would you help for my code , i need to do the multiple upload but i cant so will you help me please. i need so bad.
here's my code
i made multiple upload the form but it is not working. the output is "error array"
//HTML
<html>
<head>
<form name="Image" enctype="multipart/form-data" action="upload.php" method="POST">
<h1><font face="tahoma"> UPLOAD FILES</h1>
<label for="file">Filename:</label>
<input type="file" name="Photo[]" accept="image/*" multiple="multiple"/><br/><br/>
<input type="hidden" id="pageName" name="pageName">
<script type="text/javascript">
//get page name from parent
var value = window.opener.pageName
document.getElementById("pageName").value = value;
</script>
<INPUT type="submit" class="button" name="Submit" value=" Upload ">
<INPUT type="reset" class="button" value="Cancel"><br/><br/>
</form>
</head>
</html>
//PHP this is were upload is do.
<?php
include('global.php');
?>
<?
$uploadDir = 'directory/'; //Image Upload Folder
if(isset($_POST['Submit']))
{
$fileName = $_FILES['Photo']['name'][0];
$fileName1 = $_FILES['Photo']['name'][1];
$tmpName = $_FILES['Photo']['tmp_name'];
$fileSize = $_FILES['Photo']['size'];
$fileType = $_FILES['Photo']['type'];
$filePath = $uploadDir . $fileName . $fileName1;
//upload error
if ($_FILES["Photo"]["error"] > 0)
{
echo "Error: " . $_FILES["Photo"]["error"] . "<br />";
}
//photo already exixts
else
//insert image into DB
{
move_uploaded_file($tmpName, $filePath);
$filePath = addslashes($filePath);
$filePath = stripslashes($filePath);
$filePath = mysql_real_escape_string($filePath);
$query = "INSERT INTO images (image , category ) VALUES ('$filePath', '$pageName')";
mysql_query($query) or die('Error, query failed');
echo" Upload Successful. <br/> <br/>";
echo "Stored in: " . "directory/" . $_FILES["Photo"]["name"];
?>
<br/><br/>
<img width="300" height="400" src="directory /<?=$_FILES["Photo"]["name"]?>"><br/>
<?
}
}
?>
Error: Array is telling you that the error being returned is actually an Array object, not a string. If you want to see the actual error message, you need to view the full contents of the array. Something like print_r($_FILES["Photo"]["error"]); or looping through the array like so
foreach($_FILES["Photo"]["error"] as $err) {
echo "error: " . $err . "<br>";
}
Or you can just print the first error returned just as you have returned the first file in your name array echo $_FILES["Photo"]["error"][0];
Related
I have created a basic car sales website. I have used the following PHP code to upload my image
$target_folder = "Cars_Photos/";
$target_path = $target_folder . basename( $_FILES['fileToUpload']['name'] );
//echo $target_path . '<br><br><br>';
//print_r($_FILES);
print($_FILES['fileToUpload']['tmp_name']);
if(move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
So the file is being uploaded to the server and that is great (took me ages to get it to work), I am using phpliteadmin as my database manager and I have a field called car_image_url, this is where i paste the url to the image on the server, however I have added a page on the site where users can upload an image themselves, so my question is how can I get this to work.
I am using the following to convert the url to display an image.
echo "<td id='img'><img src=\"". $row["car_image_url"] . "\" /></td>";
However uploading the file is a different story, what code do I use on my website to get the uploaded file to link to the image url.
