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PHP - How To Upload Website Created With Php And Mysql
Hi, I'm not sure if this is the right forum to post this question, but I'll still post it. I'm a newbie programmer, have 1 year experience of php, mysql and html and know how to make a basic website, but I've never actually uploaded the already written code on the internet.So my question is, if you use vertrigo serv and have folders full of php files, what do you do after getting a domain? I read that dreamweaver does that for you, but i also read that it is a closed source software with limitations if you get the free version.The other choice is using FileZilla(which I don't fully understand yet ) i presume.
Similar TutorialsI'm using the following code to create a folder from a text field and works fine. I would like to upload 3 files in the created folder. <?php // set our absolute path to the directories will be created in: $path = $_SERVER['DOCUMENT_ROOT'] . '/uploads/'; if (isset($_POST['create'])) { // Grab our form Data $dirName = isset($_POST['dirName'])?$_POST['dirName']:false; // first validate the value: if ($dirName !== false && preg_match('~([^A-Z0-9]+)~i', $dirName, $matches) === 0) { // We have a valid directory: if (!is_dir($path . $dirName)) { // We are good to create this directory: if (mkdir($path . $dirName, 0775)) { $success = "Your directory has been created succesfully!<br /><br />"; }else { $error = "Unable to create dir {$dirName}."; } }else { $error = "Directory {$dirName} already exists."; } }else { // Invalid data, htmlenttie them incase < > were used. $dirName = htmlentities($dirName); $error = "You have invalid values in {$dirName}."; } } ?> <html> <head><title>Make Directory</title></head> <body> <?php echo (isset($success)?"<h3>$success</h3>":""); ?> <h2>Make Directory on Server</h2> <?php echo (isset($error)?'<span style="color:red;">' . $error . '</span>':''); ?> <form name="phpMkDIRForm" method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>"> Enter a Directory Name (Alpha-Numeric only): <input type="text" value="" name="dirName" /><br /> <input type="submit" name="create" value="Create Directory" /> </form> </body> </html> Ok i am stuck, i created a php script that draws data from a mysql dB into a php page. How can I send the page as the body in email? I do not need to preview the page, but can send it without ever viewing if that helps. When I put my code into a variable it does nothing. Is it caching? Should I load the results into mysql then send?please advise.. Hello, I read somewhere that it is possible for users to be able to upload videos to a website via youtube api intergration. I have no idea how this works, and have tryed the tutorial on the API site. Could anyone give me an example or an easier way that users can upload videos to my website, thanks. Hey guys i have a problem, I have a form to upload 2 image files however sometimes i want to only upload the one file, but if i only upload one image, it tells me that it can't upload because there is no name for the second image. I tried fixing it with if (isset($fieldname2)) however i just realised that $feildname2 is always set so im not quite sure how to fix? Code: [Select] ($action == "add"){ $add = $_POST['add']; $reference = $_POST['reference']; $category = $_POST['category']; $sub_cat = $_POST['sub_cat']; $fieldname = 'file'; $fieldname2 = 'file2'; error_reporting(E_ALL); ini_set('display_errors',1); is_uploaded_file($_FILES[$fieldname]['tmp_name']); getimagesize($_FILES[$fieldname]['tmp_name']); if (isset($fieldname2)) { is_uploaded_file($_FILES[$fieldname2]['tmp_name']); getimagesize($_FILES[$fieldname2]['tmp_name']); } $now = time(); while(file_exists($uploadFilename = $uploadsDirectory.$_FILES[$fieldname]['name'])) { $now++; } if (isset($fieldname2)) { while(file_exists($uploadFilename2 = $uploadsDirectory.