02 aa-1 1 2014-10-03 aa-2 2 2014-10-03 aa-3 5 2014-10-03 ab-1 4 2014-10-03 ab-2 3 2014-10-03 bb-1 1 2014-10-03 bb-2 8 2014-10-03 bb-3 2 2014-10-03 I wrote code as mentioned below. <?php $accounts=mysql_connect("localhost", "root", "") or die("could not connect"); mysql_select_db("shops",$accounts) or die("could not find db!"); if(isset($_POST['search']) && ($_POST['from']) && ($_POST['to'])){ $searchq=$_POST['search']; $searchq=preg_replace("#[^0-9a-z]#i", "" , $searchq); $from=$_POST['from']; $to=$_POST['to']; $dateInput = explode('-',$from); $fdate = $dateInput[2].'-'.$dateInput[1].'-'.$dateInput[0]; $dateInput = explode('-',$to); $tdate = $dateInput[2].'-'.$dateInput[1].'-'.$dateInput[0]; for ($date=$fdate; $date<=$tdate; $date++) { $sql = "SELECT item, SUM(CASE WHEN `date` = '$date' THEN Qty ELSE 0 END) FROM shop WHERE item LIKE '%$searchq%' GROUP BY item"; $query = mysql_query($sql) or die("could not search!"); echo "<table border='1'>"; echo "<tr> <td>Item </td> <td>$date</td> </tr>" ; while ($row=mysql_fetch_array($query)) { echo "<tr> <td>". $row[0] ." </td> <td>" . $row[1] . "</td> </tr>"; } echo "</table>"; } } ?> I am getting result like this as I asked for three days. Item 2014-10-01 aa-1 2 aa-2 5 aa-3 1 Item 2014-10-02 aa-1 1 aa-2 2 aa-3 5 Item 2014-10-03 aa-1 1 aa-2 2 aa-3 5 But I need result like below Item 2014-10-01 2014-10-02 2014-10-03 aa-1 2 1 1 aa-2 5 2 2 aa-3 1 5 5 Can anybody help me to write PHP code to display result as needed. Please help. Thanks in advance. Hi, I am trying to write the code the read a search form however it is not reading my table 'productdbase' to return any results. I can echo from this table so I know it works but so far it just returns "Your search for 'Keyword' returned no results" as per below. Can anyone advise please. Code: [Select] $results = "SELECT 'description', 'fulldescription' FROM 'productdbase' WHERE $where"; Code: [Select] <?php if (isset($_POST['keywords'])){ $keywords = mysql_real_escape_string (htmlentities(trim($_POST['keywords']))); } $errors = array(); if (empty($keywords)) { $errors[] = 'Please enter a search term'; } else if (strlen($keywords)<3) { $errors[] = 'Your search must be three or more characters'; } else if (search_results($keywords) === false) { $errors[] = 'Your search for '.$keywords.' returned no results'; } if (empty($errors)) { search_results ($keywords); } else{ foreach($errors as $error) { echo $error, '</br>'; } } ?> <?php function search_results ($keywords) { $returned_results = array(); $where = ""; $keywords = preg_split('/[\s]+/', $keywords); $total_keywords = count($keywords); foreach($keywords as $key=>$keyword) { $where .= "'keywords' LIKE '%$keyword%'"; if ($key != ($total_keywords - 1)) { $where .= " AND "; } } $results = "SELECT 'description', 'fulldescription' FROM 'productdbase' WHERE $where"; $results_num = ($results = mysql_query($results)) ? mysql_num_rows($results) : 0; if ($results_num === 0) { return false; }else{ echo 'something found.'; } } ?> |