|
PHP - Next Link To See More Images...
I have a personal website that retrieves images from directories on the server. However, one of my directories has 1000+ images; which I don't want to be displayed on one page. Is there any way that I can have it so that the first 100 images are displayed on the page with a next link for the next 100 images. Kind of like the way that clothes are displayed on websites that have a large number of products.
I am sorry if this is vague. Any help would be appreciated. Thanks Similar TutorialsI built a CMS system using CKEditor and KCFinder that store information od a databse via textarea/php. So far so good!
The issue comes to when I want to store and display images that link to themselves. The way I am storing images is exactly the same: There is a textarea where I insert an image via KCFinder/CKEditor. The image is uploaded to the server and the path stored at the database. Later I try to pick up that path from the database to display the image and because I want the image to link to itself, I try to use the same method to insert the url on the link. Problem? The link is missing and the images are not displaying.
Can anyone point me the error and suggest any solution? I would be so thankful! CODE: try { $DBH = new PDO('mysql:host=localhost;dbname=yourdb;charset=utf8', 'user', 'password'); $DBH->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $STH = $DBH->prepare('SELECT * FROM php_maskiner ORDER BY timestamp DESC'); $STH->execute(); $STH->setFetchMode(PDO::FETCH_OBJ); while($row = $STH->fetch()) { $title = $row->title; $entry = $row->entry; $images = $row->images; $img_url = $row->images; $img_pack = '<div class="mask3 span3"> <a rel="prettyPhoto" href="'.$img_url.'"><img src="'.$images.'"></a> </div>'; } $DBH = null; } catch (PDOException $e) { echo '<div class="alert alert-standard fade in"> <a class="close" data-dismiss="alert" href="#">×</a> <strong>Can\'t read the database!</strong> </div><br />'.$e; } <?php echo '<article class="span12 post"> '.$img_pack.' <div class="inside"> <div class="span8 entry-content"> <div class="span12"> <h2>'.$title.'</h2> <p>'.$entry.'</p> </div> </div> </div> </article>'; ?> Thanks in advance! Hi I've got this database I created with fields ProductId ProductName Image I've managed to get it to list the ID,productname, and Image urls in a list. My next step is to have the image field actually display an image and make it clickable: heres what I've done so far: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("productfeed", $con); $result = mysql_query("SELECT * FROM productfeeds"); echo "<table border='0'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Image</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; echo "<td>" . $row['ImageURL'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Heres what I want to do: Code: [Select] while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; // my changes beneath echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Can you guys point me in the right direction? Many thanks Hello.
I have a bit of a problem. When I fetch the link field from the database.i don't see an actual link on the page.
One more thing, what type of field should I use to store the link in the database? Probably there is where I went wrong.
All help is
I 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
Hi Support, I have a form, where it collects user description input. I can collect the inputs and store it with newline. The issue is - how to collect the http link to actual hyperlink ref during display. The following is my code: <textarea name="description" cols="50" rows="10" id="description"><?php echo str_replace("<br>", "\n", $description);?></textarea></td> For example, User input: Hi, Check it out - http://www.google.com/ I would like to display google link as href so that Viewers can click the link and go to the page. Right now, it is not href and user need to copy the link to new tabs or pages and then it can come. Thanks for your help. Regards, Ahsan here's my code that i've used to send an email. Code: [Select] $link = "<a href=\"http://www.example.com/" . $num . "\">" . $num . "</a>"; $query = "SELECT content FROM emails"; $result = mysql_query($query) or die(); $email_content = mysql_result($result, 0); $email = sprintf($email_content, $first, $name, $from, $link, $record, $rec, $inc, $max); $email_body = stripslashes(htmlentities($email, ENT_QUOTES, 'UTF-8')); // this is sent to another php script via post.... $subject = $_POST['subject']; $message = nl2br(html_entity_decode($_POST['email_body'])); $to = "me@whatever.com"; $charset='UTF-8'; $encoded_subject="=?$charset?B?" . base64_encode($subject) . "?