I have a Signature Pad source code, this system save the signatures in one folder (with .png extension), but work only via md5.
<?php
$result = array();
$imagedata = base64_decode($_POST['img_data']);
$filename = md5(date("dmYhisA"));
//Location to where you want to created sign image
$file_name = './doc_signs/'.$filename.'.png';
file_put_contents($file_name,$imagedata);
$result['status'] = 1;
$result['file_name'] = $file_name;
echo json_encode($result);
?>
In the homepage I have the signature pad, and one "input" in html with name "nome_cognome_cliente"
<input name="nome_cognome_cliente" type="text" required="true" placeholder="Nome e Cognome"><br><br>
I would like save the signatures files with the name that user write in it.
I tested to write this in $filename variable. But doesn't work... where am i wrong? Thanks for helping.
$filename = $_POST['nome_cognome_cliente'];
Also, I would like save this signatures (images) in one Database with the date of entry (DD/MM/YYYY).
Related
I have a table CRUD that display my database.
When I upload an image in my image folder I generate a random name of numbers with that function rand()
Here is what I have coded :
My upload function
function importer_image()
{
if(isset($_FILES["image"]))
{
$extension = explode('.', $_FILES['image']['name']);
$new_name = rand() . '.' . $extension[1];
$destination = './upload/' . $new_name;
move_uploaded_file($_FILES['image']['tmp_name'], $destination);
return $new_name;
}
}
My upload php
if($_POST["operation"] == "Ajouter")
{
$image = '';
if($_FILES["image"]["name"] != '')
{
$image = importer_image();
}
The problem is then when I code the update function, the substitued image stays in my folder and the new one has a new name generated. In order to avoid this, I would like to create a condition that says if $image !='' 1/ erase the old file 2/ upload the new file and keep the same name than the deleted image.
So I'm trying to create an update php process that would 1/ delete unlink() 2/ upload the new image with the name of the previous image.
In order to delete the old image and maintain new name of image as previous, you have to take a hidden input in your form which contain your uploaded image name.
For e.g :
<input type="hidden" name="old-image" value="here is your previous image">
Now when your upload function will hit then you can get previous image name by request and can delete or maintain new image as previous.
if($_FILES['image']['name'] != '') {
$old_image = $_POST['old-image']; // get old image
$new_name = $old_image; // make new image name as previous
unlink('/upload/'.$old_image); // remove old image from folder
$destination = './upload/' . $new_name;
move_uploaded_file($_FILES['image']['tmp_name'], $destination);
return $new_name;
} else {
//if there is no image uploaded in the form then it will maintain old image
return $new_name= $_POST['old-image'];
}
Hope it will help you.
Your code to get extension works only if you have just one dot in the uploaded filename, it won't work for example on image_a.1.jpg. Use instead:
$extension = strrchr($_FILES['image']['name'],'.');
Also, using rand() for naming will eventually give you a headache when it generates the same number the second time and the uploaded image will overwrite the old one or not get saved, possibly producing error showing the user your full server save path. I'd use some hashing, like:
$new_name = md5($_FILES['image']['name'] . time());
If you need to save the file name as int, use crc32() instead of md5(). You could convert md5() result to int, but that would produce a very long int (128bit) that might not fit in your database or even PHP code.
I would like to add user ID to file name in file upload. I tried this:
$date=date("Y-m-d");
$id=mysql_insert_id();
//Abstract File codes
$info = pathinfo($_FILES['abstract']['name']);
$ext = $info['extension']; // get the extension of the file
$newname = $id."_Abstract.".$ext;
$target = "application/abstracts/"; $target = $target .$newname;
if (empty($ext))
{ $abstract='' ;}
else $abstract=$newname;
But I only get 0 for ID.
I can display $name but $id is not working. Is it because I do not have an input for $id on my form? Thank you. I should also mention that the user is filling out an application form (for the first time) and attaching files on the same page. I had the file upload after the application was submitted I don;t think I would have problem call the $id. But since everything is on one page I might be calling an empty $id.
Today i am looking for help. This is my first time asking so sorry in advance if I make a few mistakes
I am trying to code a small web application that will display images.Originally I used the blob format to store my images in a database, however from researching on here People suggest to use a file system. My issue is I cannot display an image. It could be a very small error or even a bad reference to a files location however I cannot make it work.
This is a small project that I hope to be able to improve on and hopefully create into a sort of photo gallery. I am running this application on a localhost.
I am having an issue with displaying images from a filesystem.
// index.php
<form action="process.php" method="post" enctype="multipart/form-data">
<input type="file" name="image" />
<input type="submit" name="submit" value="Upload" />
</form>
My form then leads to a process page where the request is dealt with.
<?php
// process.php
// connect to the database
include 'connection.php';
// take in some file data
$filename =$_FILES['image']['name'];
// get the file extension
$extension = strtolower(substr($filename, strpos($filename, '.')+1));
// if the file name is set
if(isset($filename)){
// set save destination
$saved ='images/';
// rename file
$filename = time().rand().".".$extension;
$tmp_name=$_FILES['image']['tmp_name'];
// move image to the desired folder
if(move_uploaded_file($tmp_name, $saved.$filename)){
echo "Success!";
// if success insert location into database
$insert="INSERT INTO stored (folder_name,file_name) VALUES('$saved', '$filename')";
// if the query is correct
if($result=mysqli_query($con,$insert)){
echo "DONE";
echo"</br>";
// attempt to print image
echo "<img src=getimage.php?file_name=$filename>";
}
}
}
else{
echo "Please select a photo!!";
}
?>
Now as you can see I have an < img > tag. To try and learn, I was trying to just display the recently uploaded image. To try and do this I created a getimage file.