$_FILES[$fieldname2]['name'])) { $now++; } } move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadFilename); if (isset($fieldname2)) { move_uploaded_file($_FILES[$fieldname2]['tmp_name'], $uploadFilename2); } $sqlname = $_FILES[$fieldname]['name']; if (isset($fieldname2)) { $sqlname2 = $_FILES[$fieldname2]['name']; } $id = mysql_insert_id(); $query = "INSERT INTO cakes (id, reference, category, sub_cat, image, image2) VALUES ('$id', '$reference', '$category', '$sub_cat','$sqlname','$sqlname2')"; $sql = mysql_query($query) or die (mysql_error()); } echo("<form name='add' method='post' enctype='multipart/form-data' action='?action=add'>"); echo("<input type='hidden' name='MAX_FILE_SIZE' value='3000000"); echo("<input type='hidden' name='?action=add'>"); echo("<table class=main cellspacing=0 cellpadding=2 width=60%>"); echo("<tr><td>Reference: </td><td align='right'><input type='text' size=50 name='reference' value='$reference'></td></tr>"); echo("<tr><td>Category: </td><td align='center'><select name='category'><option>Occasional</option><option>Special</option><option>Figures</option><option>Novelty</option><option>Wedding</option></SELECT></td></tr>"); echo("<tr><td>Sub Category: </td><td align='center'><select name='sub_cat'><option></option><option>Birthday</option><option>Anniversary</option><option>Christmas</option><option>Christening</option><option>Valentine</option><option>Mothers Day</option><option>Fathers Day</option><option>Easter</option></select></td></tr>"); echo("<tr><td>Cake Image: </td><td align='center'><input id='file' type='file' name='file'></td></tr>"); echo("<tr><td>Cake Image 2: </td><td align='center'><input id='file2' type='file' name='file2'></td></tr>"); echo("Before uploading an image please resize the image to 290px x 218px"); echo("<tr><td></td><td><div align='right'><input type='Submit'></div></td></tr>"); echo("</table>"); Hi guys, I am trying to create a coding which allows users to upload and download files into mysql server, but I came across some errors. The coding itself works fine on the surface, but when I uploaded the file into mysql and download them, the contents of the downloaded file was different from the original file that was upload. One good example would be when I upload txt file or doc file. I was stuck on this for quite a few days without any leads. Here are the codes: index.html <form action="add_file.php" method="post" enctype="multipart/form-data"> <input type="file" name="uploaded_file"><br> <input type="submit" value="Upload file"> </form> <p> <a href="list_files.php">See all files</a> </p> add_file.php <?php // Check if a file has been uploaded if(isset($_FILES['uploaded_file'])) { // Make sure the file was sent without errors if($_FILES['uploaded_file']['error'] == 0) { // Connect to the database $dbLink = new mysqli('localhost', 'root', 'pwd', 'myTable'); if(mysqli_connect_errno()) { die("MySQL connection failed: ". mysqli_connect_error()); } // Gather all required data $name = $dbLink->real_escape_string($_FILES['uploaded_file']['name']); $mime = $dbLink->real_escape_string($_FILES['uploaded_file']['type']); $data = $dbLink->real_escape_string(file_get_contents($_FILES ['uploaded_file']['tmp_name'])); $size = intval($_FILES['uploaded_file']['size']); // Create the SQL query $query = " INSERT INTO `file` ( `name`, `mime`, `size`, `data`, `created` ) VALUES ( '{$name}', '{$mime}', {$size}, '{$data}', NOW() )"; // Execute the query $result = $dbLink->query($query); // Check if it was successfull if($result) { echo 'Success! Your file was successfully added!'; } else { echo 'Error! Failed to insert the file' . "<pre>{$dbLink->error}</pre>"; } } else { echo 'An error accured while the file was being uploaded. ' . 'Error code: '. intval($_FILES['uploaded_file']['error']); } // Close the mysql connection $dbLink->close(); } else { echo 'Error! A file was not sent!'; } // Echo a link back to the main page echo '<p>Click <a href="index.html">here</a> to go back</p>'; ?