Forum PHP.pl > Proszeo poprawienie skryptu (FORM)

Pomoc - Szukaj - Użytkownicy - Kalendarz

Pełna wersja: Proszeo poprawienie skryptu (FORM)

Gość_janusz_23_*

18.08.2006, 11:45:11

wITAM!
mam problem z ograniczeniem typu przyesłynaych na serwer plikow (chcialbym zeby mozna bylo wysylac tylko jpg, gif, jpeg, doc i pdf) gdzie mam wpisac te ograniczenia w poniższym skrypcie i w jakiej formie tzn. odzielnie czy w jedym polecieniu.
Prosze o pomoc
P.S. czy ten skrypt jest bezpieczny??

<?php

$site_name = $_SERVER['HTTP_HOST'];
$url_dir = "http://".$_SERVER['HTTP_HOST'].dirname($_SERVER['PHP_SELF']);
$url_this = "http://".$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF'];

$upload_dir = "upload_files/";
$upload_url = $url_dir."/upload_files/";
$message ="";

//create upload_files directory if not exist
//If it does not work, create on your own and change permission.
if (!is_dir("upload_files")) {
die ("upload_files directory doesn't exist");
}

if ($_FILES['userfile']) {
$message = do_upload($upload_dir, $upload_url);
}
else {
$message = "Invalid File Specified.";
}

print $message;

function do_upload($upload_dir, $upload_url) {

$temp_name = $_FILES['userfile']['tmp_name'];
$file_name = $_FILES['userfile']['name'];
$file_type = $_FILES['userfile']['type'];
$file_size = $_FILES['userfile']['size'];
$result = $_FILES['userfile']['error'];
$file_url = $upload_url.$file_name;
$file_path = $upload_dir.$file_name;

//File Name Check
if ( $file_name =="") {
$message = "Invalid File Name Specified";
return $message;
}
//File Size Check
else if ( $file_size > 5000000) {
$message = "The file size is over 5 MB.";
return $message;
}
//File Type Check
else if ( $file_type == "*.jpg" ) {
$message = "Sorry, You cannot upload any script file" ;
return $message;
}

$result = move_uploaded_file($temp_name, $file_path);
$message = ($result)?"File url <a href=$file_url>$file_url</a>" :
"Somthing is wrong with uploading a file.";

return $message;
}
?>
<html>
<head>
<title>Untitled</title>
</head>
<body>
<form name="upload" id="upload" ENCTYPE="multipart/form-data" method="post">
Upload Image<input type="file" id="userfile" name="userfile">
<input type="submit" name="upload" value="Upload">
</form>

</body>
</html>

Ja_Szczur

18.08.2006, 11:51:34

[PHP] pobierz, plaintext 
<?php
//File Type Check
else if ( $file_type != "image/pjpeg" && $file_type != "image/gif" && ... ) {
$message = "Sorry, You cannot upload any files except jpg,gif,pdf... " ;
return $message;
}
?>
[PHP] pobierz, plaintext

janisz_23

18.08.2006, 12:14:04

Dziękuje za promoc!

Przypomnialo mi się jeszcze jedno. Chce zeby w formularzu było 5 pól do załadowania zdjęc. Czy wystarczy jedynie w kodzie HTMl dodać kilka pól inpute czy trzeba też zmieniać sam skrypt php??

<?php

$site_name = $_SERVER['HTTP_HOST'];
$url_dir = "http://".$_SERVER['HTTP_HOST'].dirname($_SERVER['PHP_SELF']);
$url_this = "http://".$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF'];

$upload_dir = "upload_files/";
$upload_url = $url_dir."/upload_files/";
$message ="";

//create upload_files directory if not exist
//If it does not work, create on your own and change permission.
if (!is_dir("upload_files")) {
die ("upload_files directory doesn't exist");
}

if ($_FILES['userfile']) {
$message = do_upload($upload_dir, $upload_url);
}
else {
$message = "Invalid File Specified.";
}

print $message;

function do_upload($upload_dir, $upload_url) {

$temp_name = $_FILES['userfile']['tmp_name'];
$file_name = $_FILES['userfile']['name'];
$file_type = $_FILES['userfile']['type'];
$file_size = $_FILES['userfile']['size'];
$result = $_FILES['userfile']['error'];
$file_url = $upload_url.$file_name;
$file_path = $upload_dir.$file_name;

//File Name Check
if ( $file_name =="") {
$message = "Invalid File Name Specified";
return $message;
}
//File Size Check
else if ( $file_size > 5000000) {
$message = "The file size is over 5 MB.";
return $message;
}
//File Type Check
else if ( $file_type == "*.jpg" ) {
$message = "Sorry, You cannot upload any script file" ;
return $message;
}

$result = move_uploaded_file($temp_name, $file_path);
$message = ($result)?"File url <a href=$file_url>$file_url</a>" :
"Somthing is wrong with uploading a file.";

return $message;
}
?>
<html>
<head>
<title>Untitled</title>
</head>
<body>
<form name="upload" id="upload" ENCTYPE="multipart/form-data" method="post">
Upload Image<input type="file" id="userfile" name="userfile">
<input type="submit" name="upload" value="Upload">
</form>

Ja_Szczur

19.08.2006, 17:17:59

krypto-reklama ! ;]

To jest wersja lo-fi głównej zawartości. Aby zobaczyć pełną wersję z większą zawartością, obrazkami i formatowaniem proszę kliknij tutaj.