PHP - Displaying Information When Image Is Clicked
Hi
I'm trying to create a page where there is a grid of images and if you click on one of those images it will expand. Once the image has expanded and you click on the i icon that is below it, the i icon will display the information about that image. This isn't like a popup or anything like that, because when you click on the image and it expands, it expands in the same spot and it doesn't popup. Here is the html $str.='<div class="portfolio-banner-wrapper'.$langClass.'" style="position: relative;">'; $str.= '<a class="portfolio-arrow-left'.$langClass.' portfolio-navigate'.$langClass.'" href="javascript:void(0);"><img src="images/left-arrow.png" onmouseover="this.src=\'images/blue-left-arrow.png\'" onmouseout="this.src=\'images/left-arrow.png\'" /></a>'; $str.= '<a class="portfolio-arrow-right'.$langClass.' portfolio-navigate'.$langClass.'" href="javascript:void(0);"><img src="images/right-arrow.png" onmouseover="this.src=\'images/blue-right-arrow.png\'" onmouseout="this.src=\'images/right-arrow.png\'"/></a>'; $str.= '<div class="portfolio-banner-inner">'; $str.= $this->getPortfolioTest(); $str.= '</div>'; $str.='</div>'; $str.='<div class="overlay-bg'.$langClass.'">'; $str.= '<div class="additional-navigation-wrapper'.$langClass.'">'; $str.= '<div class="info'.$langClass.'" style="display: none;">'; $str.= '<a class="border-bottom-white padding-level-one inactive additional-nav-info1" href="javascript:void(0);">'; $str.= '<img class="icon" src="images/i_icon.png" />'; $str.= '<img class="nav-arrow no-action floatright" src="images/nav-arrow-white.png" />'; $str.= '<span class="clearboth"></span>'; $str.= '</a>'; $str.= '<div class="additional-nav-info-wrapper1">'; $str.= $this->i_icon(); $str.= '</div>'; $str.= '</div>'; $str.= '</div>';Here is the php function getPortfolioTest(){ global $_PRODUCTS_TABLE, $_HTTP_ADDRESS, $_PRODUCTS_IMAGES_DIR; $i=0; $cpt = 1; $str = ''; $whichCount = 1; $jobSearch=''; $query = mysql_query("SELECT * FROM $_PRODUCTS_TABLE WHERE `active`='1' AND `image` LIKE '%.%'".$jobSearch." ORDER BY `client` ASC"); $combineArr = mysql_num_rows($query); while( $result = mysql_fetch_object($query) ){ $product = new Product($result->id); $product->setFromDatabase(); $linkOut = getSEOLink($product->id); $target = ""; if(trim($product->linkout) != ""){ $linkOut = $product->linkout; $target = ' target="_blank"'; } if($whichCount == 1){ $portfolioClass="portfolio-active"; $style = "position: absolute; left:0%; top:0; width:100%;"; }else{ $portfolioClass="portfolio-inactive"; $style = "position: absolute; left:-100%; top:0; width:100%;"; } if($whichCount == 1){ $str.='<div id="portfolio-slide'.$i.'" class="portfolio-slide '.$portfolioClass.'" style="'.$style.'">'; $str.= '<div class="portfolio-slide-inner">'; $str.= '<div class="portfolio-banner-content portfolio-banner-left">'; $str.= '<div class="portfolio-banner-header">'; $str.= '</div>'; $str.= '<div class="portfolio-banner-copy">'; $str.= '<ul id="gallery">'; } $str.= '<li>'; $str.= '<a'.$target.' href="javascript:void(0);">'; $str.= '<img src="'.$_HTTP_ADDRESS.'products_images/'.$result->image.'">'; echo "this is getPortfolioTest"; $str.= '</a>'; $str.= '<span>'; $str.= '<h3>'.$result->name.'</h3>'; $str.= $result->description; $str.= '</span>'; $str.= '</li>'; if($whichCount % 9 == 0 && $whichCount < $combineArr){ $i++; $str.= '<div class="clearboth"></div>'; $str.= '</ul>'; $str.= '</div>'; $str.= '</div>'; $str.= '</div>'; $str.='</div>'; $str.='<div id="portfolio-slide'.$i.'" class="portfolio-slide '.$portfolioClass.'" style="'.$style.'">'; $str.= '<div class="portfolio-slide-inner">'; $str.= '<div class="portfolio-banner-content portfolio-banner-left">'; $str.= '<div class="portfolio-banner-header">'; $str.= '</div>'; $str.= '<div class="portfolio-banner-copy">'; $str.= '<ul id="gallery">'; } if($whichCount == $combineArr){ $str.= '<div class="clearboth"></div>'; $str.= '</ul>'; $str.= '</div>'; $str.= '</div>'; $str.= '</div>'; $str.='</div>'; } $whichCount++; $cpt++; } return $str; } function i_icon(){ global $_PRODUCTS_TABLE, $_HTTP_ADDRESS, $_PRODUCTS_IMAGES_DIR; $str = ''; $i = 0; $query = mysql_query("SELECT * FROM $_PRODUCTS_TABLE WHERE `active`='1'"); while($result = mysql_fetch_object($query)){ $str.= '<div class="additional-nav-info-inner'.$i.' overlay-bg" style="display:none;">'; $str.= '<ul>'; $str.= '<h3>'.$result->name.'</h3>'; $str.= '<p>'; $str.= $result->overview; $str.= '</p>'; $str.