> list_files.php <?php // Connect to the database $dbLink = new mysqli('localhost', 'root', 'pwd', 'myTable'); if(mysqli_connect_errno()) { die("MySQL connection failed: ". mysqli_connect_error()); } // Query for a list of all existing files $sql = 'SELECT `id`, `name`, `mime`, `size`, `created` FROM `file`'; $result = $dbLink->query($sql); // Check if it was successfull if($result) { // Make sure there are some files in there if($result->num_rows == 0) { echo '<p>There are no files in the database</p>'; } else { // Print the top of a table echo '<table width="100%"> <tr> <td><b>Name</b></td> <td><b>Mime</b></td> <td><b>Size (bytes)</b></td> <td><b>Created</b></td> <td><b> </b></td> </tr>'; // Print each file while($row = $result->fetch_assoc()) { echo " <tr> <td>{$row['name']}</td> <td>{$row['mime']}</td> <td>{$row['size']}</td> <td>{$row['created']}</td> <td><a href='get_file.php?id={$row['id']}'>Download</a></td> </tr>"; } // Close table echo '</table>'; } // Free the result $result->free(); } else { echo 'Error! SQL query failed:'; echo "<pre>{$dbLink->error}</pre>"; } // Close the mysql connection $dbLink->close(); ?> get_file.php <?php // Make sure an ID was passed if(isset($_GET['id'])) { // Get the ID $id = intval($_GET['id']); // Make sure the ID is in fact a valid ID if($id <= 0) { die('The ID is invalid!'); } else { // Connect to the database $dbLink = new mysqli('localhost', 'root', 'pwd', 'myTable'); if(mysqli_connect_errno()) { die("MySQL connection failed: ". mysqli_connect_error()); } // Fetch the file information $query = " SELECT `mime`, `name`, `size`, `data` FROM `file` WHERE `id` = {$id}"; $result = $dbLink->query($query); if($result) { // Make sure the result is valid if($result->num_rows == 1) { // Get the row $row = mysqli_fetch_assoc($result); // Print headers header("Content-Type: ". $row['mime']); header("Content-Length: ". $row['size']); header("Content-Disposition: attachment; filename=". $row['name']); // Print data echo $row['data']; } else { echo 'Error! No image exists with that ID.'; } // Free the mysqli resources @mysqli_free_result($result); } else { echo "Error! Query failed: <pre>{$dbLink->error}</pre>"; } @mysqli_close($dbLink); } } else { echo 'Error! No ID was passed.'; } ?> I am not too sure where the error could be but I suspect that it could be somewhere at the add_file or get_file. Seriously hope someone could help me with this bug and thanks for the trouble. Regards Jasmine Hi all, creating a MYSQL, PHP & XHTML site designed to support local rugby clubs. Just putting the final touches to the functionality now so thanks for your help so far. I would like to provide site administrators with the ability to assign photos to a members profile when initially registering their account, I have no experience of dealing with what presumably will be a function that will upload a photo to a location and then linking it some how to data in the database. Thanks for your help, Tom Hey everyone ... This should be a pretty easy question for someone who is used to using fgetcsv: I have a script that needs to import a csv file into a mysql table ... I keep getting stuck in a loop that only ends when the script hits the 30 second timeout I have set on the server ... It never actually inserts a record, and it never spits out a message indicating and error ... Can someone help figure out why I get caught in this loop? Here's the code: Code: [Select] function import_csv() { if(isset($_POST['submit_csv'])) { $filename=$_POST['filename']; $handle = fopen("$filename", "r"); while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) { $import="INSERT INTO `{mydatabase}`.`{my table}` (`name`, `qrl`, `begin`) VALUES ('$data[0]','$data[1]','$data[2]')"; mysql_query($import) or die(mysql_error()); } fclose($handle); print "Import done"; } else { print "<form action='results.php' method='post'><input type='file' name='filename'><br /><input type='submit' value='submit' name='submit_csv' /></form>"; } } Hi Guys, I am trying to use a script that is available on the net to upload an image into a mysql using BLOB. However it wont upload it keep getting Warning: fread() [function.fread]: Length parameter must be greater than 0 in /home/theacidf/public_html/attendance/upload.php on line 17 Here is the code... <?php // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","theacidf_admin","password")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("theacidf_attendanc"); // Insert any new image into database if ($_REQUEST[completed] == 1) { 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("../images/logo_png.png","rb"); $bytes = fread($instr,filesize("../images/logo_png.png")); } // 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><br> <?