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Picture upload


Boxerman

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hi guys im in need of a script that when a user uploads it will display the picture uploaded and add to a page where other users have uploaded too!

 

this is what i got so far..

 

<?php
//define a maxim size for the uploaded images in Kb
define ("MAX_SIZE","100"); 

//This function reads the extension of the file. It is used to determine if the file  is an image by checking the extension.
function getExtension($str) {
         $i = strrpos($str,".");
         if (!$i) { return ""; }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
}

//This variable is used as a flag. The value is initialized with 0 (meaning no error  found)  
//and it will be changed to 1 if an errro occures.  
//If the error occures the file will not be uploaded.
$errors=0;
//checks if the form has been submitted
if(isset($_POST['Submit'])) 
{
	//reads the name of the file the user submitted for uploading
	$image=$_FILES['image']['name'];
	//if it is not empty
	if ($image) 
	{
	//get the original name of the file from the clients machine
		$filename = stripslashes($_FILES['image']['name']);
	//get the extension of the file in a lower case format
  		$extension = getExtension($filename);
		$extension = strtolower($extension);
	//if it is not a known extension, we will suppose it is an error and will not  upload the file,  
//otherwise we will do more tests
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) 
		{
	//print error message
			echo '<h1>Unknown extension!</h1>';
			$errors=1;
		}
		else
		{
//get the size of the image in bytes
//$_FILES['image']['tmp_name'] is the temporary filename of the file
//in which the uploaded file was stored on the server
$size=filesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
echo '<h1>You have exceeded the size limit!</h1>';
$errors=1;
}

//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname="images/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied) 
{
echo '<h1>Copy unsuccessfull!</h1>';
$errors=1;
}}}}

//If no errors registred, print the success message
if(isset($_POST['Submit']) && !$errors) 
{
	echo "<h1>File Uploaded Successfully! Try again!</h1>";
}

?>

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Well, since I haven't seen your code. I assume you could do something like

 

$sql = "SELECT * FROM pictures ORDER BY id DESC LIMIT 1";

 

In this way you get last (id) picture in your database and that one is shown.

 

Again, I haven't seen your code so I just asume what you could do.

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The code you have above, that's not usersimages.php is it?

 

We would need to see the code of usersimages.php because that is the code that is relevant to the displaying of the image.

 

Also, you just upload the image. You don't store it's name somewhere for later reference or anything? How are you supposed to know which image to load up? You need a database or perhaps a flat file which stores the id of the user who is uploading, along with the name of the image.

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You shouldn't upload pictures to a database. Instead, you store the name of the uploaded image to the database. Assign it a unique ID, and perhaps the ID of the user who uploaded. Then you can call on the image later and get the name of the image from the database.

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If you want to just display all uploaded images, without caring of the storage of the image information you can run a basic script like this:

 

foreach(glob("images/*.*") as $image){
echo "<img src='" . $image . "' /><br />";
}

 

Of course if you have any .php files in there it'll try and put them in there, so you need to add in more error checking.

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