check script ni sat.....

slier · #1 (**permalink**) 15-09-2006, 07:39 PM

hello guys...ni aku try buat image gallery kat web page aku ni..malang sekali tak berjaya....
aku try ambil semua gambar dalam folder d:\data.....but bila aku run script tu...alu dapat error ni:

Warning: sort() expects parameter 1 to be array, null given in C:\xamp\xampp\htdocs\projek2\gallery.php on line 19

Warning: Invalid argument supplied for foreach() in C:\xamp\xampp\htdocs\projek2\gallery.php on line 20

<?php
echo"
<html>
<head><title>Image gallery</title></head>
<body>";
$dir="d:/data/";

$do=opendir($dir);

while($filename=readdir($do)){
$filepath=$dir.$filename;

if(is_file($filepath) and ereg("\.jpeg$",$filename)){
$gambar[]=$filepath;
}
}

sort($gambar);
echo"Senarai Gambar Dalam Gallery Kami...<br>";
foreach($gambar as $listgambar){
echo"<img src='$listgambar'><br>";
}
?>
</body>
</html>

aku harap korang dapat tolong aku....

mysticmind · #2 (**permalink**) 16-09-2006, 04:41 PM

copy folder data ( yang ada gambo tuh kan? )
pindah ke :
C:\xamp\xampp\htdocs\projek2\data

dan ubahkan :

PHP Code:


			
$dir="data";

sebab pc localhost lain dengan hosting..
cube fikir camner hosting punya directory style..
ok, try dulu.

slier · #3 (**permalink**) 16-09-2006, 09:55 PM

erm...sorry lambat reply....mmg kena buat mcm tu ker....
x independence la klu mcm tu....klu folder tu kat luar public_html ,,,,so x leh run ker script ni....

x pa nanti aku try ko nyer suggestionn..

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