= $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> </body> </html> Why wont it run properly? Thanks in advance... S Hi All I am trying to upload a csv file to a database using php html form however, the results shows none of the data passsing through to the database and it produces 1000's of empty fields here is the form Code: [Select] form action="lib/import.php" method="post" enctype="multipart/form-data"> Type file name to import: <input type="file" name="filename" size="20"><br /> <input type="submit" name="submit" value="submit"></form> and here is the passing info Code: [Select] $file = $_FILES['filename']; $sqlstatement="LOAD DATA INFILE '$file' into TABLE shop FIELDS TERMINATED BY ',' (id, merchant_name, product_name, description, category_name, Affiliate_deep_link, Affiliate_image_url, price)" ; mysql_query($sqlstatement); echo "it is done!"; can anyone help please? Hey guys! I'm not that good with PHP, and I really need to finish this script. I'm having a hard time finding the problem..it keeps saying Error, query failed and I've also checked the database info, connectivity info and even tried a connection and database selection verification to see if that was working. Is there any other error you guys can find in the script that may be causing the message? Edit: The files I want to be able to upload should be pdf's, txts, docs, etc. Would that be possible with this script? Thanks!! Code: [Select] <form method="post" enctype="multipart/form-data"> <table width="350" border="0" cellpadding="1" cellspacing="1" class="box"> <tr> <td width="246"> <input type="hidden" name="MAX_FILE_SIZE" value="104857600"></td> <td width="80"> </td> </tr> </table> <p>Archivo: <input name="userfile" type="file" id="userfile" /> </p> <p>Nomb <label for="name"></label> <input type="text" name="nombrelista" id="nombrelista" /> </p> <p>Password (para luego borrar si necesario): <input type="text" name="pass" id="pass" /> </p> <p> <input name="upload" type="submit" class="box" id="upload" value=" Upload " /> </p> <?php if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0) { $fileName = $_FILES['userfile']['name']; $tmpName = $_FILES['userfile']['tmp_name']; $fileSize = $_FILES['userfile']['size']; $fileType = $_FILES['userfile']['type']; $filePass = $_POST['pass']; $nombreLista = $_POST['nombrelista']; $fp = fopen($tmpName, 'r'); $content = fread($fp, filesize($tmpName)); $content = addslashes($content); fclose($fp); $host_db = "localhost"; $user_db = "melkien_rudesa"; $pass_db = "jorlan2407"; $base_db = "melkien_rudesa"; $link = mysql_connect($host_db, $user_db, $pass_db); if (!$link) { die('Could not connect: ' . mysql_error()); mysql_close($link); } mysql_select_db($base_db, $link) or die(mysql_error()); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } $query = "INSERT INTO 'upload'('nombrelista', 'name', 'pass', 'size', 'type', 'content' ) VALUES ('$nombreLista', $fileName', '$filePass', $fileSize', '$fileType', '$content')"; mysql_query($query) or die('Error, query failed'); echo "<br>File $fileName uploaded<br>"; mysql_close($link); } ?> </form> Hello freaks! Im new to this forum, but im not all that new to PHP and MySQL. Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right Let's go on-topic: Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us. I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database. In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database. Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this. I guess there are several ways.. What would the expert do (That's you right?)? Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload? Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table? Any thoughts would be appreciated! <!