PHP - Image Display In A Div Reading Image Filename From Database
Hello,
Five images will be displayed inside a division. There will be a previous and next button/link. If someone click the next button the next image will be added in that div and the first image will be gone from that div. The previous button/link will do the same thing. Is it possible with php? I am confused if it's a javascript or ajax question. Thanks. Similar TutorialsHi everyone. I have a small problem. I have a database that shows student images, in the format of 1234.jpg but the file name extension is in both upper case and lower case. My server OS is linux ubuntu. apache2 php5. Linux see these as 2 different file so will not show the upper case ones. 1234.JPG What i have already is working but for lower case only, i need something to detect when the extension is in uppercase and change accordingly. This is what i have now that works for lower case only. <img src=../images/mbr_images/<?php echo H($mbr->getBarcodeNmbr());?>.jpg height=150 width=100 border=1px </img This is what i have tried but gives me a filename of "1234jpg" see the "." is missing? and is still lower case. <img src=../images/mbr_images/<?php echo H($mbr->getBarcodeNmbr()); $image = jpg; $image2 = JPG; if ($image = $image) { echo $image; } else { echo $image2; };?> height=150 width=100 border=1px </img> can anyone offer any help please? TIA Peter Hi
I am winning I think
I have got the records displayed for the current user logged in so basically they can only see their submitted listings and just working on the edit of them so the current user logged in can update their listing and is all working apart from the update of the images
the images are stored by the file name on the database and then gets moved onto the server so the actual images are not stored on the database only the file names are and the images are moved onto the server, hope that makes sense
what I can't do at the mo is work out how to update the file names of the images on the database and update on the server
I get the following error
Notice: Undefined index: photo on line 209 and line 211
the coding for them lines are below
$pic1= basename($_FILES['photo']['name'][0]); $pic2= basename($_FILES['photo']['name'][1]);The rest of the coding below that is if(!empty($_FILES['photo']['tmp_name'])) { // Number of uploaded files $num_files = count($_FILES['photo']['tmp_name']); /** loop through the array of files ***/ for($i=0; $i < $num_files;$i++) { // check if there is a file in the array if(!is_uploaded_file($_FILES['photo']['tmp_name'][$i])) { $messages[] = 'No file uploaded'; } else { // move the file to the specified dir if(move_uploaded_file($_FILES['photo']['tmp_name'][$i],$target.'/'.$_FILES['photo']['name'][$i])) { $messages[] = $_FILES['photo']['name'][$i].' uploaded'; } else { // an error message $messages[] = 'Uploading '.$_FILES['photo']['name'][$i].' Failed'; } } }The update query is below // save the data to the database mysql_query("UPDATE privatelistings SET listingtitle='$listingtitle', make='$model', model='$model', exteriorcolour='$exteriorcolour', enginesize='$enginesize', fueltype='$fueltype', yearregistered='$yearregistered', transmission='$transmission', mileage='$mileage', nodoors='$nodoors', bodystyle='$bodystyle', price='$price', photo='$pic1', photo1='$pic2' WHERE id='$id'") or die(mysql_error());I am going to be updating to mysqli just want to get it working first My HTML form coding is below <form action="" method="post" enctype="multipart/form-data"> <input type="hidden" name="id" value="<?php echo $id; ?>"/> <div> <strong>Listing Title: *</strong> <input type="text" name="listingtitle" value="<?php echo $listingtitle; ?>"/> <br/> <strong>Make: *</strong> <input type="text" name="make" value="<?php echo $make; ?>"/> <br/> <strong>Model: *</strong> <input type="text" name="model" value="<?php echo $model; ?>"/> <br/> <strong>Exterior Colour: *</strong> <input type="text" name="exteriorcolour" value="<?php echo $exteriorcolour; ?>"/> <br/> <strong>Engine Size: *</strong> <input type="text" name="enginesize" value="<?php echo $enginesize; ?>"/> <br/> <strong>Fuel Type: *</strong> <input type="text" name="fueltype" value="<?php echo $fueltype; ?>"/> <br/> <strong>Year Registered: *</strong> <input type="text" name="yearregistered" value="<?php echo $yearregistered; ?>"/> <br/> <strong>Transmission: *</strong> <input type="text" name="transmission" value="<?php echo $transmission; ?