Hi i have a code which is uploading file and saving its path in database. Now i want to change its path to its corresponding id with which it is saving in database i.e i have uploaded an image and its id is '4' in database and its file path should also be 4. and if i upload an other image and if its id is 5 then in its file path column there should also be 5 and so on. I have searched for a while but i'm not able to find the proper answer. Kindly help me here.
Here is my code
directory-image.php
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>image in directory</title>
</head>
<body>
<form method="post" action="directory-imagedatabase.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" id="upload" />
</form>
</body>
</html>
directory-imagedatabase.php
<?php
$target_Folder = 'images/';
$uid = $_POST['id'];
$target_Path = $target_Folder.basename( $_FILES['uploadimage']['name'] );
$savepath = $target_Path.basename( $_FILES['uploadimage']['name'] );
$file_name = $_FILES['uploadimage']['name'];
if(file_exists('officework/php-startup/images/'.$file_name))
{
echo "That File Already Exisit";
}
else
{
// Database
$con=mysqli_connect("localhost","root","sal123","test"); //Change it if required
//Check Connection
if(mysqli_connect_errno())
{
echo "Failed to connect to database" . mysqli_connect_errno();
}
$sql = "INSERT INTO directoryimage (id,image, image_name)
VALUES ('','$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'];
}
}
?>
Get the record's insert ID $id = mysqli_insert_id($con) after the INSERT
Get file extension: $ext = preg_replace("/\.(gif|jpg|etc)$/", ".$1", $file_name);
Rename the file with rename() rename("$target_Folder$file_name", $id . $ext);
Update the record with something like $sql = "UPDATE directoryimage SET image = CONCAT('{$target_Folder}', id, '{$ext}') WHERE id = $id";
Related
Hello i'm having this issue here that i have a task here i need to upload images in the folder and save the path of the images in database. Now i have this code that path is getting save in database and if i try to upload the same file which i uploaded before it gives me the message of file exists already but in the destination folder no image is showing. kindly help me here. Here is my code
directory-image.php
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>image in directory</title>
</head>
<body>
<form method="post" action="directory-imagedatabase.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" id="upload" />
</form>
</body>
</html>
directory-imagedatabase.php
<?php
$target_Folder = '/officework/php-startup/images/';
$uid = $_POST['id'];
$target_Path = $target_Folder.basename( $_FILES['uploadimage']['name'] );
$savepath = $target_Path.basename( $_FILES['uploadimage']['name'] );
$file_name = $_FILES['uploadimage']['name'];
if(file_exists('/officework/php-startup/images/'.$file_name))
{
echo "That File Already Exisit";
}
else
{
// Database
$con=mysqli_connect("localhost","root","sal123","test"); //Change it if required
//Check Connection
if(mysqli_connect_errno())
{
echo "Failed to connect to database" . mysqli_connect_errno();
}
$sql = "INSERT INTO directoryimage (id,image, image_name)
VALUES ('','$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
if your main file same folder then update
$target_Folder = '/officework/php-startup/images/';
to
$target_Folder = 'officework/php-startup/images/';
then this is your total code
<?php
$target_Folder = 'officework/php-startup/images/';
$uid = $_POST['id'];
$target_Path = $target_Folder.basename( $_FILES['uploadimage']['name'] );
$savepath = $target_Path.basename( $_FILES['uploadimage']['name'] );
$file_name = $_FILES['uploadimage']['name'];
if(file_exists('officework/php-startup/images/'.$file_name))
{
echo "That File Already Exisit";
}
else
{
// Database
$con=mysqli_connect("localhost","root","sal123","test"); //Change it if required
//Check Connection
if(mysqli_connect_errno())
{
echo "Failed to connect to database" . mysqli_connect_errno();
}
$sql = "INSERT INTO directoryimage (id,image, image_name)
VALUES ('','$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
$sql = "INSERT INTO directoryimage (id,image, image_name)
VALUES ('','$target_Folder$file_name','$file_name') ";
Change it to:
$sql = "INSERT INTO directoryimage (id,image, image_name)
VALUES ('','".$target_Folder."/".$file_name."','".$file_name."') ";
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);
?>
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];