-- TechThat --> hi- i'm trying to browse for a CSV file and then upload it to my mysql database. i found the code below and not working. i don't get any errors. the connection to the database is ok bcz i get the existing results but not the ones from the CSV file added to the db.. any ideas? Code: [Select] <?php ob_start(); require_once('../../connections/congif.php'); mysql_select_db($dbname, $db); $sql_get_project="SELECT * FROM gifts_tbl ORDER BY autoID DESC LIMIT 25"; $get_project = mysql_query($sql_get_project, $db) or die(mysql_error()); $row_get_project = mysql_fetch_assoc($get_project); //database connect info here //check for file upload if(isset($_FILES['csv_file']) && is_uploaded_file($_FILES['csv_file']['tmp_name'])){ //upload directory $upload_dir = "csv_dir/"; //create file name $file_path = $upload_dir . $_FILES['csv_file']['name']; //move uploaded file to upload dir if (!move_uploaded_file($_FILES['csv_file']['tmp_name'], $file_path)) { //error moving upload file echo "Error moving file upload"; } //open the csv file for reading $handle = fopen($file_path, 'r'); //turn off autocommit and delete the product table mysql_query("BEGIN"); while (($data = fgetcsv($handle, 1000, ',')) !== FALSE) { //Access field data in $data array ex. $name = $data[0]; //Use data to insert into db $sql = sprintf("INSERT INTO gifts_tbl (player_id) VALUES ('%s)", mysql_real_escape_string($name) ); mysql_query($sql) or (mysql_query("ROLLBACK") and die(mysql_error() . " - $sql")); } unlink($file_path); } ob_flush(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Gifts</title> </head> <body> <h1>Testing for CSV upload</h1> <form action="" method="post"> <input type="file" name="csv_file"> <input type="submit" name="csv_submit" value="Upload CSV File"> </form> <h2>Results</h2> <?php do { ?> <ul> <li><?php echo $row_get_project['player_id']; ?></li> </ul> <?php } while ($row_get_project = mysql_fetch_assoc($get_project)); ?> </body> </html> Hello! I want to make the users in the website able only to upload, for example, 1 GB, so that when a user finishes the 1 GB available for his files, he cannot upload more files. I know how to set up upload limit for a single file in the upload page, or even globally for anyone to upload. But I would like to know how to make each user has specific capacity, and how it is updated so that if he uploads 0.5 GB he has only 0.5 GB left. I thought of creating a column in the `users` or something concerning upload limit that I will set it (one GB for example). If I will do that, how will I be able to determine how much he uploaded? Any help would be appreciated. Thank you! I receive files/images and loop through an array as follows:
$files=rearrfiles( $_FILES['image']);
foreach($files as $key=>$item)
{
if(is_array($item) && !empty($item['name']) && !empty($item['tmp_name'])){
//$_POST['imageFile']=$item['name'];
//echo $item['name'];
if($error!=true && !uploadImageB(array('image'=>'image'), DB_EXTENTION."mod_projects_images", $naming, $search = $editId, $dir, 'photo'))
{
$error=true;
$action='Edit';
$message.='<br>Images were unable to be uploaded.';
}
}
}
The function uploadImageB() is as follows:
function uploadImageB($info, $table, $naming, $search, $path, $add='uploaded', $maxw=1280, $maxh=1280, $dynamic=true, $id='Id', $thumbMaxWidth = 120)
{
global $connection;
if(!$_POST){ global $HTTP_POST_VARS; @$_POST=$HTTP_POST_VARS;}
if(!$_FILES){ global $HTTP_POST_FILES; @$_FILES=$HTTP_POST_FILES;}
if(is_array($info) && !empty($info) && $table!='' && !empty($path))
{
$error=false;
$d=$naming;
foreach($info as $key=>$val)
{
$names[$val]=$dynamic==true?$add.'_'.$result[$id].'_'.$d.'_.'.strtolower(mygetext($_FILES[$val]['name'])):$add.strtolower(mygetext($_FILES[$val]['name']));
if(is_array(@$_FILES[$val]) && !empty($_FILES[$val]['name']) && !empty($names[$val]))
{
if(!singMove($path, $names[$val], $_FILES[$val], array(), array('gif', 'jpeg', 'jpg', 'png', 'bmp', 'svg')))
{
$error=true; break;
}
else{
$valid[$val]=$names[$val];
resizeImage($path.$names[$val], $maxw, $maxh);
make_thumb($path.$names[$val],$names[$val], $path, $thumbMaxWidth);
}
}
}
if($error==true) return false;
elseif(!empty($valid))
{
$connection->info=$info; $connection->input=$valid; $connection->id=array('field'=>$id, 'val'=>addslashes($result[$id]));
$connection->makenew($table); $connection->return_db($connection->query);
}
}
return true;
}
I get the error:
Images were unable to be uploaded.