<?php
//getimage.php
// set the page to display images
header("Content-Type: image/jpeg");
include "connection.php";
// get requested filename
$name = ($_GET['file_name']);
$query = "SELECT * FROM stored WHERE file_name=$name";
$image = mysqli_query($con,$query);
$row = mysqli_fetch_array($image,MYSQLI_ASSOC);
$img = $row['file_name'];
echo $img;
?>
My database structure is as follows:
database name = db_file.
table name = stored.
columns = folder_name, file_name
Again, this is just a small project so I know I will have to alter the database if I wish to create a larger more efficient application.
It seems you use the database lookup to get just the file name, but you already have the file name. Try adding the folder name, create a valid path.
change
$img = $row['file_name'];
to
$img = $row['folder_name'] . '/' . $row['file_name'];
check your <img>tag to see if the correct url is present. You may or may not need the '/', it depends on how you stored the folder name. You may need to add the domain name. There is just not enough information know what is needed.
Your <img> should look like this
<img href="http://www.yourdomain.com/folder name/file name">
in the end
Hello I am having trouble getting imagepng to not only create a image on my server but name it a specific name according to a POST input from a form on another page.
It will display the finished image on screen just fine but depending on which method I try (tried many), it either creates a file name ".png" or does not create one at all.
Here is the "form.html" the info is pulling from:
<form action="test.php" method="post"><BR><BR>
Type Forum Name:
<input type="text" name="name" value=""/><BR><BR>
<input type="submit" name="Submit1"><BR><BR>
</form><BR>
And here is "test.php":
<?php
$name = $_POST["name"];
$input1 = $_POST["fsbg"];
$input2 = $_POST["rank"];
header ("Content-type: image/png");
$background = imagecreatefrompng($fsbg);
$pkt = imagecreatefromgif($rank);
imagecopymerge($background,$pkt,260,136,0,0,55,55,100);
imagepng($background);
$save = strtolower($name) .".png";
imagepng($background, $save);
imagedestroy($background);
imagedestroy($pkt);
header('location:link.php');
?>
I have set folder/sub folder permissions to 777, tried saving in the same path as the file, as well as saving in a sub folder called "pvt".
Any help/Advice would be greatly appreciated, Thanks!
You do not have a "name" field in the form. You have a unnamed field whose value is "name":
<input type="text" value="name"/><BR><BR>
Needs to be:
<input type="text" name="name" value="name"/><BR><BR>
Also, when you do this, sanitize the input. What if I sent you name="../../etc/passwd" or something like that? (of course this example won't work. But still.).
At the very least use basename to strip all path components from the name, and realpath to ensure you're writing where you should.
I finally figured this out.
Turns out even though I was changing the Permissions in my FTP, I still needed to add this in to get it to save a file: chmod($name,0755);
Completed code looks like:
$background = imagecreatefrompng($fsbg);
$pkt = imagecreatefromgif($rank);
//$pkt2 = imagecreatefromgif($rank);
imagecopymerge($background,$pkt,260,136,0,0,55,55,100);
//imagecopymerge($background,$pkt2,290,136,0,0,55,55,100);
header( "Content-type: image/png" );
imagepng($background);
$name = $filename.".png";
chmod($name,0755);
imagepng($background, $name);
imagedestroy($background);
imagedestroy($pkt);
Thanks to all who tooks the time to reply.
I am new to php and trying to upload an image file in mysql database using php.I tried various tutorial but it didnot work for me.
Code Snippet:-
<?php
//connect to database. Username and password need to be changed
mysql_connect("localhost", "root", "");
//Select database, database_name needs to be changed
mysql_select_db("yelldb");
if (!$_POST['uploaded']){
//If nothing has been uploaded display the form
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"
ENCTYPE="multipart/form-data">
Upload:<br><br>
<input type="file" name="image"><br><br>
<input type="hidden" name="uploaded" value="1">
<input type="submit" value="Upload">
</form>
<?php
}else{
//if the form hasn't been submitted then:
//from here onwards, we are copying the file to the directory you made earlier, so it can then be moved
//into the database. The image is named after the persons IP address until it gets moved into the database
//get users IP
$ip=$_SERVER['REMOTE_ADDR'];
//don't continue if an image hasn't been uploaded
if (!empty($image)){
//copy the image to directory
copy($image, "./temporary/".$ip."");
//open the copied image, ready to encode into text to go into the database
$filename1 = "./temporary/".$_SERVER['REMOTE_ADDR'];
$fp1 = fopen($filename1, "r");
//record the image contents into a variable
$contents1 = fread($fp1, filesize($filename1));
//close the file
fclose($fp1);
//encode the image into text
$encoded = chunk_split(base64_encode($contents1));
//insert information into the database
mysql_query("INSERT INTO servicelist (ImgData)"."VALUES ('$encoded')");
//delete the temporary file we made
//unlink($filename1);
}
}
?>
We don't save out the whole image in our database usually. We go through inserting the permanent of picture in our database. Use this php function
move_uploaded_file(file,newloc)
This will move from your temporary directory to permanent directory. Then, get path from there and insert that to the database.
Typically, you wouldn't save an entire image into an SQL database. Instead, you store the on disk path or some other 'pointer' to the actual file.
Change your code to read something like the following:
//don't continue if an image hasn't been uploaded
if (isset($_POST['image'])){
$image = $_POST['image'];
//copy the image to directory
$path = "/some/path";
move_uploaded_file($image,$path);
//store the name and path. PS: you will want to validate your input, and look
//at using prepared statements.
//Concentating values like this is NOT safe, or ideal
$location = $path . "/" . $image
mysql_query("INSERT INTO servicelist (ImgData) VALUES (" . $location . ")");
}
If however, you still wish to store the image in the SQL database, look into the blob storage type, not encoded text.
PHP move_uploaded_file