これが私が達成しようとしていることです。ユーザーが写真を読み込んで、プロフィール ページに表示できるようにする必要があります。私のSQLデータベースには、ユーザー名、パスワード、名、姓、写真のようなフィールドを持つ「メンバー」という名前のテーブルが1つあります。写真を除くすべてのフィールドが登録フォームに入力されています。ユーザーがプロフィール ページに移動すると、プロフィールに写真をアップロードするためのフォームが表示されます。upload_form.php のこのコードを取得しました.. (コードは以下にリストされています) と別のファイル upload_processor.php (コードは後にリストされています)
このコードは、ファイルを自分のフォルダー Upload_files に正常にロードし、ファイルの名前を次のように変更します... 1140732936-filename.jpg ファイルが一意であることを確認します。SQLテーブルの「写真」フィールドに保存された1140732936-filename.jpgの名前を取得するにはどうすればよいですか? 何か方法はありますか？助けてください....
upload_form のコード
<?php
// filename: upload.form.php
// first let's set some variables
// 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 location of the upload handler script
$uploadHandler = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'upload.processor.php';
// set a max file size for the html upload form
$max_file_size = 3000000; // size in bytes
// now echo the html page
?>
これは同じファイルのhtmlフォームです
<form id="Upload" action="<?php echo $uploadHandler ?>" enctype="multipart/form-data" method="post">
<h1>
Upload form
</h1>
<p>
<input type="hidden" name="MAX_FILE_SIZE" value="<?php echo $max_file_size ?>">
</p>
<p>
<label for="file">File to upload:</label>
<input id="file" type="file" name="file">
</p>
<p>
<label for="submit">Press to...</label>
<input id="submit" type="submit" name="submit" value="Upload me!">
</p>
</form>
</body>
フォームを処理するファイルのコードを次に示します。
<?php
// filename: upload.processor.php
// first let's set some variables
// 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 file
$uploadsDirectory = $_SERVER['DOCUMENT_ROOT'] . $directory_self . 'uploaded_files/';
// make a note of the location of the upload form in case we need it
$uploadForm = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'updateprofile.php';
// make a note of the location of the success page
$uploadSuccess = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'upload.success.php';
// fieldname used within the file <input> of the HTML form
$fieldname = 'file';
// Now let's deal with the upload
// possible PHP upload errors
$errors = array(1 => 'php.ini max file size exceeded',
2 => 'html form max file size exceeded',
3 => 'file upload was only partial',
4 => 'no file was attached');
// check the upload form was actually submitted else print the form
isset($_POST['submit'])
or error('the upload form is neaded', $uploadForm);
// check for PHP's built-in uploading errors
($_FILES[$fieldname]['error'] == 0)
or error($errors[$_FILES[$fieldname]['error']], $uploadForm);
// check that the file we are working on really was the subject of an HTTP upload
@is_uploaded_file($_FILES[$fieldname]['tmp_name'])
or error('not an HTTP upload', $uploadForm);
// validation... since this is an image upload script we should run a check
// to make sure the uploaded file is in fact an image. Here is a simple check:
// getimagesize() returns false if the file tested is not an image.
@getimagesize($_FILES[$fieldname]['tmp_name'])
or error('only image uploads are allowed', $uploadForm);
// make a unique filename for the uploaded file and check it is not already
// taken... if it is already taken keep trying until we find a vacant one
// sample filename: 1140732936-filename.jpg
$now = time();
while(file_exists($uploadFilename = $uploadsDirectory.$now.'-'.$_FILES[$fieldname]['name']))
{
$now++;
}
// now let's move the file to its final location and allocate the new filename to it
@move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadFilename)
or error('receiving directory insuffiecient permission', $uploadForm);
// If you got this far, everything has worked and the file has been successfully saved.
// We are now going to redirect the client to a success page.
header('Location: ' . $uploadSuccess);
// The following function is an error handler which is used
// to output an HTML error page if the file upload fails
function error($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
?>