= '</ul>'; $str.= '</div>'; $i++; } return $str; }Here is the Jquery function galleryInit(){ $('#gallery li').hover( function(){$('span',this).slideToggle('fast');}, function(){$('span',this).slideToggle('fast'); }); $(".portfolio-banner-inner li").click(function(e){ if($(".activeExpand").length > 0){ $(".portfolio-active").css({"left":"0%"}); $(".portfolio-active").prevAll().css({"left":"-100%"}); $(".portfolio-active").nextAll().css({"left":"100%"}); $(".portfolio-banner-inner li").removeClass("inactiveExpand").removeClass("activeExpand").removeAttr("style").find("img").removeAttr("style"); $(".portfolio-arrow-left-scroll").addClass("portfolio-arrow-left").removeClass("portfolio-arrow-left-scroll"); $(".portfolio-arrow-right-scroll").addClass("portfolio-arrow-right").removeClass("portfolio-arrow-right-scroll"); return false; } $this = $(this); $(".portfolio-banner-inner li").addClass("inactiveExpand") $this.removeClass("inactiveExpand").addClass("activeExpand"); $(".portfolio-slide").css({"left":"0"}) $this.parents(".portfolio-slide:first").prevAll().find(".portfolio-slide-inner li").css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"-100%"}).find("img").css({"width":"952px","height":"502px"}); $this.parents(".portfolio-slide:first").nextAll().find(".portfolio-slide-inner li").css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"100%"}).find("img").css({"width":"952px","height":"502px"}); $this.siblings().prevAll().css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"-100%"}).find("img").css({"width":"952px","height":"502px"}); $this.siblings().nextAll().css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"100%"}).find("img").css({"width":"952px","height":"502px"}); $this.css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"0"}); $this.find("img").animate({ width: '952px',//what the width of the image to be expanded is height: '502px'//what the width of the image to be expanded is }, 200); $(".portfolio-arrow-left").addClass("portfolio-arrow-left-scroll").removeClass("portfolio-arrow-left"); $(".portfolio-arrow-right").addClass("portfolio-arrow-right-scroll").removeClass("portfolio-arrow-right"); }); } function imageSlider(direction){ $activeExpandWhere = $(".activeExpand"); if($(".activeExpand").css("left") != "0px") return false; if(direction == "right"){ $(".inactiveExpand").css({"left":"100%"}); $(".activeExpand").animate({"left":-100+"%"},500,"easeInCubic",function(){ $(this).removeClass("activeExpand").addClass("inactiveExpand"); }); if($(".activeExpand").next("li").length == 0){ if($activeExpandWhere.parents(".portfolio-slide:first").next().length == 0){ $(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive"); $(".portfolio-slide:first").removeClass("portfolio-inactive").addClass("portfolio-active"); $(".portfolio-slide:first").find(".portfolio-slide-inner li").eq(0).animate({"left":0+"%"},500,"easeInCubic",function(){ $(this).removeClass("inactiveExpand").addClass("activeExpand"); }); return false; } $(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive"); $activeExpandWhere.parents(".portfolio-slide:first").next().removeClass("portfolio-inactive").addClass("portfolio-active"); $activeExpandWhere.parents(".portfolio-slide:first").next().find(".portfolio-slide-inner li").eq(0).animate({"left":0+"%"},500,"easeInCubic",function(){ $(this).removeClass("inactiveExpand").addClass("activeExpand"); }); return false; } $(".activeExpand").next("li").animate({"left":0+"%"},500,"easeInCubic",function(){ $(this).removeClass("inactiveExpand").addClass("activeExpand"); }); return false; } $(".inactiveExpand").css({"left":"-100%"}); $(".activeExpand").animate({"left":100+"%"},500,"easeInCubic",function(){ $(this).removeClass("activeExpand").addClass("inactiveExpand"); }); if($(".activeExpand").prev("li").length == 0){ if($activeExpandWhere.parents(".portfolio-slide:first").prev().length == 0){ $(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive"); $(".portfolio-slide:last").removeClass("portfolio-inactive").addClass("portfolio-active"); $(".portfolio-slide:last").find(".portfolio-slide-inner li").eq(($(".portfolio-slide:last").find(".portfolio-slide-inner li").length-1)).animate({"left":0+"%"},500,"easeInCubic",function(){ $(this).removeClass("inactiveExpand").addClass("activeExpand"); }); return false; } $(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive"); $activeExpandWhere.parents(".portfolio-slide:first").prev().removeClass("portfolio-inactive").addClass("portfolio-active"); $activeExpandWhere.parents(".portfolio-slide:first").prev().find(".portfolio-slide-inner