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PHP - Display Images With A Variable
I want to retrieve an image id from a db and show the images.
I cant get the syntax right for the image tag.Any help appreciated. Code: [Select] function display_covers() { global $wpdb; $query = "select * from wp_cover"; $result = mysql_query($query)or trigger_error("Query: $query\n<br />MySQL Error: " . mysql_error()); echo mysql_error(); if (!$result) return false; echo'<div class="wrap"><p>choose from one of the covers below</p></div>'; echo'<div id="main">'; echo'<table class="main" cellpadding="2">'; //echo"<caption>Please choose a book cover</caption>"; ?> <thead><tr><td colspan="5" ><h6 class="main">Book Covers</h6></td></tr> </thead> <?php $i=0; $size=3; echo "<tbody>"; echo "<tr>"; while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) { /* display picture */ ?> <td class="main"> <?php echo"<img src="/Applications/MAMP/htdocs/wordpress_3/wp-content/plugins/Authors2/jackets/"{.$row['pix'].}""/>"; echo"</td>"; $i++; if($i==$size) { echo "</tr><tr>"; $i=0; } } } Similar TutorialsI have this script from http://lampload.com/...,view.download/ (I am not using a database) I can upload images fine, I can view files, but I want to delete them. When I press the delete button, nothing happens
http://www.jayg.co.u...oad_gallery.php
<form>
<?php $dir = dirname(__FILENAME__)."/images/gallery" ; $files1 = scandir($dir); foreach($files1 as $file){ if(strlen($file) >=3){ $foil = strstr($file, 'jpg'); // As of PHP 5.3.0 $foil = $file; $pos = strpos($file, 'css'); if ($foil==true){ echo '<input type="checkbox" name="filenames[]" value="'.$foil.'" />'; echo "<img width='130' height='38' src='images/gallery/$file' /><br/>"; // for live host //echo "<img width='130' height='38' src='/ABOOK/SORTING/gallery-dynamic/images/gallery/ $file' /><br/>"; } } }?> <input type="submit" name="mysubmit2" value="Delete"> </form>
any ideas please?
thanks
I want to display thumbnails for a picture gallery and when user clicks on image the larger image opens in a new window. I can't find a straight answer on the php.net website. Can someone help? I read everywhere where it says not to stoe images in MYSQL but I have a code where I'm trying to display the images. They are all jpegs that are stored in the table. Here is the code Code: [Select] <?php include "dbaptsConfig.php"; include "searchaptsstyle.css"; // test id, you need to replace this with whatever id you want the result from $id = "1"; // what you want to ask the db $query = "SELECT * FROM `apartments` WHERE `id` = ".$id; // actually asking the db $res = mysql_query($query, $ms); // recieving the answer from the db (you can only use this line if there is always only one result, otherwise will give error) $result = mysql_fetch_assoc($res); // if you uncomment the next line it prints out the whole result as an array (prints out the image as weird characters) // print_r($result); // print out specific information (not the whole array) echo "<br/>"; echo "<div id='title'>".$result['title']."<br/></div>"; echo "<br/>"; echo "<div id='description'>".$result['description']."<br /></div>"; echo "<br/>"; echo "<div id='table'><tr>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Provider's Phone Number: ".$result['phone']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Provider: ".$result['service']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Borough: ".$result['county']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Town: ".$result['town']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Bedrooms: ".$result['rooms']."</td>"; echo "<td> </td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Bathrooms: ".$result['bath']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Square Footage: ".$result['square']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Rent: ".$result['rent']."<br /></td>"; echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Listed On: ".$result['time']."<br /></td>"; echo "</tr></div>"; header("Content-type: image/jpeg"); echo "<td bgcolor='#FFFFFF' style='color: #000' align='center'> Listed On: ".