Here is my PHP code that makes a new car on the main site:
<?php
try {
# Connect to SQLite database
$dbh = new PDO("sqlite:../Car_Sales_Network");
$make = $_POST['Make'];
$model = $_POST['Model'];
$badge = $_POST['Badge'];
$price = $_POST['Price'];
$trans = $_POST['Transmission'];
$ppl = $_POST['P_Plate_Legal'];
$sth = $dbh->prepare('INSERT INTO Cars_On_Network
("car_make","car_model","car_badge","price","trans","P_Plate_Legal")
VALUES
(?, ?, ?, ?, ?, ?)');
$sth->execute(array($make, $model, $badge, $price, $trans, $ppl));
$id = $dbh->lastInsertId();
//header("Location: ../Carsales_Network.php");
}
catch(PDOException $e) {
echo $e->getMessage();
}
$target_folder = "Cars_Photos/";
$target_path = $target_folder . basename( $_FILES['fileToUpload']['name'] );
//echo $target_path . '<br><br><br>';
//print_r($_FILES);
print($_FILES['fileToUpload']['tmp_name']);
if(move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
Here is the HTML:
<!DOCTYPE html>
<html>
<head>
<title>New Vehicle</title>
<link type="text/css" rel="stylesheet" href="New_Car_Form.css"/>
</head>
<body>
<div id="main">
<form action="Insert_Car.php" method="post" enctype="multipart/form-data">
Make:<br>
<input type="text" name="Make">
<br>
Model:<br>
<input type="text" name="Model">
<br><br>
Badge:<br>
<input type="text" name="Badge">
<br>
Price:<br>
<input type="text" name="Price">
<br>
Transmission: <br>
<input type="radio" name="Transmission" value="Manual" checked>Manual
<br>
<input type="radio" name="Transmission" value="Auto">Automatic
<br><br>
P Plate Legal: <br>
<select name="P_Plate_Legal">
<option value="Yes">Yes</option>
<option value="No">No</option>
</select>
<br>
Choose a Picture: <br>
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
<br>
<br>
<input class="submit" type="submit" value="Submit">
<br>
Let's go back!
<br>
</div>
</body>
</html>
</form>
</body>
</html>
I am sure it is a similar process however I just can't think of how to do it.
Cheers.
it seems to me like youre having trouble with the upload script.
Here is a way that comes in handy (It creates an unique name for all images)
Crate your sql within the WRITE TO DATABASE comment i made:
if(isset($_FILES['fileToUpload'])){
$File = $_FILES['fileToUpload'];
//File properties:
$FileName = $File['name'];
$TmpLocation = $File['tmp_name'];
$FileSize = $File['size'];
$FileError = $File['error'];
//Figure out what kind of file this is:
$FileExt = explode('.', $FileName);
$FileExt = strtolower(end($FileExt));
//Allowed files:
$Allowed = array('jpg', 'png', 'jpeg', 'gif');
//Check if file is allowed:
if(in_array($FileExt, $Allowed)){
//Does it return an error?
if($FileError==0){
//Create new filename:
$NewName = uniqid('', true) . '.' . $FileExt;
//Destination
$UploadDestination = $_SERVER['DOCUMENT_ROOT'] . '/Cars_Photos/' . $NewName;
//Move file to location:
if(move_uploaded_file($TmpLocation, $UploadDestination)){
//Filename = $NewName
//WRITE TO DATABASE:
//encode url:
$Image = urlencode($UploadDestination);
//Redirect:
header("Location: /complete.php?image=$Image&cat=Cars");
}
//File didnt upload:
else{
echo "File didnt upload...";
}
}
//An error occured
else{
echo "An error occured while uploading...";
}
}
//Filetype not allowed
else{
echo "Sorry, the file you tried to upload is not allowed.";
}
}
Complete.php:
<?php
//Your file is here:
echo urldecode($_GET['image']);
I'm trying to allow an admin upload pictures of products in to the database, but I only want to store the link/url of the picture in the database and then store the uploaded file in a folder.
This is what I've got so far, and I keep getting "Sorry there was a problem uploading your file".
Here is the PHP code:
if ($_FILES['product_image']['error'] == 0) { // checking the file for any errors
$imgName = mysql_real_escape_string($_FILES['product_image']['name']); //returns the name of the image and stores it in variable $imgName
$imgData = mysql_real_escape_string(file_get_contents($_FILES["product_image"]["tmp_name"])); // returns the content of the file and stores it in $imgData
$imgType = mysql_real_escape_string($_FILES["product_image"]["type"]); //returns image/whatever the image type is
$targetFolder = "ProductImages/"; //directory where images will be stored...