li").eq(0).animate({"left":0+"%"},500,"easeInCubic",function(){ $(this).removeClass("inactiveExpand").addClass("activeExpand"); }); return false; } $(".activeExpand").prev("li").animate({"left":0+"%"},500,"easeInCubic",function(){ $(this).removeClass("inactiveExpand").addClass("activeExpand"); }); } $(document).ready(function(e) { galleryInit(); $(".additional-nav-info1").click(function(){ if($(".additional-nav-info-inner1").css("display") == "block"){ $(this).removeClass("active"); $(this).addClass("inactive"); $(".additional-nav-info-inner1").stop().slideToggle(250); } else { $(this).removeClass("inactive"); $(this).addClass("active"); $(".additional-nav-info-inner1").stop().slideToggle(250); } }); $(".additional-nav-info-inner1 a").hover(function(){ $(this).find("img").eq(0).fadeOut(250); $(this).find("img").eq(1).fadeIn(250); }, function(){ $(this).find("img").eq(0).fadeIn(250); $(this).find("img").eq(1).fadeOut(250); }); } Similar TutorialsHi guys, hopefully someone can help walk me through this.
Basically I want to:
1) Send parameters to remote sites form
2) Receive information/failsafe
3) display data
Sounds easier than it is, can someone explain to a php newb how this can be done?
Site will be given appropriate credit.
I want to make the news feed on my website easier to maintain. I just created a form that I will use to add the information. The form saves the files that it creates as "entry-date-time.txt" and places it in a folder for the year and a sub folder for the month. Such as "2011\09". I have displayed data from a single flat file before, but I am unsure how to go about displaying the data from multiple files. Can anyone show me an example? I want to be able to paginate the data, displaying a number of updates per page from newest to oldest. I don't have MySQL access on the server or I'd just use that, instead. i need to display the fullname and email of the logged in user. <?php session_start(); mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("db_register") or die(mysql_error()); $query = "SELECT` fullname`, `email` FROM `members` WHERE `username`='".$_SESSION['user']."' LIMIT 1"; if($doQuery = mysql_query($query)) { if(mysql_num_rows($doQuery)) { $user = mysql_fetch_assoc($doQuery); print_r($user); } else { echo 'No result returned for the query: '.$query; } } else { echo 'The following query failed: '.$query; } $id = $user['id']; $fullname = $user['fullname']; $email = $user['email']; } ?> <br> Fullname : <?php echo $fullname; ?> <br> Email : <? echo $email; ?> ?> HELP please . I feel like a moron asking this question as it seems to be one of the most common things done with php but I cannot find a tutorial (probably because I don't know the correct wording to search under) on this specific thing.. Anyhow. I'm spitting out a list of the Titles of my test table like so (and its working as expected): <?php $posting_sql = "SELECT * FROM postings"; $posting_results = (mysqli_query($cxn, $posting_sql)) or die("Was not able to grab the Postings!"); while($posting_row = mysqli_fetch_array($posting_results)) { echo "<li><a href='posting_details.php'>$posting_row[title]</a></li>"; } ?> Now as you can see: <a href='posting_details.php'> I am calling a new page, and that page contains this: <h2><?php echo "$posting_row[title]"; ?></h2> <p><?php echo "$posting_row[description]"; ?></p> It's no surprise it's not working (with many variations etc), but I am only familiar with using <form> GET or POST, and these of course are not form elements, so I cant seem to call them the same way. My suspicion (and from the research data I could find) is that I need to pass values in my link: <a href='posting_details.php'> ... I could not get it to work, but I was trying variations like: <a href='posting_details.php?get[title&get[description]]'> but it seems like I am screwing things up even more by doing this.. Anyhow. If anyone could show me a tutorial that covers this specifically or some suggestions on the best approach to this would be much appreciated.. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=317810.0 Hi, I'm not super fluent in PHP or whatnot, but I currently have a website set up with an index.php to load the content pages within it. As part of the template, it consists of a header image, links, a photo, and a video. The content of each page loads within the specified area below the "template". Here's my question: My client wants to see a specific picture when they click one of hte links. For example, if they clicked on spot removal, the image beside the navigation should show spot removal and stay there on the index page while the content page, spot removal is shown. Does this make sense? I need help and fast. Screenshot included in attachment. And like I said, I'm not super fluent in PHP/MYSQL, so be very detailed when talking about what to do. I'd appreciate the help. Thanks, Please i need help on popup message when an image is clicked. Some one should help me with the code to place in my html document. Hi Everyone, I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening..... Here is MY code.......... <?php session_start(); $sid = $_SESSION['id']; $_SESSION['imageDispCnt'] = 0; $myQuery = "SELECT * from image"; $conn = mysql_connect("localhost","User","Passwd"); mysql_select_db("database_Name",$conn); $result = mysql_query($myQuery); $img =Array(); $id =Array(); $i =0; $imagepath = 'http://localhost/images/'; while ($row = mysql_fetch_array($result)) { $img[$i] = $imagepath.$row['img_name']; $id[$i] = $row['imageid']; $i = $i + 1; } ?> </head> <script language="JavaScript1.2"> var scrollerwidth='800px'; var scrollerheight='600px'; var scrollerbgcolor='white'; var pausebetweenimages=3200; var s; var sec; var d; var j; var imgid; var milisec = 0; var seconds = 0; var flag = 1; var ses_id = '<?php echo $sid;?>'; var count = 0; var i = 0; var imgname; var imgid; var k =0; var slideimages=new Array(); var img_id = new Array(); var index; <?php $l =0; $count = array(); $j = rand(0,199); while($l < 200) { while(in_array($j, $count)) { $j = rand(0,199); } $count[$l] = $j; $l++; }?> <?php $k = 0; for($k = 0;$k<count($count);$k++){ ?> index = <?php echo $k;?>; <?php $indx = $count[$k];?> if(index == 0){ slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>'; img_id[0] = '<input type="hidden" value="0" id="imgId" />'; } else if(index > 0) { slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>'; img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />'; } <?php } ?> Can Any one plese help me Appreciate your help... Thanks I am using this code Code: [Select] <?php echo IMAGES_HEADER . "header_02.jpg"; ?> Instead of displaying the actual image on my site I am getting the path of the image. It is displaying "images/header/header_02.jpg" instead. Thank You in advance i have a db which store jpg image thru the upload prg code in php.But the image is not displayed properly in the <img> and also when i echo the blob data. how to correct this code.i am pasting the code below $result = mysql_query("select cover from Movies where movie_id=4"); $row = mysql_fetch_row($result); $data = base64_decode($row[0]); $im = imagecreatefromstring($row[0]); imagejpeg($im); header('Content-type: ' . $mime); // 'image/jpeg' for JPEG images echo $data; <img src=<?php echo $data;?>> Now I am trying to display the images from my table and I have almost got it working, except for one small thing --- it seems to be displaying all the records, but instead of displaying the right image for each record, it's displaying the same image across all the records. Can anyone tell me what I have done wrong? Code: [Select] this is the gallery <?php $dbhost = 'localhost'; $dbuser = 'webdes17_lizkula'; $dbpass = 'minimoon'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'webdes17_jewelry'; mysql_select_db($dbname); $all_records = "SELECT * FROM gallery"; $all_records_res = mysql_query($all_records); $image = mysql_result($all_records_res, 0, 'image'); while($nt=mysql_fetch_array($all_records_res)){ echo "<img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/$image>"; } ?> I'm guessing I have the $image variable in the wrong place, but when I tried placing it within the while statement, the page never loaded, and instead acted like it was loading forever. What am I doing wrong? Here is the link to the page so you can see what is happening: http://webdesignsbyliz.com/wdbl_wordpress/test-submit/ Hi guys, I have followed a tutorial and made a members only area using sessions. The user can upload an image and which gets renamed as their username. I was hoping to display all the users images that are logged in. I know how to do it with a single image by just setting the img src as the session username but I don't know how I would display multiple images if more than one person were logged in. Is it even possible? I must be doing something incredibly daft, because I'm incredibly new at this. I have an image stored in a DB under a table called 'images' and I want to display it on my website but instead of that image I get the error: Warning: Cannot modify header information - headers already sent by (output started at /home/... This is how I'm trying to achieve it. Any ideas where I'm doing wrong? Thanks. Code: [Select] <?php $user="###"; $password="###"; $database="###"; $con = mysql_connect(localhost,$user,$password); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db($database, $con); ?