=\n"; $headers="From: " . $userEmail . "\n" . "Content-Type: text/html; charset=$charset; format=flowed\n" . "MIME-Version: 1.0\n" . "Content-Transfer-Encoding: 8bit\n" . "X-Mailer: PHP\n"; mail($to,$encoded_subject,$message,$headers); in the db, emails.content is of the text type and contains several lines of text with %4$s which inserts the value of $link into the body. when the email arrives, there is a link and it appears fine, with the value of $num hyperlinked. however when you click on it it doesn't go anywhere. when copying the link location from the email it gives me x-msg://87/%22http://www.example.com/16 what is x-msg? how can i get this to work properly? hi hope you all are fine. i have been working on a Email Form (like user fills up the form which send the information to our email) but i was having problem with (URL field i created) link of form is (http://services.shadowaura.com/allquotations/static.php) field which is not working is "Inspirational Website:" when i submit the form it says (Forbidden You don't have permission to access /allquotations/staticworking.php on this server. Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.) Can some one help me out ????????????? code behind this form is: Code: [Select] <?php /* Email Variables */ $emailSubject = 'Shadow Aura Contact Info!'; $webMaster = '*****@shadowaura.com'; /* Data Variables */ $Name = $_POST['Name']; $email = $_POST['email']; $Cell = $_POST['Cell']; $Phone = $_POST['Phone']; $CompanyName = $_POST['CompanyName']; $TypeOfBusiness = $_POST['TypeOfBusiness']; $Address = $_POST['Address']; $YourBudget = $_POST['YourBudget']; $HaveDomain = $_POST['HaveDomain']; $RunningWeb = $_POST['RunningWeb']; $WebLink = $_POST['WebLink']; $Inspiration1 = $_POST['Inspiration1']; $Inspiration2 = $_POST['Inspiration2']; $NumberPages = $_POST['NumberPages']; $UseFlash = $_POST['UseFlash']; $TimeFrame = $_POST['TimeFrame']; $Provided = $_POST['Provided']; $Comments = $_POST['Comments']; $body = <<<EOD <h1> Static Website Quotation </h1> <br> <b>Name of Client:</b>$Name<br> <b>Your Email:</b>$email<br> <b>Cell Number:</b>$Cell<br> <b>Line Phone Number:</b>$Phone<br> <b>Company Name:</b>$CompanyName<br> <b>Type of Business:</b>$TypeOfBusiness<br> <b>Address:</b>$Address<br> <b>Your Budget:</b>$YourBudget <br> <b>Do you have Domain:</b>$HaveDomain<br> <b>Your Site is Running:</b>$RunningWeb <br> <b>Website Link:</b><a href="$WebLink">$WebLink</a><br> <b>Inspiration:</b>$Inspiration1<br> <b>2nd Inspiration:</b>$Inspiration2<br> <b>Number of Pages:</b>$NumberPages<br> <b>Use Flash:</b>$UseFlash <br> <b>Time Frame:</b>$TimeFrame<br> <b>You will provide:</b>$Provided<br> <b>Comments:</b>$Comments<br> EOD; $headers = "From: $email\r\n"; $headers .= "Content-type: text/html\r\n"; $success = mail($webMaster, $emailSubject, $body, $headers); /* Results rendered as HTML */ $theResults = <<<EOD <html> <head> <title>sent message</title> <meta http-equiv="refresh" content="3;URL=http://services.shadowaura.com/"> <style type="text/css"> <!-- body { background-color: #8CC640; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 20px; font-style: normal; line-height: normal; font-weight: normal; color: #fec001; text-decoration: none; padding-top: 200px; margin-left: 150px; width: 800px; } --> </style> </head> <div align="center">Your email will be answered soon as possible! You will return to <b>Shadow Aura Services</b> in a few seconds !</div> </div> </body> </html> EOD; echo "$theResults"; ?> Hi guys, I'm using a twitter script that grabs the title and publishings it like so: "Title - Read More at..." I was wondering how i would be able to post the direct link into twitter.. like news.php?id=1 for example. Code: [Select] $tweet->post('statuses/update', array('status' => ''.$_POST[title].' - more at MY URL')); This part is in the script to publish automatically when the users adds to the news database. How am i able to get the ID just after the posting of the news? Thanks! I have a line like this it prints text link but I prefer image link how should i edit it I would appreciate some feedback Code: [Select] $templates['etiket'] = array('name' => t('ETİKET'), 'module' => 'uc_invoice_pdf', 'path' => $templates_uc_invoice_pdf_path, 'pdf_settings' => $pdf_settings); Ok, I have a GD image that is then rewrote as a png. My question is, is it possible to find out what site is requesting the image and then use it on the php image. Example, if I didn't want a domain to use my images, could I somehow me able to restirict that site? Hi guys I just want to know something. With php you can handle excel files by saving them as .csv files. Then the php script reads it line by line, and commas separating the