How to delete previous image after changing it?

[scdogas321]

Hello everyone, i have one question.
So, I've made avatar upload in my website and there is one problem.
My code is:

$path = "images/avatars/".$HTTP_POST_FILES['image']['name'];
rename($HTTP_POST_FILES['image']['tmp_name'], $path);


$query = mysql_query("UPDATE user SET imageloc='$path' WHERE username='$username'");

As you can see my image (avatar) is copied to: images/avatars/.
Now, when I'm changing my image (avatar) it's copying new image (avatar) to: image/avatar/.
but lieving the old one too... And it's just takes a lot of takes a lot of after a lot of images(avatars).

Is there are way to delete old image(avatar) when changing it to the new one?
Thank you. 🙂

[bradgrafelman]

Retrieve the name of the old file and then use [man]unlink/man to remove it.

[dalecosp]

Or always write the same name for a user's avatar ... unless you want to offer them historical versions (for which I can see no reason).

So, if I'm user #1234, my avatar is user_avatar_1234.png ... and when I upload a new one it's still "user_avatar_1234.png" ... there there's no "old file" because the new file overwrote the old one ....

[scdogas321]

bradgrafelman;11030329 wrote:

Retrieve the name of the old file and then use [man]unlink/man to remove it.

So I've done that:

$location = "images/avatars/$name";
				$path = "images/avatars/".$HTTP_POST_FILES['image']['name'];
				rename($HTTP_POST_FILES['image']['tmp_name'], $path);

			$trinti = ("SELECT imageloc FROM user WHERE username='$username'");
			unlink();

			if ($trinti) {
			$query = mysql_query("UPDATE user SET imageloc='$location' WHERE username='$username'");
			echo "<script type='text/javascript'>    
										window.location.reload('../index.php');    
										window.close();
										</script>   ";
				}

And I don't know if I've done everything right, but magically it's working 😮, but when I go to: images/avatars there is no avatars, but when I log out and log in It was saved. As I said i don't know if I've done everything right and if I did, then i need to fix it 'cause I need to see images/avatars/image.jpg, .png, .jpeg, .gif.
Thanks. 🙂

bradgrafelman;11030329 wrote:

Retrieve the name of the old file and then use [man]unlink/man to remove it.

I don't know how to do that, but I think it's a good idea too! Maybe you can tell me how to do that? 🙂
Thanks. 🙂

[scdogas321]

dalecosp;11030331 wrote:

Or always write the same name for a user's avatar ... unless you want to offer them historical versions (for which I can see no reason).

So, if I'm user #1234, my avatar is user_avatar_1234.png ... and when I upload a new one it's still "user_avatar_1234.png" ... there there's no "old file" because the new file overwrote the old one ....

Sorry, I've made mistake by Quoting the wrong person. 🙂 I don't know how to do that, but I think it's a good idea too! Maybe you can tell me how to do that?
Thanks. 🙂

[scdogas321]

scdogas321;11030337 wrote:

So I've done that:

$location = "images/avatars/$name";
				$path = "images/avatars/".$HTTP_POST_FILES['image']['name'];
				rename($HTTP_POST_FILES['image']['tmp_name'], $path);

			$trinti = ("SELECT imageloc FROM user WHERE username='$username'");
			unlink();

			if ($trinti) {
			$query = mysql_query("UPDATE user SET imageloc='$location' WHERE username='$username'");
			echo "<script type='text/javascript'>    
										window.location.reload('../index.php');    
										window.close();
										</script>   ";
				}

And I don't know if I've done everything right, but magically it's working 😮, but when I go to: images/avatars there is no avatars, but when I log out and log in It was saved. As I said i don't know if I've done everything right and if I did, then i need to fix it 'cause I need to see images/avatars/image.jpg, .png, .jpeg, .gif.
Thanks. 🙂

I looked in my avatar folder today and if i've done right then this code not working :/

[bradgrafelman]

That's because you don't do anything with $trinti, such as 1) execute it as a SQL query, 2) check to see if it succeeded, and 3) retrieve the result. Furthermore, you didn't give [man]unlink/man anything to... unlink.

EDIT: Having said that, have you considered dalecosp's suggestion above? All you would need to do is simply use some unique value (such as the username or a user ID) for each user as that user's avatar filename.

[scdogas321]

bradgrafelman;11030363 wrote:

That's because you don't do anything with $trinti, such as 1) execute it as a SQL query, 2) check to see if it succeeded, and 3) retrieve the result. Furthermore, you didn't give [man]unlink/man anything to... unlink.

EDIT: Having said that, have you considered dalecosp's suggestion above? All you would need to do is simply use some unique value (such as the username or a user ID) for each user as that user's avatar filename.

I would be really appreciated if you fix my code to your suggestion or dalecosp. 🙂
I'm not so good on php, sorry 🙁

<?php

	if ($_POST['submit']) {
	require("./connect.php");

$name = $_FILES['image']['name'];
$tmp_name = $_FILE['image']['tmp_name'];

	if ($name) {

			$extensions = array('.png', '.gif', '.jpg', '.jpeg');  
			$extension = strrchr($_FILES['image']['name'], '.');  

			if(!in_array($extension, $extensions)){ 

				echo "blgoas formatas";

			}

		else {
		$file_size = $_FILE['image']['tmp_name'];
		$limit_size=50000;

			if($file_size >= $limit_size){
			echo "Your file size is over limit";
			}
			else {


			$location = "images/avatars/$name";
			$path = "images/avatars/".$HTTP_POST_FILES['image']['name'];
			rename($HTTP_POST_FILES['image']['tmp_name'], $path);


			$query = mysql_query("UPDATE user SET imageloc='$location' WHERE username='$username'");
			echo "<script type='text/javascript'>    
										window.location.reload('../index.php');    
										window.close();
										</script>   ";


			}
		}
	}
	else
		echo "J&#363;s turite pasirinkti nuotrauk&#261;";


mysql_close();
}

echo "<form action='redaguoti.php' method='post' enctype='multipart/form-data'>

			File:
			<input type='file' name='image'> <input type='submit' name='submit' value='Upload'>
			</form>";			

?>

And by the way, sorry for other language used in my code, it's my language 🙂.
Thanks