$result['image1']."<br /></td>"; ?> Thanks I'm creating a funeral home site and on the side of the page I would like to display the last 5 funeral obituaries that are in mySQL database (name and certain sized image (images are in "image" field) w/ a link to an "obituaries detail page for the individual deceased person"). I am able to do this successfully listing only the names....how can I list the image associated with the name? This is what I have so far: Code: [Select] <?php include("connect.php"); ?> Code: [Select] <?php $query = "SELECT id, deceased_name, deceased_date, DATE_FORMAT(deceased_date, '%M %D, %Y') as datetext"; $query .= " FROM scales_obits ORDER BY deceased_date DESC LIMIT 5"; $listings = mysql_query($query); if (!listings) { echo("<p>Error retrieving listings from lookup table<br>". "Error: " . mysql_error()); exit(); } echo("<table border=\"0\" width=\"100%\" class=\"obit\">"); while ($listing = mysql_fetch_array($listings)) { $deceased_name = $listing["deceased_name"]; $deceased_date = $listing["datetext"]; $id = $listing["id"]; echo("<tr><td width=\"100%\"><a href=\"obitdetail.php?id=".$id."\"><strong>".$deceased_name."</strong></a></td><td> </td>"); } echo("</table>"); ?> I am looking to display image paths in a row separated by commas. There are 6 images that goes to each user and I would like only the 6 images at be in each " " Like this: "images/listings/listing_516013019A-only.jpg,images/listings/listing_848813453A-1.jpg,images/listings/listing_664613453A-2.jpg,images/listings/listing_520313453A-3.jpg,images/listings/listing_690513453A-4.jpg,images/listings/listing_125113453A-5.jpg,images/listings/listing_641013453A-6.jpg," "images/listings/listing_736913186A-1.jpg,images/listings/listing_822713186A-2.jpg,images/listings/listing_136513186A-3.jpg,images/listings/listing_700313186A-4.jpg,images/listings/listing_716013186A-5.jpg,images/listings/listing_213113186A-6.jpg," "images/listings/listing_292113254A..-1.jpg,images/listings/listing_854413254A..-2.jpg,images/listings/listing_446013254A..-3.jpg,images/listings/listing_676313254A..-4.jpg,images/listings/listing_563413254A..-5.jpg,images/listings/listing_341513254A..-6.jpg," Right now it is displaying them like this "images/listings/listing_516013019A-only.jpg," "images/listings/listing_848813453A-1.jpg," "images/listings/listing_664613453A-2.jpg," "images/listings/listing_520313453A-3.jpg," "images/listings/listing_690513453A-4.jpg," "images/listings/listing_125113453A-5.jpg," "images/listings/listing_641013453A-6.jpg," "images/listings/listing_736913186A-1.jpg," "images/listings/listing_822713186A-2.jpg," "images/listings/listing_136513186A-3.jpg," "images/listings/listing_700313186A-4.jpg," "images/listings/listing_716013186A-5.jpg," "images/listings/listing_213113186A-6.jpg," "images/listings/listing_292113254A..-1.jpg," "images/listings/listing_854413254A..-2.jpg," "images/listings/listing_446013254A..-3.jpg," "images/listings/listing_676313254A..-4.jpg," "images/listings/listing_563413254A..-5.jpg," "images/listings/listing_341513254A..-6.jpg," Here is the code I have: Code: [Select] <?php // Make a MySQL Connection mysql_connect("localhost", "xxxxxxxx", "xxxxxxxx") or die(mysql_error()); mysql_select_db("xxxxxxxx") or die(mysql_error()); $result = mysql_query("SELECT * FROM listimages ORDER BY listimages.listingid DESC ") or die(mysql_error()); while($row = mysql_fetch_array($result)) { echo "\""; echo "$row[imagepath],"; echo "\""; echo "<br>"; } ?