fields. But is it possible that I can insert an image (small jpg or png) file in one of these fields...and then handle the image in the php script? Just for intrest I inserted an image in excel worksheet and it did not insert it into a cell, it float. Is there maybe another to insert it in the cell and then control it with php?? Thanx in advance Hey every i'm just trying to change my 'Read More' text after a blog post on my site into am image. Got some code that lets me change the actual text to something else, just can't find how to change the text into an image. Code: [Select] add_filter( 'the_content_more_link', 'child_content_more_link', 10, 2 ); function child_content_more_link( $more_link, $more_link_text ) { return str_replace( $more_link_text, 'my new read more text', $more_link ); } Thanks, Matty44m Ok I have finally got rid of my syntax errors with this upload script. Now im getting the error uploading your image. My folders are called project_files and project_thumbs. This is my script <?php $results = ""; error_reporting(E_ALL); ini_set("display_errors", 1); //Auto Thumbnail Function function create_thumbnail($source,$destination, $thumb_width) { $size = getimagesize($source); $width = $size[0]; $height = $size[1]; $x = 0; $y = 0; if($width> $height) { $x = ceil(($width - $height) / 2 ); $width = $height; } elseif($height> $width) { $y = ceil(($height - $width) / 2); $height = $width; } $new_image = imagecreatetruecolor($thumb_width,$thumb_width) or die('Cannot initialize new GD image stream'); $extension = get_image_extension($source); if($extension=='jpg' || $extension=='jpeg') $image = imagecreatefromjpeg($source); if($extension=='gif') $image = imagecreatefromgif($source); if($extension=='png') $image = imagecreatefrompng($source); imagecopyresampled($new_image,$image,0,0,$x,$y,$thumb_width, $thumb_width,$width,$height); if($extension=='jpg' || $extension=='jpeg') imagejpeg($new_image,$destination); if($extension=='gif') imagegif($new_image,$destination); if($extension=='png') imagepng($new_image,$destination); } //Get Extension Function function get_image_extension($name) { $name = strtolower($name); $i = strrpos($name,"."); if (!$i) { return ""; } $l = strlen($name) - $i; $extension = substr($name,$i+1,$l); return $extension; } //Random Name Function function random_name($length) { $characters = "abcdefghijklmnopqrstuvwxyz01234567890"; $name = ""; for ($i = 0; $i < $length; $i++) { $name .= $characters[mt_rand(0, strlen($characters) - 1)]; } return "image-".$name; } $images_location = "/project_files/"; $thumbs_location = "/project_thumbs/"; $thumb_width = "100"; $maximum_size = "5000000"; //Check And Save Image if($_POST) { if($_FILES['image']['name'] == ""){ $results = "Please select the image you would like to upload by clicking browse"; } else{ $size=filesize($_FILES['image']['tmp_name']); $filename = stripslashes($_FILES['image']['name']); $extension = get_image_extension($filename); if($size > $maximum_size) { $results = "Your file size exceeds the maximum file size!"; } else if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { $results = "Only the following extensions are allowed. 'jpg', 'jpeg', 'png', 'gif'."; } else { $image_random_name=random_name(15).".".$extension; $copy = @copy($_FILES['image']['tmp_name'], $images_location.$image_random_name); if (!$copy) { $results = "Error while uploadin your image! Please try again!"; } else{ create_thumbnail($images_location.$image_random_name,$thumbs_location.$image_random_name, $thumb_width); $results = "Your image has been uploaded!"; } } } } ?> and the form Code: [Select] <html> <head> <title>test</title> </head> <body> <?php echo $results; ?> <form action="#" method="post" enctype="multipart/form-data"> <input type="hidden" name="MAX_FILE_SIZE" value="5000000" /> <input type="file" name="image" /> <input type="submit" value="Upload Image" /> </form> </body> </html> Hi guys I have got this code here, giving quite a few errors and was wondering if anyone could help me sort it out, the script (should) run straight away just have to edit the database connect, I've tried manually creating the table and taking out the create table but when I try and upload an image I get these errors: Warning: move_uploaded_file(latest.img) [function.move-uploaded-file]: failed to open stream: Permission denied in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 23 Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move '/Applications/XAMPP/xamppfiles/temp/php2lgspw' to 'latest.img' in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 23 Warning: fopen(latest.img) [function.fopen]: failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 24 Warning: filesize() [function.filesize]: stat failed for latest.img in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 25 Warning: fread() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/php/image.php on line 25 here is the code: Code: [Select] <?php // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); // First run ONLY - need to create table by uncommenting this // Or with silent @ we can let it fail every subsequent time ;-) $q = < < <CREATE create table pix ( pid int primary key not null auto_increment, title text, imgdata longblob) CREATE; @mysql_query($q); // Insert any new image into database if ($_REQUEST[completed] == 1) { // Need to add - check for large upload. Otherwise the code // will just duplicate old file ;-) // ALSO - note that latest.img must be public write and in a // live appliaction should be in another (safe!) directory. move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($image) < 149000) { mysql_query ("insert into pix (title, imgdata) values (\"". $_REQUEST[whatsit]. "\", \"". $image. "\")"); } else { $errmsg = "Too large!"; } } // Find out about latest image $gotten = @mysql_query("select * from pix order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } // If this is the image request, send out the image if ($_REQUEST[gim] == 1) { header("Content-type: image/jpeg"); print $bytes; exit (); } ?> <html><head> <title>Upload an image to a database</title> <body bgcolor=white><h2>Here's the latest picture</h2> <font color=red><?= $errmsg ?></font> <center><img src=?gim=1 width=144><br> <b><?= $title ?></center> <hr> <h2>Please upload a new picture and title</h2> <form enctype=multipart/form-data method=post> <input type=hidden name=MAX_FILE_SIZE value=150000> <input type=hidden name=completed value=1> Please choose an image to upload: <input type=file name=imagefile><br> Please enter the title of that pictu <input name=whatsit><br> then: <input type=submit></form><br> <hr> By Graham Ellis - graham@wellho.net </body> </html> Ok, so i've managed to upload images, but now i'm trying to rename them to avoid replacement and i've done something like this: mysql_query( "INSERT INTO `softmarket`.`internet_securitate` (`internet_securitateID`, `nume_produs`, `descriere`, `versiune`,`poza`, `pret`, `disponibilitate_produs`) VALUES ('', '$_POST[titlu]', '$_POST[descriere]', '$_POST[versiune]', '$foto', '$_POST[pret]', '$_POST[disponibilitate]');"); Code: [Select] function findexts ($filename) { $filename = strtolower($filename) ; $exts = split("[/\\.]", $filename) ; $n = count($exts)-1; $exts = $exts[$n]; return $exts; } $ext = findexts ($_FILES['foto']['name']) ; $ran2 = $foto."."; $target = "../images/"; $target = $target . $ran2.$ext; while (file_exists($target)) { $ran2 = 'copyof'.$ran2; $target = $target . $ran2.$ext; } if(move_uploaded_file($_FILES['foto']['tmp_name'], $target)) { echo "The file has been uploaded as ".$ran2.$ext; } else { echo "Sorry, there was a problem uploading your file."; } header('Location: succes.php'); but it renames them in a wierd way...what is wrong? I have a site where I need to have lets call it image1 displayed, then I want to change this image based on a php if statement, for instance: if $var == $var2 change the image ....blah blah so I was also going to have the names of my images stored in my database, i.e. image1.jpg and image2.jpg in my database. the image is in its own div tag set as the background image of the div tag if that makes any diference. Thanks Hi guys, i have a form that uploads a school photo. All i want to do is resize the image to its widest dimension of a width of 480 px and then obviously constrain the height. can anyone help me, ive tried to do this myself but when it comes to image properties i really struggle etc. heres my upload code. Code: [Select] <?php include("includes/connection.php"); // Where the file is going to be placed $schoolimage = "SchoolImages/"; //This path will be stored in the database as it does not contain the filename $currentdir = getcwd(); $path = $currentdir . '/' . $schoolimage; // Get the schoolid for the image and school linker table $schoolid = $_POST['schoolid']; //Get the school name $query = "SELECT * FROM school WHERE school_id = ".$schoolid; $result = mysql_query($query) or die("Error getting school details"); $row = mysql_fetch_assoc($result); $schoolname = $row['name']; //Use this path to store the path of the file in the database. $filepath = $schoolimage . $schoolname; //Create the folder if it does not already exist if(!file_exists('SchoolImages')) { if(mkdir('SchoolImages')) { echo 'Folder ' . 'SchoolImages' . ' created.'; } else { echo 'Error creating folder ' . 