0

Hello

I'm trying to name a uploaded image to my website with a random number / letter combination
and if the number it generates is taken it makes a new one untill it finds a free one, then i need to insert the url of the image into a table called photos -> url
how would i name it, and go about it?

Thanks in advanced

2
Contributors
2
Replies
3
Views
6 Years
Discussion Span
Last Post by MackeyJenkins
0
// make a note of the current working directory, relative to root. 
$directory_self = str_replace(basename($_SERVER['PHP_SELF']), '', $_SERVER['PHP_SELF']); 

// make a note of the directory that will recieve the uploaded files 
$uploadsDirectory = $_SERVER['DOCUMENT_ROOT'] . $directory_self . 'photos/'; 

// make a note of the location of the upload form in case we need it 
$uploadForm = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'uploadform.php'; 

// make a note of the location of the success page 
$uploadSuccess = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'uploaded.php'; 

// name of the fieldname used for the file in the HTML form 
$fieldname = 'file'; 

//echo'<pre>';print_r($_FILES);exit; 



// Now let's deal with the uploaded files 

// possible PHP upload errors 
$errors = array(1 => 'php.ini max file size exceeded', 
                2 => 'html form max file size exceeded', 
                3 => 'file upload was only partial', 
                4 => 'no file was attached'); 

// check the upload form was actually submitted else print form 
isset($_POST['submit']) 
    or error('the upload form is neaded', $uploadForm); 
     
// check if any files were uploaded and if 
// so store the active $_FILES array keys 
$active_keys = array(); 
foreach($_FILES[$fieldname]['name'] as $key => $filename) 
{ 
    if(!empty($filename)) 
    { 
        $active_keys[] = $key; 
    } 
} 

// check at least one file was uploaded 
count($active_keys) 
    or error('No files were uploaded', $uploadForm); 
         
// check for standard uploading errors 
foreach($active_keys as $key) 
{ 
    ($_FILES[$fieldname]['error'][$key] == 0) 
        or error($_FILES[$fieldname]['tmp_name'][$key].': '.$errors[$_FILES[$fieldname]['error'][$key]], $uploadForm); 
} 
     
// check that the file we are working on really was an HTTP upload 
foreach($active_keys as $key) 
{ 
    @is_uploaded_file($_FILES[$fieldname]['tmp_name'][$key]) 
        or error($_FILES[$fieldname]['tmp_name'][$key].' not an HTTP upload', $uploadForm); 
} 
     
// validation... since this is an image upload script we 
// should run a check to make sure the upload is an image 
foreach($active_keys as $key) 
{ 
    @getimagesize($_FILES[$fieldname]['tmp_name'][$key]) 
        or error($_FILES[$fieldname]['tmp_name'][$key].' not an image', $uploadForm); 
} 
     
// make a unique filename for the uploaded file and check it is 
// not taken... if it is keep trying until we find a vacant one 
foreach($active_keys as $key) 
{ 
    $now = time(); 
    while(file_exists($uploadFilename[$key] = $uploadsDirectory.$now.'-'.$_FILES[$fieldname]['name'][$key])) 
    { 
        $now++; 
    } 
} 

// now let's move the file to its final and allocate it with the new filename 
foreach($active_keys as $key) 
{ 
    @move_uploaded_file($_FILES[$fieldname]['tmp_name'][$key], $uploadFilename[$key]) 
        or error('receiving directory insuffiecient permission', $uploadForm); 
} 

echo $_FILES[$fieldname]['tmp_name'], $uploadFilename[$key];
$update = mysql_query('UPDATE photos SET url=');
     
// If you got this far, everything has worked and the file has been successfully saved. 
// We are now going to redirect the client to the success page. 
//header('Location: ' . $uploadSuccess); 

// make an error handler which will be used if the upload fails 
function error($error, $location, $seconds = 5) 
{ 
    header("Refresh: $seconds; URL=\"$location\""); 
    echo '<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"'."\n". 
    '"http://www.w3.org/TR/html4/strict.dtd">'."\n\n". 
    '<html lang="en">'."\n". 
    '    <head>'."\n". 
    '        <meta http-equiv="content-type" content="text/html; charset=iso-8859-1">'."\n\n". 
    '        <link rel="stylesheet" type="text/css" href="stylesheet.css">'."\n\n". 
    '    <title>Upload error</title>'."\n\n". 
    '    </head>'."\n\n". 
    '    <body>'."\n\n". 
    '    <div id="Upload">'."\n\n". 
    '        <h1>Upload failure</h1>'."\n\n". 
    '        <p>An error has occured: '."\n\n". 
    '        <span class="red">' . $error . '...</span>'."\n\n". 
    '         The upload form is reloading</p>'."\n\n". 
    '     </div>'."\n\n". 
    '</html>'; 
    exit; 
} // end error handler

i have no idea how i can get the url, ive tryed some things, youll see theres a query there, missing the url id put in the database

This topic has been dead for over six months. Start a new discussion instead.
Have something to contribute to this discussion? Please be thoughtful, detailed and courteous, and be sure to adhere to our posting rules.