Any help?
I've looked around at a bunch of sample code and cannot seem to get this to work. I'm trying to create a table with the values provided in a form along with an uploaded image. Here is what I have. Code: [Select] <?php //Set variables from form $title = $_POST["title_textfield"]; $description = $_POST["description_textbox"]; $price = $_POST["price_textfield"]; $time = date("ymdHis"); $me = $myuserID /Create Table $con = mysql_connect("mydatabase.mysql.com","databaseAdmin","Adminpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } //check if table exists // Create table $time = date("ymdHis"); $tablename = "1_$me$time"; mysql_select_db("db_01", $con); $sql = "CREATE TABLE $tablename ( ID int NOT NULL AUTO_INCREMENT, imageData LONGBLOB NOT NULL , PRIMARY KEY(comicID), Owner varchar(15), Status varchar(15), AgeRestriction varchar(25), Title varchar(25), Description varchar(100), Price varchar(15) )"; // Execute query mysql_query($sql,$con); //input info into table mysql_query("INSERT INTO $tablename (Owner, Status, Title, Description, Price) VALUES ('$myid', 'Pending', '$title', '$description', '$price')"); //Image control!!!! $maxFileSize = "2000000"; // 2 MB file size //Initializing the image types. $image_array = array("image/jpeg","image/jpg","image/gif","image/bmp","image/pjpeg","image/png"); // valid image type //store the file type of the uploading file. $fileType = $_FILES['userfile']['type']; $nname= $_FILES['userfile']['name']; echo "$nname"; //Initializing the msg variable. Although , not necessary. $msg = ""; // if Submit button is clicked if(@$_POST['Submit']) { // check for the image type , before uploading if (in_array($fileType, $image_array)) { // check whether the file is uploaded if(is_uploaded_file($_FILES['userfile']['tmp_name'])) { // check whether the file size is below 2 mb if($_FILES['userfile']['size'] < $maxFileSize) { // reading the image data $imageData =addslashes (file_get_contents($_FILES['userfile']['tmp_name'])); // inserting into the database $sql = "INSERT INTO $tablename (imageData) VALUES ('$imageData')"; mysql_query($sql) or die(mysql_error()); $msg = " Data successfully uploaded"; } else { $msg = " Error : File size exceeds the maximum limit "; } } } else { $msg = "Error: Not a valid image "; } } mysql_close($con); ?> All the fields populate just fine except the imageData field which shows "[BLOB - 0 Bytes]" Any help would be GREATLY appreciated. Thank you for even taking the time to look. Cordially, TcS Okay, so I found this code on-line that uploads images via a form, inserts them in to MySQL and displays the images on the page. It works great! Problem is, I spent a long time hacking this script to make it work with my site just to realize it did not re-size the image after it uploaded (dumb me). I have found other code that do this but I really do not want to start over at this point and my level of expertise with PHP are a step below what I need to figure this out. I am attaching the "original" code. If anyone knows how to add the necessary lines of code to make every image re-size (reduce) itself. I would very much appreciate it. Thanks <?php $con=mysql_connect("localhost", "root", "rootwdp")or die("cannot connect"); mysql_select_db("student",$con)or die("cannot select DB"); if(isset($_POST['upload'])) { $img=$_FILES["image"]["name"]; foreach($img as $key => $value) { $name=$_FILES["image"]["name"][$key] ; $tname=$_FILES["image"]["tmp_name"][$key]; $size=$_FILES["image"]["size"][$key]; $oext=getExtention($name); $ext=strtolower($oext); $base_name=getBaseName($name); if($ext=="jpg" || $ext=="jpeg" || $ext=="bmp" || $ext=="gif"){ if($size< 1024*1024){ if(file_exists("upload/".