$targetFolder = $targetFolder . basename($imgName); //adds the image name to the directory
}
$sql = "INSERT INTO products " . "(product_name,product_model,product_price,product_width,product_height,product_weight,product_quantity,product_category,product_subcategory, product_image, product_description,date_added) " . "VALUES('$product_name','$product_model','$product_price','$product_width','$product_height','$product_weight','$product_quantity', '$product_category', '$product_subcategory', '$imgName', '$product_description', NOW())";
//echo $sql;
mysql_select_db('online_store');
$result = mysql_query($sql, $conn);
$itemResult = "";
if (!$result) {
die('Could not enter data: ' . mysql_error());
}
$itemResult = "Product has been added";
if (move_uploaded_file($imgData, "$targetFolder" . $imgName)) { // writes/stores the image in the targetfolder->ProductImages
echo "The file " . basename($imgName) . "has been uploaded!";
} else {
echo "Sorry, there was a problem uploading your file!";
}
and the HTML form:
<form id="product_form" name="product_form" enctype="multipart/form-data" action="inventory_list.php" method="post">
<label for="product_image">Product Image*:</label> <input type="file" name="product_image"id="product_image"/>
</div>
<div>
<button name="add" id="add">Add Item</button>
</div>
</form
Use Sql Query Below.
$sql = "INSERT INTO products(`product_name`,`product_model`,`product_price`,`product_width`,`product_height`,`product_weight`,`product_quantity`,`product_category`,`product_subcategory`,`product_image`,`product_description`,`date_added`) VALUES('".$product_name."','".$product_model."','".$product_price."','".$product_width."','".$product_height."','".$product_weight."','".$product_quantity."', '".$product_category."', '".$product_subcategory."', '".$imgName."', '".$product_description."','".date("Y-m-d H:i:s")."')";
Also Change below line for upload image $imgData = mysql_real_escape_string(file_get_contents($_FILES["product_image"]["tmp_name"])); to $imgData = $_FILES["product_image"]["tmp_name"];
Try this Hope this helps.Not tested
<form id="product_form" name="product_form" enctype="multipart/form-data" method="post" action="" >
<label for="product_image">Product Image*:</label> <input type="file" name="product_image" id="product_image" />
</div>
<div>
<button name="add" id="add">Add Item</button>
</div>
</form>
PHP code :
<?php
if ($_FILES['product_image']['error'] == 0) { // checking the file for any errors
$imgName = mysql_real_escape_string($_FILES['product_image']['name']); //returns the name of the image and stores it in variable $imgName
$imgData = mysql_real_escape_string(file_get_contents($_FILES["product_image"]["tmp_name"])); // returns the content of the file and stores it in $imgData
$imgType = mysql_real_escape_string($_FILES["product_image"]["type"]); //returns image/whatever the image type is
$targetFolder = "ProductImages/"; //directory where images will be stored...
$targetFolder = $targetFolder . basename($imgName); //adds the image name to the directory
}
$sql = "INSERT INTO products " . "(product_name,product_model,product_price,product_width,product_height,product_weight,product_quantity,product_category,product_subcategory, product_image, product_description,date_added) " . "VALUES('$product_name','$product_model','$product_price','$product_width','$product_height','$product_weight','$product_quantity', '$product_category', '$product_subcategory', '$imgName', '$product_description', NOW())";
//echo $sql;
mysql_select_db('online_store');
$result = mysql_query($sql, $conn);
$itemResult = "";
if (!$result) {
die('Could not enter data: ' . mysql_error());
}
$itemResult = "Product has been added";
if (move_uploaded_file($imgData, $targetFolder)) { // writes/stores the image in the targetfolder->ProductImages
echo "The file " . basename($imgName) . "has been uploaded!";
} else {
echo "Sorry, there was a problem uploading your file!";
}
?>
First of all in HTML form action="post" is incorrect, the action attribute should contain a path. The method attribute should contain post or get like this: method="get" or method="post".
I need help in writing this code to upload and save uploaded image in data base.