'SchoolImages'; } } //Store the folder for the course title. if(!file_exists( $filepath )) { if(mkdir( $filepath )) { echo 'Folder ' . $schoolname . ' created.'; } else { echo 'Error creating folder ' . $schoolname; } } // Where the file is going to be placed $target_path = $filepath; // Add the original filename to our target path. Result is "uploads/filename.extension" echo $target_path = $target_path . '/' . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['uploadedfile']['name'])." has been uploaded"; $filename = $_FILES['uploadedfile']['name']; //Store the filename, path other criteria in the database echo $query = "INSERT INTO image(image_id, name, path) VALUES(0, '$filename', '$filepath')"; //Perform the query $add = mysql_query($query, $conn) or die("Unable to add the image details to the database"); $imageid = mysql_insert_id(); //Store the filename, path other criteria in the database echo $query = "INSERT INTO image_school( image_id, school_id ) VALUES('$imageid', '$schoolid')"; //Perform the query $add = mysql_query($query, $conn) or die("Unable to add the image details to the database"); $message = 'Upload Successful'; } else { $message = 'There was an error uploading the file, please try again!'; } //Close the connection to the database mysql_close($conn); header("Location: add_school_photo_form.php? message={$message}&schoolid={$schoolid}"); //header("Location: add_school_photo_form.php? message=$message, schoolid=$schoolid"); exit(); ?> Id be eternially gratefull. Kind Regards Dean Hello I am using this script right here http://www.nearby.org.uk/sphinx/search-example5-withcomments.phps and am trying to show images from my db but I can only get it to show the first image. Does anyone know how to make it show all images if images even exist for that particular search result? I changed the query to $CONF['mysql_query'] = ' SELECT l.link_id AS id, l.title AS title, l.fulltxt AS body, l.url AS url, m.media_id AS im_id, m.title AS im_title, m.thumb_link AS im_t_link FROM search1_links AS l LEFT JOIN search1_media AS m ON (m.link_id = l.link_id) WHERE l.link_id IN ($ids) '; and added if ($row['im_id']) { echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> '; } right here //Actully display the Results print "<ol class=\"results\" start=\"".($currentOffset+1)."\">"; foreach ($ids as $c => $id) { $row = $rows[$id]; $link = htmlentities(str_replace('$id',$row['id'],$CONF['link_format'])); print "<li><a href=\"$link\">".htmlentities($row['title'])."</a><br/>"; if ($CONF['body'] == 'excerpt' && !empty($reply[$c])) print ($reply[$c])."</li>"; else if ($row['im_id']) { echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> '; } print htmlentities($row['body'])."</li>"; } print "</ol>"; if ($numberOfPages > 1) { print "<p class='pages'>Page $currentPage of $numberOfPages. "; printf("Result %d..%d of %d. ",($currentOffset)+1,min(($currentOffset)+$CONF['page_size'],$resultCount),$resultCount); print pagesString($currentPage,$numberOfPages)."</p>"; } Can anyone help me to be able to display all images if they exist? Thanks. Hi, I have this upload script that uploads a file and writes path to DB: Code: [Select] <?php $uploadDir = '../uploads/'; if(isset($_POST['upload'])) { foreach ($_FILES as $file) { $fileName = $file['name']; $tmpName = $file['tmp_name']; $fileSize = $file['size']; $fileType = $file['type']; if($fileName==""){ $filePath = '../img/none.jpg'; } else{ $filePath = $uploadDir . $fileName; } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } $fileinsert[]=$filePath; } ?> this works fine for a single image, but I eed to upload 2 images with a single submission....any ideas how to do this? Thanks Hello... I have some simple code but no ideas for upload another image?? PHP stores image name in database but there is no second picture uploaded in the images/ folder. Pls help. Thanks if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/png") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 2000000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "<strong>Slika:</strong> " . $_FILES["file"]["name"] . "<br />"; echo "<strong>Format slike:</strong> " . $_FILES["file"]["type"] . "<br />"; echo "<strong>Velicina:</strong> " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "<strong>Temp file:</strong> " . $_FILES["file"]["tmp_name"] . "<br />"; // here is mistake??? move_uploaded_file($_FILES["file"]["tmp_name"], "images/" . $_FILES["file"]["name"]); // $sql="INSERT INTO test (name, something, slika, slika2) VALUES ('".$_POST['name']."', '".$_POST['something']."', '".$_FILES['file']['name']."', '".$_FILES['slika1']['name']."')"; if (mysql_query($sql)) { echo "Sucess"; } else { echo "Error" . mysql_error(); } } } else { echo "Invalid file"; } ?> |
|||||||