$name)){ move_uploaded_file($tname,"upload/".$name); $result = 1; list($width,$height)=getimagesize("upload/".$name); $qry="select id from img where `img_base_name`='$base_name' and `img_ext`='$ext'"; $res=mysql_fetch_array(mysql_query($qry)); $id=$res['id']; $qry="UPDATE img SET `img_base_name`='$base_name' ,`img_ext`='$ext' ,`img_height`='$height' ,`img_width`='$width' ,`size`='$size' ,`img_status`='Y' where id=$id"; mysql_query($qry); echo "Image '$name' updated<br />"; } else{ move_uploaded_file($tname,"upload/".$name); $result = 1; list($width,$height)=getimagesize("upload/".$name); $qry="INSERT INTO `img`(`id` ,`img_base_name` ,`img_ext` ,`img_height` ,`img_width`, `size` ,`img_status`)VALUES (NULL , '$base_name', '$ext', '$height', '$width', '$size', 'Y');"; mysql_query($qry); echo "Image '$name' uploaded<br />"; } } else{ echo "Image size excedded.<br />File size should be less than 1Mb<br />"; } } else{ echo "Invalid file extention '.$oext'<br />"; } } } function getExtention($image_name){ return substr($image_name,strrpos($image_name,'.')+1); } function getBaseName($image_name){ return substr($image_name,0,strrpos($image_name,'.')); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <script type="text/javascript"> function addItems() { var table1 = document.getElementById('tab1'); var newrow = document.createElement("tr"); var newcol = document.createElement("td"); var input = document.createElement("input"); input.type="file"; input.name="image[]"; newcol.appendChild(input); newrow.appendChild(newcol); table1.appendChild(newrow); } function remItems() { var table1 = document.getElementById('tab1'); var lastRow = table1.rows.length; if(lastRow>=2) table1.deleteRow(lastRow-1); } </script> <style type="text/css"> <a class="tooltip" and them url href="/http.www.anyurl.com" </a> a.tooltip:hover span{display:inline; position:absolute; border:2px solid #cccccc; background:#efefef; color:#333399;} a.tooltip span {display:none; padding:2px 3px; margin-left:8px; width:150px;} </style> </head> <body> <form method="post" action="" enctype="multipart/form-data"> <table align="center" border="0" id="tab1"> <tr> <td width="218" align="center"><input type="file" name="image[]" /></td> <td width="54" align="center"><img src="Button-Add-icon.png" alt="Add" style="cursor:pointer" onclick="addItems()" /></td> <td><img src="Button-Delete-icon.png" alt="Remove" style="cursor:pointer" onclick="remItems()" /></td> </tr> </table> <table align="center" border="0" id="tab2"> <tr><td align="center"><input type="submit" value="Upload" name="upload" /></td></tr> </table> </form> <table border="0" style="border:solid 1px #333; width:800px" align="center"><tr><td align="center"> <iframe style="display:none" name="if1" id="if1"></iframe> <? $qry="select * from img where img_status='Y' order by id"; $res=mysql_query($qry); $i=0; if(mysql_num_rows($res)){ ?> <div align="center"><ul style="width:650px; border: 0px"> <? while($fetch=mysql_fetch_array($res)){ $hratio=120/$fetch['img_height']; $wratio=120/$fetch['img_width']; $ratio=($hratio < $wratio) ? $hratio : $wratio; $hth=$fetch['img_height']*$ratio; $wth=$fetch['img_width']*$ratio; ?> <li style="width:120px; height:180px; border:0px solid #333;float:left;list-style:none outside none; padding-right:5px;"><img src="upload/<? echo $fetch['img_base_name'].'.'.$fetch['img_ext']; ?>" width="<? echo $wth; ?>" height="<? echo $hth; ?>" title="<? echo "image : ".$fetch['img_base_name'].".".$fetch['img_ext']; ?>" /><br /> <? if($i==0) $fp=fopen("fileInfo.txt",'w'); else $fp=fopen("fileInfo.txt",'a'); fwrite($fp,"Image : ".++$i ."\r\n"); fwrite($fp,"Name : ".$fetch['img_base_name'].".".