>"/> <br/> <strong>Mileage: *</strong> <input type="text" name="mileage" value="<?php echo $mileage; ?>"/> <br/> <strong>Number of Doors: *</strong> <input type="text" name="nodoors" value="<?php echo $nodoors; ?>"/> <br/> <strong>Body Style: *</strong> <input type="text" name="bodystyle" value="<?php echo $bodystyle; ?>"/> <br/> <strong>Price: *</strong> <input type="text" name="price" value="<?php echo $price; ?>"/> <br/> <strong>Photo One:</strong> <input type='hidden' name='size' value='350000'><input type='file' name='photo[]'> <br> <strong>Photo Two:</strong> <input type='hidden' name='size' value='350000'><input type='file' name='photo[]'> <br> <input type="submit" name="submit" value="Submit"> </div> </form>sorry was not sure what other I need to show, so you guys get the idea? I'm trying to create a form that has a drag and drop image with it. I did a tutorial for the drag and drop and now I'm trying to put it together with a form. The problem I'm having is that I'm not able to get the filename, so that I can insert it into the database. I've tried doing this if(isset($_POST['add_product'])) { $filename = $_FILES['files']['name']; echo print_r($filename); die(); }
Array ( [0] => ) 1
if(isset($_FILES['files'])) { $filename = $_FILES['files']['name']; echo print_r($filename); die(); }
if(isset($_FILES['files'])) { $filename = $_FILES['files']['name']; } if(isset($_POST['add_product'])) { echo print_r($filename); die(); }
Array ( [0] => ) 1
if(isset($_POST['add_product])) Here is my form <form action="" method="POST" enctype="multipart/form-data"> <div class="form-group"> <label for="title">Title</label> <input type="text" class="form-control" name="title"> </div> <div class="form-group"> <label for="content">content</label> <textarea name="content" id="content" class="form-control" cols="30" rows="10"></textarea> </div> <input type="file" name="files[]" id="standard-upload-files" multiple> <input type="submit" value="Upload files" id="standard-upload"> <div class="wrapper"> <div class="upload-console"> <h2 class="upload-console-header"> Upload </h2> <div class="upload-console-body"> <div class="upload-console-drop" id="drop-zone"> just drag and drop files here </div> <div class="bar"> <div class="bar-fill" id="bar-fill"> <div class="bar-fill-text" id="bar-fill-text"></div> </div> </div> <div class="hidden" id="uploads-finished"> <h3>Process files</h3> <div class="upload-console-upload"> <a href="#">filename.jpg</a> <span>Success</span> </div> </div> </div> </div> </div> <div class="form-group"> <button type="submit" class="btn btn-dark btn-lg btn-block" name="add_product">Save</button> </div> </form>
main.js (function(){ "use strict"; var dropZone = document.getElementById('drop-zone'); var barFill = document.getElementById('bar-fill'); var barFillText = document.getElementById('bar-fill-text'); var uploadsFinished = document.getElementById('uploads-finished'); var startUpload = function(files){ // console.log(files); app.uploader({ files: files, progressBar: barFill, progressText: barFillText, processor: 'index.php', finished: function(data){ // console.log(data); var x; var uploadedElement; var uploadedAnchor; var uploadedStatus; var currFile; for(x = 0; x < data.length; x = x + 1) { currFile = data[x]; uploadedElement = document.createElement('div'); uploadedElement.className = 'upload-console-upload'; uploadedAnchor = document.createElement('a'); uploadedAnchor.textContent = currFile.name; if(currFile.uploaded) { uploadedAnchor.href = 'uploads/'+currFile.file; } uploadedStatus = document.createElement('span'); uploadedStatus.textContent = currFile.uploaded ? 'Uploaded' : 'Failed'; uploadedElement.appendChild(uploadedAnchor); uploadedElement.appendChild(uploadedStatus); uploadsFinished.appendChild(uploadedElement); } uploadsFinished.className = ''; }, error: function(){ console.log('there was an error'); } }); }; //drop functionality dropZone.ondrop = function(e){ e.preventDefault(); this.className = 'upload-console-drop'; startUpload(e.dataTransfer.files); }; dropZone.ondragover = function(){ this.className = 'upload-console-drop drop'; return false; }; dropZone.ondragleave = function(){ this.className = 'upload-console-drop'; return false; }; }());