File 1:
<?php
<form action="upload.php" method="post" enctype="multipart/form-data" target="upload_target" onsubmit="startUpload();" >
<label>File:
<input name="myfile" type="file" size="30" />
</label>
<input type="submit" name="submitBtn" class="sbtn" value="Upload" />
<iframeid="upload_target"name="upload_target"src="#"style="width:0;height:0;border:0px solid #fff;"></iframe>
</form>
</div>
File 2:
<?php
// Edit upload location here
$destination_path = getcwd().DIRECTORY_SEPARATOR;
$target_path="my/";
$result = 0;
$name=$_FILES['myfile']['name'];
$target_path = $target_path . basename( $_FILES['myfile']['name']);
if(#move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path)) {
list($width, $height, $type, $attr) = getimagesize($target_path);
echo "Image width " .$width;
echo "<BR>";
echo "Image height " .$height;
echo "<BR>";
echo "Image type " .$type;
echo "<BR>";
echo "Attribute " .$attr;
$result = 1;
}
// sleep(1);
$link=mysql_connect('localhost','root','');
if(!$link)
{die('you cannot connect to database...');}
$db=mysql_select_db('final');
if(!$db)die('smile');
if (isset($_FILES['myfile']) && $_FILES['myfile']['size'] > 0) {
// define the posted file into variables
$name = $_FILES['myfile']['name'];
$tmp_name = $_FILES['myfile']['tmp_name'];
$type = $_FILES['myfile']['type'];
$size = $_FILES['myfile']['size'];
// if your server has magic quotes turned off, add slashes manually
//if(!get_magic_quotes_gpc()){
//$name = addslashes($name);
//}
// open up the file and extract the data/content from it
$extract = fopen($tmp_name, 'r');
$content = fread($extract, filesize($tmp_name));
$content = addslashes($content);
fclose($extract);
// connect to the database
// the query that will add this to the database
$s=mysql_query("SET NAMES 'utf8'");
$sql = "INSERT INTO `final`.`products` (`Productcode`, `Productname`, `Price`,`Descriptionofgood`, `image`) VALUES ('','','','','".$target_path."') WHERE `products`.`Productcode`='1371' ";
$results = mysql_query($sql);
if(!$result)die('not');
}
?>
Technically, if it is a small project.
You should not store images files in "Database" ; rather only their link (you may not even need that). Image or any media files are stored on server along with your other files (html,css,php). Of course, you need to put them on a dedicated folder for it.
The reason for not storing on database : because they are meant for data retrieval only and more importantly they are of smaller size (bigger sizes do exist, i am speaking in case of a small project which requires least possible resources. Storing media files on database is just not efficient.
Looking at your code, i can tell you are trying to store the files on your server.
They have used a very simple script for uploading here . Try it on your localhost before trying on a server.
if(#move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path)) is incorrect.
It should be if(move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path))
Remove the #.
I have created three page for it
index.php (Form for uplad image)
upload.php (Save the image in upload folder and its path in database)
3.showimage.php (Show the images from database)
Database Structure
id int(4) auto increment - image varchar(100) - image_name varchar(50)
here is the code:
index.php
<form method="post" action="upload.php" enctype="multipart/form-data">
<label>Choose File to Upload:</label><br />
<input type="hidden" name="id" />
<input type="file" name="uploadimage" /><br />
<input type="submit" value="upload" />
</form>
uplad.php
<?php
$target_Folder = "upload/";
$uid = $_POST['id'];
$target_Path = $target_Folder.basename( $_FILES['uploadimage']['name'] );
$savepath = $target_Path.basename( $_FILES['uploadimage']['name'] );
$file_name = $_FILES['uploadimage']['name'];
if(file_exists('upload/'.$file_name))
{
echo "That File Already Exisit";
}
else
{
// Database
con=mysqli_connect("localhost","user_name","password","database_name"); //Change it if required
//Check Connection
if(mysqli_connect_errno())
{
echo "Failed to connect to database" . mysqli_connect_errno();
}
$sql = "INSERT INTO image (id,image, image_name)
VALUES ('','$target_Folder$file_name','$file_name') ";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added successfully in the database";
echo '<br />';
mysqli_close($con);
// Move the file into UPLOAD folder
move_uploaded_file( $_FILES['uploadimage']['tmp_name'], $target_Path );
echo "File Uploaded <br />";
echo 'File Successfully Uploaded to: ' . $target_Path;
echo '<br />';
echo 'File Name: ' . $_FILES['uploadimage']['name'];
echo'<br />';
echo 'File Type: ' . $_FILES['uploadimage']['type'];
echo'<br />';
echo 'File Size: ' . $_FILES['uploadimage']['size'];
}
?>
Show Image
showimage.php
<?php
$con=mysqli_connect("localhost","user_name","password","test"); //Change it if required
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM image " );
while($row = mysqli_fetch_array($result))
{
echo '<img src="' . $row['image'] . '" width="200" />';
echo'<br /><br />';
}
mysqli_close($con);
?>
I am using a form for users to upload files to my website. I want to allow them to upload multiple photos at once, so I am using the HTML5 "multiple" attribute.