$fetch['img_ext']."\r\n"); fwrite($fp,"width X height : ".$fetch['img_width']." X ".$fetch['img_height']."\r\n"); fwrite($fp,"Size : ".round($fetch['size']/1024,1)."Kb\r\n"); fwrite($fp,"____________________________________\r\n"); fclose($fp); echo $fetch['img_base_name'].".".$fetch['img_ext'].'<br />'; echo $fetch['img_width'].' X '; echo $fetch['img_height'].'<br />'; echo round($fetch['size']/1024,1) .'Kb'; ?> </li> <? }?> </ul> </div><? }?> </td></tr></table> </body> </html> what's wrong with this ? i can't upload This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=328623.0 I am in the process of writing a CMS for a friend that is looking to add a classified section to his site. He will be the only one that will ever use it. I got the code to work with one image when he asked me if I could do it so he could post 6 images for each item. I am unable to figure out how to do this. I was told to do this with a while loop. This is the code that I have written so far. <?php //Check to make sure title is filled in. If not redirects to show_add.php if (!$_POST[title]) { header("Location: show_add.php"); exit; } else { //check and see if a session has started session_start(); } //if session has not been properly started redirects back to the administration menu if ($_SESSION[valid] != "yes") { header("Location: admin_menu.php"); exit; } include('includes/connection.php'); //check and see if the type of uploaded file is an image function is_valid_type($file) { $valid_types = array("image/jpg", "image/jpeg", "image/gif", "image/bmp"); if (in_array($file['type'], $valid_types)) return 1; return 0; } //Set Constants $TARGET_PATH = "/home/content/m/i/k/mikedmartiny/html/db_images/"; $title = $_POST['title']; $year = $_POST['year']; $make = $_POST['make']; $model = $_POST['model']; $descript = $_POST['descript']; $image = $_FILES['image']; //Sanitize the inputs $title = mysql_real_escape_string($title); $year = mysql_real_escape_string($year); $make = mysql_real_escape_string($make); $model = mysql_real_escape_string($model); $descript = mysql_real_escape_string($descript); $image['name'] = mysql_real_escape_string($image['name']); //$target_path full string $TARGET_PATH .= $image['name']; //make sure that all fields from form are filled in if ( $title == "" || $year == "" || $make =="" || $model == "" || $descript == "" || $image['name'] == "") { $_SESSION['error'] = "ALL FIELDS ARE REQUIRED!"; header ("Location: show_add.php"); exit; } //check to make sure it has the right file type if (!is_valid_type($image)){ $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header ("Location: show_add.php"); exit; } //check to see if a file with that name exsists if (file_exists($TARGET_PATH)){ $_SESSION['error'] = "A FILE WITH THAT NAME ALL READY EXIST!"; header ("Location: show_add.php"); exit; } //move the image - write path to database while($image <=2) { move_uploaded_file($image['tmp_name'], $TARGET_PATH) } else { // Make sure you chmod the directory to be writeable $_SESSION['error'] = "COULD NOT UPLOAD FILE. CHECK WRITE/REWRITE PERMISSIONS ON THE FILE DIRECTORY!"; header ("Location: show_add.php"); exit; } $sql = "INSERT INTO $table (id, title, year, make, model, descript, image, image_two) VALUES ('', '$_POST[title]', '$_POST[year]', '$_POST[make]', '$_POST[model]', '$_POST[descript]', '" . $image['name'] . "', '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); ?> I only have it set up to work with 2 images right now. I thought it would be easier to get it to work properly. Then I could just add in the rest of the info later. Any help would be greatly appreciated. |
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