var app = app || {}; (function(o){ "use strict"; //private methods var ajax, getFormData, setProgress; ajax = function(data){ var xmlhttp = new XMLHttpRequest(); var uploaded; xmlhttp.addEventListener('readystatechange', function(){ if(this.readyState === 4) { if(this.status === 200) { uploaded = JSON.parse(this.response); if(typeof o.options.finished === 'function') { o.options.finished(uploaded); } }else{ if(typeof o.options.error === 'function') { o.options.error(); } } } }); xmlhttp.upload.addEventListener('progress', function(e){ var percent; if(e.lengthComputable === true) { percent = Math.round((event.loaded / event.total) * 100); setProgress(percent); } }); xmlhttp.open('post', o.options.processor); xmlhttp.send(data); }; getFormData = function(source){ var data = new FormData(); var i; for(i = 0; i < source.length; i = i + 1) { data.append('files[]', source[i]); } return data; }; setProgress = function(value){ if(o.options.progressBar !== undefined) { o.options.progressBar.style.width = value ? value + '%' : 0; } if(o.options.progressText !== undefined) { o.options.progressText.textContent = value ? value + '%' : ''; } }; o.uploader = function(options){ o.options = options; if(options.files !== undefined) { ajax(getFormData(o.options.files)); // getFormData(o.options.files); } } }(app));
Hi, I've read a lot of places that it's not recommended to store binary files in my db. So instead I'm supposed to upload the image to a directory, and store the link to that directory in database. First, how would I make a form that uploads the picture to the directory (And what kinda directories are we talking?). Secondly, how would I retrieve that link? And I guess I should rename the picture.. I'd appreciate any help, or a good tutorial (Haven't found any myself). Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance. BTW, is using php only doable? Is there a simpler or more elegant way to do this? Thanks all! hello friends, while clicking the form all the information goes to database, I have one image upload field, when cliking the submit button, i would like 'image name' to go in database and file to go in /upload folder, i have tried this for hours and gave up, if anyone help me in this, i would be very greatful Hi guys, I am wirting a script to upload an image file, but I get the error: copy() [function.copy]: Filename cannot be empty in Obviouly it has something to do with the copy() function - (copy($HTTP_POST_FILES['product_img']['tmp_name'], $path)) I think it is not copying from my temp folder or something. Can anyone help me out? Code is: Code: [Select] //IMAGE FILE $file_name = $_FILES['product_img']['name']; echo '<p>File Name:' . $file_name . '</p>'; //IMAGE UPLOAD // random 4 digit to add to file name $random_digit=rand(0000,9999); //combine random digit to file name to create new file name //use dot (.) to combile these two variables $new_image_name = $random_digit.$file_name; echo $new_image_name; //SET DESINATION FOLDER $path= "uploads/".$new_image_name; echo '<p>Path:' . $path . '</p>'; if($product_img !=none) { if(copy($HTTP_POST_FILES['product_img']['tmp_name'], $path)) { echo "Successful<BR/>"; echo "File Name :".$new_file_name."<BR/>"; echo "File Size :".$HTTP_POST_FILES['product_img']['size']."<BR/>"; echo "File Type :".$HTTP_POST_FILES['product_img']['type']."<BR/>"; } else { echo "Image upload Error<BR/>"; } } Any help would be greatly appreciated! Thanks I have a very simple piece of code to create a grid based gallery. The thumbnails are loaded from a single directory and are name 1.jpg... 2.jpg... 3.jpg etc At the moment the images appear to be loaded in randomly. I want them to load in numerically in terms of their filename. I know I may need to use 'sort' or 'natsort' but where in the code? Thanks for any help you can give me Code: [Select] <? $images = "Images/Bag Thumbs/"; # Location of small versions $big = "ruxxwomens.php?id="; # Location of big versions (assumed to be a subdir of above) $cols = 4; # Number of columns to display if ($handle = opendir($images)) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != ".." && $file != rtrim($big,"/")) { $files[] = $file; } } closedir($handle); } $colCtr = 0; echo '<table width="800" cellspacing="0" cellpadding="0" border="0"><tr>'; foreach($files as $file) { if($colCtr %$cols == 0) echo '</tr><tr>'; echo '<td align="center"><a href="' . $big . $file . '"><img src="' . $images . $file . '" cellspacing="0" cellpadding="0" border="0"></a></td>'; $colCtr++; } echo '</table>' . "\r\n"; ?> Well the subject line is pretty explicit. I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database. Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work. Code: [Select] //authenticate user //Start session session_start(); //Connect to database require ('config.php'); $sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($sql); $imagebytes = $row['image']; header("Content-type: image/jpeg"); print $imagebytes; Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which. Can anyone help?? Hey there I am having a problem where I am pulling the records for images from my database and then trying to display them. What happens is there are three images in the database and I want to display them. My code displays the images but it only shows two of the three. I know there are three images in the database I put them in and checked just to make sure. It just seems to only want to display two images instead of three. Here's the code:$getImages = "SELECT * FROM Images WHERE SKU = '".$sku."'"; $resultGetImages = mysql_query($getImages); $records = mysql_fetch_assoc($resultGetImages); ?> while($records = mysql_fetch_assoc($resultGetImages)){ echo "<a onclick=\"document.getElementById('placeholder').src='../../".$records['Path']."'\" target=\"_blank\"><img src=\"../../".$records['ThumbPath']."\" alt=\"".$records['Description']."\" height=\"".$records['ThumbWidth']."\" /></a>\n"; } ?> I have tried a for loop as well but it displayed more images than were in the database and just a single image over and over again. I know that was a flaw in my code and I'm not even sure a for loop would even be a reasonable way to go. I have a script that uploads my images to a folder and database which works fine. I am looking to display my images on the fly to resize them. How do I intergrate my image output string with the fly string. Current output Code: [Select] echo '<img src="' . $dir . '/' . $image_id . '.jpg">'; Fly string Code: [Select] <img src="/scripts/thumb.php?src=/images/whatever.jpg&h=150&w=150&zc=1" alt="" /> Thanks for your help Hey there, my problem is that I am trying to pull three images from a database and then displaying them with a large image above them. I want to then be able to click the smaller ones and then they become the larger one(kind of like on amazon). I am having 2 major problems with it at this point. the first is my while statement is only pulling 2 of the three images from the database, and secondly my image replacement javascript replaces with the smaller image. With the javascript it dosen't have to be done that way or in javascript so a link or suggestions on how to do that better would be great. So here's my code: <?php include('../../php/includes/openDbConn.php'); $sku = 'c-111111'; $getImages = "SELECT * FROM Images WHERE SKU = '".$sku."'"; $resultGetImages = mysql_query($getImages); $records = mysql_fetch_assoc($resultGetImages); ?> <img src="../../<?php echo $records['Path'] ?>" id="placeholder" alt="<?php echo $record['Description']; ?>" height="<?php echo $record['Width']; ?>" /> <div id="thumbImages"> <?php while($records = mysql_fetch_assoc($resultGetImages)){ echo "<a onclick=\"document.getElementById('placeholder').src='../../".$records['ThumbPath']."'\" target=\"_blank\"><img src=\"../../".$records['ThumbPath']."\" alt=\"".$records['Description']."\" height=\"".$records['ThumbWidth']."\" /></a>\n"; } ?> and a bit on the table with the images. It stores the sku and the Path of a thumbnail, regular and large sized image. All help is much appreciated. Don't know if it is php or html but the image won't display. Code: [Select] <!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" xml:lang="en" lang="en" > <head> <meta name="google-site-verification" content="y_kspZkCZOQM-WGNB_lQ8BhJS8p7-B66hkHI_UhzKbs" /> <meta http-equiv="Content-type" content="text/html; charset=utf-8" /> <meta http-equiv="Content-language" content="en" /> <link type="text/css" rel="stylesheet" href="css/reset.css" /> <link type="text/css" rel="stylesheet" href="css/960.css" /> <link type="text/css" rel="stylesheet" href="css/custom.css" /> <title>Bay Area Remote Control Society</title> </head> <body> <div id="wrapper" class="container_12"> <div id="header" class="grid_12"> <div id="left-header" class="grid_5 alpha"><?php include('content/left-header.php'); ?></div> <div id="newsflash" class="grid_7 omega"><?php include('content/newsflash.php'); ?></div> </div> <div id="leftmenu" class="grid_3"><?php include('content/menu2.php'); ?