> <!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/xhtml"> <head> <title>MySite</title> </head> <body> <div id="container"> <?php include("../navbar.php"); ?> <div id="left-content"></div> <div id="right-content"> <?php $item = $_GET['item']; $query = "SELECT * FROM main WHERE Ref='$item'"; $result = mysql_query($query) or die("Oops" .mysql_error()); $row = mysql_fetch_array($result,MYSQL_BOTH) or die("Oops" .mysql_error()); extract($row); $query2 = "SELECT image FROM main WHERE Ref='$item'"; // the result of the query $result2 = mysql_query($query2) or die("Invalid query: " . mysql_error()); header("Content-type: image/jpg"); echo mysql_result($result2, 0,'image'); echo "<p><strong>Name: </strong>".$FirstName." - ".$SecondName."</p>"; mysql_close($con); ?> </div> <br /> <?php include("../footer.php"); ?> </div> </div> </body> </html> MOD EDIT: [code] . . . [/code] BBCode tags added. Hi, I am uploading an image to a folder and I need to display the uploaded image in a new page. This is my code for the upload: Code: [Select] <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","2000000"); //This function reads the extension of the file. It is used to determine if the // file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no // error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['upload'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and // will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Tipo de imagen no permitido.</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>La imagen es demasiado grande.</h1>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images //folder) $newname="banners/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Error al subir la imagen.</h1>'; $errors=1; }}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h1>La imagen se ha cargado correctamente</h1>"; } $query = "INSERT INTO banner_rotator.t_banners (banner_path) ". "VALUES ('$newname')"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); //echo "<br>Files uploaded<br>"; header("Location: PC_cropbanner.php"); }Sowhat I need to do is to display the uploaded image in PC_cropbanner.php. How can I do this? I just can't get anything right today! Now I am trying to have a list of images, and when the user clicks on the image, the next page will display the image along with other fields for that record. I am sending the id through the hyperlink to the next page, and I have echoed it to ensure it's comign through, but I cannot get anythign to display ont he next page. What am I doin wrong? Here is the link to the page. If you click on one of the images, you 'll see that the next page is empty: http://webdesignsbyliz.com/wdbl_wordpress/test-submit/ Here is my code: Code: [Select] this is the gallery <?php $dbhost = 'localhost'; $dbuser = 'user'; $dbpass = 'pass'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'jewelry'; mysql_select_db($dbname); $all_records = "SELECT * FROM gallery"; $all_records_res = mysql_query($all_records); $image = mysql_result($all_records_res, 0, 'image'); $id = mysql_result($all_records_res, 0, 'id'); while($nt=mysql_fetch_array($all_records_res)){ echo "<a href=http://webdesignsbyliz.com/wdbl_wordpress/test-display/?id=" .$nt['id']." ><img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/".$nt['image']." width=133 height=86></a>"; } ?> display page: Code: [Select] <?php $id = $_GET['id']; $dbhost = 'localhost'; $dbuser = 'user'; $dbpass = 'pass'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'jewelry'; mysql_select_db($dbname); $all_records = "SELECT * FROM gallery WHERE id = $_GET[id]"; $all_records_res = mysql_query($all_records); $image = mysql_result($all_records_res, 0, 'image'); $id = mysql_result($all_records_res, 0, 'id'); while($nt=mysql_fetch_array($all_records_res)){ echo "<img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/".$nt['image']." width=133 height=86></a>"; } ?