> My tables for the images is "listimages" and the columns a id (which are the auto_increments) imagepath (which shows the path/image1.jpg) mainimage (which just shows 0 or 1 depending on the picture that is the default for that listing, 1 being default) listingid (shows numbers 1 2 3 etc corresponding to the Id in the listings table to show what images go with what listing) There are up to 6 images for each listing. Any idea how to fix this? How can i display all the images in a certain directory and echo the image names. So if a new image is uploaded, it will still display without edited the gallery.php file. Cheers. Hello everyone, I need help figuring out why im not getting images displayed. I have tried everything I could think of and narrowed down the issue. What Im doing id querying urls from DB for my images. Then using md5 to encrypting the url them before display on web page. Here is the code: Code: [Select] $img = @ md5(mysql_result(mysql_query("SELECT `url` FROM `pictures` WHERE `mls`='{$row['mls']}' ORDER BY `id` LIMIT 1",$avenu->link),0)); When html is displayed I get the hash.jpg. The image failed to be loaded. Hi all, Newbie here, i am having a problem to get my images to show which are stored in mysql database as a mediumblob. I get id number to print in table ut am just getting empty square with red cross in where my image should be. Is my code incorrect or is it something else? Appreciate your help with this. I have included both of the pages codes i am using. Thanks Tony image2.php <?php include("common.php"); error_reporting(E_ALL); $link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error()); mysql_select_db(db) or die(mysql_error()); $sql = "SELECT id FROM photos"; $result = mysql_query("$sql") or die("Invalid query: " . mysql_error()); ?> <table border="1"><tr><td>id</td><td>image</td></tr> <?php while($row=mysql_fetch_assoc($result)){ print '<tr><td>'.$row['id'].'</td><td>'; print '<img src="image1.php?id='.$row['id'].'height="75" width="100"">'; } echo '</td></tr></table>' ?> image1.php <?php ob_start(); include("common.php"); mysql_connect(host,username,password) or die(mysql_error()); mysql_select_db(db) or die(mysql_error()); $query = mysql_query("SELECT imgage FROM photos WHERE id={$_GET['image_id']}"; $row = mysql_fetch_array($query); $content = $row['image']; header('Content-type: image/jpg'); echo $content; } ob_end_flush(); ?> Hi I tried to write a script: 1) to locate all directories in a directory (1 level) 2) with the function GetImages() I try to display the image(s) in the subfolder there are only images in the subfolder I guess I'm doing something wrong in the GetImages() with the glob function, can anyone check this ? Thanks in advance function GetImages($map) { $files = glob('$map/*.jpg'); //$files = glob("$map/*.*"); for ($i=0; $i<count($files); $i++) { $num = $files[$i]; echo '<img height="50" width="50" src="'.$num.'" />'." <br />"; } } if ($handle = opendir('mystuff')) { /* loop through directory. */ while (false !== ($dir = readdir($handle))) { if($dir != ".." && $dir != "."){ echo '<option value='.$dir.'>'.$dir.'</option><br>'; GetImages($dir); } } closedir($handle); } Hi everyone!! I have looked into how the upload script works and this is what i have: Code: [Select] <?php if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 20000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file"; } ?> Which is un-tested at the moment, but let's just say for talking sake it worked 100% what elements of this script would i be looking at to display the files uploaded on to another page, in my case my homepage? ive found as to yet, that the uploads have to be stored on a file somewhere on my server, which i've set up. But i thought it would be just as easy to have a field in my table named upload and display it within the table next to the other results? instead i just get whatever the file name is named.jpg. Any help in looking towards the answer? many thanks in advance guys! I'm trying to display an html <div> based on the state of a variable set during php execution . The variable is $chk and is set to either 0 or 1 with 0 meaning failed and 1 meaning pass. Here's the code:
if <?php echo "{$chk}";?> == 0
<div class="container">
<div class="row" style="color:red">
<br><br><br><br><br><br>
<center>Database Update Failed</center>
</div>
</div>
else
<div class="container">
<div class="row">
<br><br><br><br><br><br>
<center>Database Updated</center>
</div>
</div>