></div> <div id="maincontent" class="grid_9"><?php include('content/maincontent.php'); ?></div> <div id ="footer" class="grid_12"><?php include('content/footer.php'); ?></div> </div><!-- end wrapper --> </body> </html> left-header.php Code: [Select] <?php ?> <img src="Pictures_Other/concorde.jpg" alt="concorde" border="0" width="250" height="188" /> Hi im wanting to create this system where it loads all your files in a specified directory, loads all the images, although i have done that what im trying to do is have it so it prints them into a 9x9 grid automatically, in which i need to find something to restrain the echo of the variable to a limit of 9.. i need to create the <tr><th>and <td>'s automatically and also id like it so if there is more than 9, it will allow me to select a button which will load up the next section of images. heres the code i have so far: <?php $dir = "images/"; $dh = opendir($dir); while (false !== ($filename = readdir($dh))) { $files[] = $filename; } sort($files); print_r($files); I then used this to try and echo into a table :S <?php if ($handle = opendir('images/')) { while (false !== ($file = readdir($handle))) { echo "<table width='450px' height'450px' border='4px solid #000'>"; if ($file != "." && $file != ".."){ echo "<tr><td>$file<br></td>"; echo "<td><img src=images/".$file." height='20px' width='20px'></td></tr><br>"; } } echo "</table>"; closedir($handle); } ?> They are seperate as i havent merged together as i wud like to get them working first. I was trying to do the max and min on the array numbers but was stuck doing that. Any help would be greatly apprecated Many thanks Lewis Stevens Ok i have been trying for a few days now with no success to show a image from my database but as this is the first time i have tried to use databases im having no luck, my problem is this... I have an image gallery that shows a list of differnt galleries with the number of images inside it, like this.... General (10) I have a form that allows me to add new galleries to the list and upload an image to be shown in the gallery list, i simply added a field in the categories table and called it category_img. I would like to make the list of galleries look better by having an image for each one like this.... This is the code that generates and displays the list of galleries/categories but it is WAY to advanced and confusing for me to work out what to do, i have tried all sorts but only ever destroys what i already have or just does nothing at all, the closest i have got was displaying the name of the image, but no actual image. Code: [Select] <?php // initialization $result_array = array(); $counter = 0; $cid = (int)($_GET['cid']); $pid = (int)($_GET['pid']); // Category Listing if( empty($cid) && empty($pid) ) { $number_of_categories_in_row = 1; $result = mysql_query( "SELECT c.category_id,c.category_name,COUNT(photo_id) FROM gallery_category as c LEFT JOIN gallery_photos as p ON p.photo_category = c.category_id GROUP BY c.category_id" ); while( $row = mysql_fetch_array( $result ) ) { $result_array[] = '<p><a href="gallery.php?cid=' . $row[0] . '">' . $row[1] . '</a> (' . $row[2] . ')</p>'; } mysql_free_result( $result ); $result_final = "<tr>\n"; foreach($result_array as $category_link) { if($counter == $number_of_categories_in_row) { $counter = 1; $result_final .= "\n</tr>\n<tr>\n"; } else $counter++; $result_final .= "\t<td>".$category_link."</td>\n"; } if($counter) { if($number_of_categories_in_row-$counter) $result_final .= "\t<td colspan='".($number_of_categories_in_row-$counter)."'> </td>\n"; $result_final .= "</tr>"; } } i could really do with some help as trying to learn this stuff is really hard when i have to work in my living room with children running around, my head has not stopped hurting in days. Hi i am trying to display image using 'CAT_BIO' => (isset($_GET['c'])) ? '<br />' '<img src="images/' . $row['image'] . ' " width="150" height="200" />' '', but i get the error Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in i think a comma or something missing can anyone spot where i am going wrong?? the other way i tried is <img src=\"images/{$row['image']}\" width=\"150\" height=\"200\" />" this one give no error bt doesnt display the anything it might be cz wrong path how do i make to go out one folder the get inside images folder?? not sure if its the solution help This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=332434.0 |