My HTML:
<form method="post" action="save.php">
<input type="file" name="uploads[]" multiple="multiple" />
<input type="submit" name="submit" value="submit"/>
</form>
save.php:
<?php
foreach ($_FILES['uploads']['name'] as $file) {
echo $file . "<br/>";
$file= time() . $_FILES['uploads']['name'];
$target= UPLOADPATH . $file;
move_uploaded_file($_FILES['uploads']['tmp_name'], $target)
or die('error with query 2');
}
But, for some reason when I run the script, I get an error saying undefined index: uploads. And an error saying that I have an invalid argument supplied for foreach(). What could I be dong wrong?
Thanks
UPDATE
Okay, setting the enctype="mulitpart/form-data" worked. Now, I am having trouble with moving the file. I am getting the error move_uploaded_file() expects parameter 1 to be string, array given. What am I doing wrong here?
Thanks again
You need the proper enctype to be able to upload files.
<form method="post" enctype="multipart/form-data" action="save.php">
try this html code: <form method="post" action="save.php" enctype="multipart/form-data">
Then in PHP:
if(isset($_FILES['uploads'])){
foreach ($_FILES['uploads']['name'] as $file) {
echo $file . "<br/>";
$file= time() . $_FILES['uploads']['name'];
$target= UPLOADPATH . $file;
move_uploaded_file($_FILES['uploads']['tmp_name'], $target)
or die('error with query 2');
}
} else {
echo 'some error message!';
}
In order to upload files in the first place, you need enctype="multipart/form-data" on your <form> tag.
But, when you upload multiple files, every key in $_FILES['uploads'] is an array (just like $_FILES['uploads']['name']).
You need to get the array key when looping, so you can process each file. See the docs for move_uploaded_file for more deatils.
<?php
foreach ($_FILES['uploads']['name'] as $key=>$file) {
echo $file."<br/>";
$file = time().$file;
$target = UPLOADPATH.$file;
move_uploaded_file($_FILES['uploads']['tmp_name'][$key], $target)
or die('error with query 2');
}
index.html
<form method="post" action="save.php" enctype="multipart/form-data">
<input type="file" name="uploads[]" multiple="multiple" />
<input type="submit" name="submit" value="Upload Image"/>
</form>
save.php
<?php
$file_dir = "uploads";
if (isset($_POST["submit"])) {
for ($x = 0; $x < count($_FILES['uploads']['name']); $x++) {
$file_name = time() . $_FILES['uploads']['name'][$x];
$file_tmp = $_FILES['uploads']['tmp_name'][$x];
/* location file save */
$file_target = $file_dir . DIRECTORY_SEPARATOR . $file_name;
if (move_uploaded_file($file_tmp, $file_target)) {
echo "{$file_name} has been uploaded. <br />";
} else {
echo "Sorry, there was an error uploading {$file_name}.";
}
}
}
?>
Does anyone know any good tutorial on how to upload a file with php and save the files path to a sql server?
To upload a file you need at least a HTML POST form with multipart/form-data encoding. Therein you put an input type="file" field to browse the file and a submit button to submit the form.
<form action="upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="file">
<input type="submit">
</form>
In the upload.php the uploaded file is accesible by $_FILES with the field name as key.
$file = $_FILES['file'];
You can get its name as follows:
$name = $file['name'];
You need to move it to a permanent location using move_uploaded_file(), else it will get lost:
$path = "/uploads/" . basename($name);
if (move_uploaded_file($file['tmp_name'], $path)) {
// Move succeed.
} else {
// Move failed. Possible duplicate?
}
You can store the path in database the usual way:
$sql = "INSERT INTO file (path) VALUES ('" . mysqli_real_escape_string($path) . "')";
// ...
From http://www.w3schools.com/php/php_file_upload.asp
HTML
<html>
<body>
<form action="upload_file.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file" />
<br />
<input type="submit" name="submit" value="Submit" />
</form>
</body>
</html>
PHP
<?php
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
}
else
{
if (file_exists("upload/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; //<- This is it
}
}
?>
Note that to upload the file you need to specify the path to save the file. If you save the file you already know it path.