Thanks in advance, Larry Edited June 19, 2020 by larry29936additioal info This is my code Code: [Select] $file = $file . $line; fclose($fh); echo "<script language= 'JavaScript'>alert(' . $file . ');</script>"; the alert box is not coming up. Please suggest a way to print the contents of the file in a alert box. Hi, i cant seem to get something working, should be simple but its not working for me. I just need to only display a table if a variable in my table = a certain value. The column in the table is called 'option1_available' and if its value is set to 'Y' i want it to display a table. Appreciate any help, Thanks So I have these session variables being stored on login. Each page stores these session variables in local PHP variables. For some reason, all pages display that variable in the URL except one page. I have done some 'echo' work just to ensure variables have values stored and that everything was being read appropriately, no issues there. The variables are apparent everywhere on the page besides the URL. Below you can find all the code I have to make this thing work....maybe someone can point me in the right direction because I am STUMPED!!! LOL So my first page calls a popup window in javascript defined below. The link involves the following PHP Code: [Select] $customer_id = $_SESSION['cid']; $customer_email = $_SESSION['cust_email']; $customer_fname = $_SESSION['cust_first_name']; $customer_contact_id = $_SESSION['customer_contact_id']; $customer_company_id = $_SESSION['customer_company_id']; $c_project_id = $_GET['pid']; $c_project_id = mysql_real_escape_string($c_project_id ); $c_project_id = eregi_replace("`", "", $c_project_id); Then comes in the Javascript... Code: [Select] <script type="text/javascript"> <!-- $(document).ready(function(){ $(".approval").click(function(){ v = $(this).attr("id"); url = 'deny_approval.php?cid=<?php echo "$customer_id"; ?>&company_id=<?php echo "$customer_company_id"; ?>&customer_contact_id=<?php echo "$customer_contact_id"; ?>&pid=<?php echo "$customer_project_id"; ?>§ion=' + v; window.open(url, "myWindow", "status = 1, toolbar = no, scrollbars = yes, location = no, resizable = no, height = 600, width = 600, resizable = 0" ); }); }); //--> </script> And finally, the HTML.... Code: [Select] <body> <a href="#" id="deny_quote_approval" class="approval">Deny Quote Approval</a><br /><br /> </body> Any ideas?? As always, any help would be greatly appreciated. Bl4ck Maj1k Hi, i have a folder with thumbnails and original big pictures of the thumbnails, i wanna make it so that the max amount of thumbs that can be displayed on one page would be 3 Code: [Select] <body> <div id=head></div> <div id=main> <?php include("loadimages.php") ?> //<- here i take all the thumbnails that are in my thumb folder </div> <div id=foot></div> </body> i have 5 thumbnails, it displays all of them in the "main" div, i wanna add an "Next" and "Previous" buttons, i mean, i want it to show only 3 thumbs at one time, once u click "next" it would remove the first 3 thumbs and load the other 2 thumbnails.. heres a pic maybe it will explain better of what im trying to do.. I want to display 3 clickable images in a single row which repeats as long as there is data in the database, so far it is displaying a single clickable image from the database. below is all the code.. <table width="362" border="0"> <?php $sql=mysql_query("select * from `publication` GROUP BY `catsue`") or die(mysql_error()); $num=mysql_num_rows($sql); while($rowfor=mysql_fetch_array($sql)) { $cat=$rowfor['catsue']; $pic=mysql_query("select * from `category` where `catsue`='$cat'") or die(mysql_error()); $picP=mysql_fetch_array($pic); $base=basename($picP['title']); ?> <tr> <td width="352" height="88"><table width="408" border="0"> <tr> <td width="113" rowspan="5"><a href="archive_detail.php?id=<?php echo $rowfor['id'];?>&category=<?php echo $rowfor['catsue'];?>"><img src="ad/pic/<?php echo $base;?>" width="100" height="100" border="0"/></a></td> <td width="94">Title</td> <td width="179" height="1"><?php echo $rowfor['catsue'];?