> What an I doing wrong this time?? :-( I've taken this eg from php manual site for displaying txt on image wonder why it is not working Code: [Select] <?php // Create a 300x150 image $im = imagecreatetruecolor(300, 150); $black = imagecolorallocate($im, 0, 0, 0); $white = imagecolorallocate($im, 255, 255, 255); // Set the background to be white imagefilledrectangle($im, 0, 0, 299, 299, $white); // Path to our font file $font = './arial.ttf'; // First we create our bounding box for the first text $bbox = imagettfbbox(10, 45, $font, 'Powered by PHP ' . phpversion()); // This is our cordinates for X and Y $x = $bbox[0] + (imagesx($im) / 2) - ($bbox[4] / 2) - 25; $y = $bbox[1] + (imagesy($im) / 2) - ($bbox[5] / 2) - 5; // Write it imagettftext($im, 10, 45, $x, $y, $black, $font, 'Powered by PHP ' . phpversion()); // Create the next bounding box for the second text $bbox = imagettfbbox(10, 45, $font, 'and Zend Engine ' . zend_version()); // Set the cordinates so its next to the first text $x = $bbox[0] + (imagesx($im) / 2) - ($bbox[4] / 2) + 10; $y = $bbox[1] + (imagesy($im) / 2) - ($bbox[5] / 2) - 5; // Write it imagettftext($im, 10, 45, $x, $y, $black, $font, 'and Zend Engine ' . zend_version()); // Output to browser header('Content-Type: image/png'); imagepng($im); imagedestroy($im); ?> Hi all, I have this script below where I am trying to display a default image if an image can not be found. For some reason though it is not working. <?php foreach (glob('./aircraft/' . $rowX['reg'] . '[0-8].jpg') as $file) { if (file_exists($file)) { echo "<img src=\"" . $file . "\" /><br />"; } else { echo "><img src=\"aircraft/wrightflyer.jpg\" /><br />";} } ?> Hi everyone, Sorry not sure if this is a php or html problem. Im using php and a html form to upload images to my site, i have made it so that jpg, jpeg, gif and png images can be uploaded. The problem im having is displaying the image. Code: [Select] <img src="images/<?php echo $name ?>.jpg" /> That works fine if a jpg was uploaded, but what if a png or a gif was uploaded? The $name is going to be unique, there will not be more than one image with the same name, so what do i have to do to display the image regardless of what the extension is? Thanks Hi everyone, i am just trying to learn php for a bit of fun really and started making a sort of 'facebook' website. I am having trouble however trying to display different users images, for example when trying to find a correct 'friend' only the image of the last result is being shown for all people with the same name... here is my code below, if anyone can help me out that would be great file 1 $count=1; while ($numids>=$count){ echo "<form method=\"post\" action=\"friendadded.php\">"; $frienduserid=$_SESSION["passedid[$ii]"]; $friendfirstname=$_SESSION["passedfirstname[$ff]"]; $friendlastname=$_SESSION["passedlastname[$ll]"]; $_SESSION['friendsuserpicid'] = $frienduserid; echo "<table width=\"700\" height=\"50\" border=\"1\" align=\"center\">"; echo "<tr>"; echo "<th></th>"; echo "<th>First Name</th>"; echo "<th>Last Name</th>"; echo "</tr>"; echo "<tr>"; echo "<td><center>"; echo "<img border=\'0\' src=\"frienduserpic.php\" width=\"80\" height=\"80\" align=\"middle\"/>"; echo "</center></td>"; echo "<td><center>"; echo $friendfirstname; echo "</center></td>"; echo "<td><center>"; echo $friendlastname; echo "</center></td>"; echo "</tr>"; echo "</table>"; echo "<center><input type=\"submit\" value=\"Add this friend\" name=\"Add Friend\"></center><br/>"; $ii=$ii+1; $ff=$ff+1; $ll=$ll+1; $count=$count+1; echo "</form>"; } file 2 session_start(); $passeduserid=$_SESSION['friendsuserpicid']; $timespost=$_SESSION['postednum']; $host= $username= $password= $db_name= $tbl_name= mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $query = mysql_query("SELECT * FROM $tbl_name WHERE picid='".$passeduserid."'"); $row = mysql_fetch_array($query); $content = $row['image']; header("Content-type: image/jpeg"); echo $content; Thanks in advance I am having trouble with displaying an image from a url. In the following code if I echo out $test I get the url of the image. However, in the current code I am trying to display the image, but the only thing that gets displayed is a small broken image in the upper left.
$html = file_get_contents($website.$criteria); $dom = new DOMDocument; @$dom->loadHTML($html); $links = $dom->getElementsByTagName('img'); header('Content-Type: image/png'); foreach ($links as $link){ $test = $link->getAttribute('src'); echo file_get_contents($test); }image that gets displayed: http://imgur.com/epfF3wJ |