> </td> </tr> <tr> <td> </td> <td width="179" height="3"> </td> </tr> <tr> <td> </td> <td width="179" height="8"> </td> </tr> <tr> <td> </td> <td width="179" height="17"> </td> </tr> <tr> <td> </td> <td width="179" height="36"> </td> </tr> </table></td> </tr> <?php }?> </table> Hi guys its me again, I am having a problem that I cant figure out... Here is my code: <?php $sqlCommand = "SELECT image FROM background"; $query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); $sqlCommand2 = "SELECT backgroundimage FROM site"; $query2 = mysqli_query($myConnection, $sqlCommand2) or die (mysqli_error()); while ($row = mysqli_fetch_array($query)) { while ($row2 = mysqli_fetch_array($query2)) { if($row['image'] == $row2['backgroundimage']){ echo '<img src="site_background/'.$row['image'].'" width="75px" height="75px" style="border:2px solid red;" /><br /><br />'; } if($row['image'] != $row2['backgroundimage']){ echo '<img src="site_background/'.$row['image'].'" width="50px" height="50px" style="border:2px solid black;" />'; } } } mysqli_free_result($query); mysqli_free_result($query2); ?> What is will do is get the images from the "backgrounds" table and the image from the "site" table (the current image). I am then wanting to pick out the current image and give it a red border and then display the other left over images smaller with a black border. I can get the images to all display with the black border or the current image to display with a red border but the other images dont show... I have tried mixing things around but I have not been able to get all the images to display with the formatting I want. I dont know if it is a simple syntax error or I am doing things completely wrong... I have been looking at it for so long its just become one big mess of code to me lol Any help to get this working as I want would be great! Cheers Ben Ok the pagination part is all working fine. but i thought id be able to create a heap variables inside the loop then display the images in a table. the only trouble is all variables are grabbing the same img. i need them to grab the 10 different records. thanks Code: [Select] $sql = "SELECT * FROM mongrels_db.gallery ORDER BY id DESC LIMIT $offset, $rowsperpage "; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); // while there are rows to be fetched... while ($list = mysql_fetch_array($result)) { $img1=$list['img']; $img2=$list['img']; $img3=$list['img']; $img4=$list['img']; $img5=$list['img']; $img6=$list['img']; $img7=$list['img']; $img8=$list['img']; $img9=$list['img']; $img10=$list['img']; // echo data } // end while echo "<table><tr>"; echo "<td>"."<img src='../gallery/".$img1 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img2 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img3 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img4 ."' width='100' height='100''> "."</td></tr>"; echo "<tr><td>"."<img src='../gallery/".$img5 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img6 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img7 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img8 ."' width='100' height='100''> "."</td>"; echo "<tr><td>"."<img src='../gallery/".$img9 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img10 ."' width='100' height='100''> "."</td></tr>"; I have the following code which displays images and captions from a directory. I would like the images to be displayed in the order they were created. Somebody suggested I use glob(), but I have no idea how. <?php $dir = "exclusive_images/"; if ($opendir = opendir($dir)) { //read directory while(($file = readdir($opendir)) !==FALSE) { if ( file_exists($dir.'/'.$file) && in_array( strtolower(pathinfo($file,PATHINFO_EXTENSION )), array('png','jpg','jpeg','gif'))) { if (file_exists($dir.'/'.$file. '.txt')) $caption = file_get_contents($dir.'/'.$file. '.txt'); else $caption = ucwords(str_replace(array('-','_'),' ', substr($file, 0, (strlen ($file)) - (strlen (strrchr($file,'.')))))); echo '<tr><td><img src="' . $dir.'/'.$file .'" alt="'. $file .'" title="' . $file . '" width="200"></td><td>' . $caption . '</td></tr>'; } } } ?> |
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