Jumat, 26 Juli 2013

PHP Indonesia - Facebook: GAN mau tanya kenapa ketika upload malah blank yah...

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GAN mau tanya kenapa ketika upload malah blank yah...
Jul 26th 2013, 17:59, by Kurnia Elf

Kurnia Elf 5:59pm Jul 26
GAN mau tanya kenapa ketika upload malah blank yah?kira" apa yang salah??nih scriptnya gan.

ini file yang ke satu

form_upload.php

<script type="text/javascript">
function show ()
{
var n = document.myform.jumfile.value; // dwdwdw
var i ;
var string = "";

for (i=0; i<=n-1; i++)
{
string = string + "Pilih file: <input name=\"userfile"+ i+"\" type=\"file\">\<br>";
//br br diatas untuk memberi baris baru
}
document.getElementById('selectfile').innerHTML=string;
document.myform.n.value=n;
}
</script>

<form name="myform" enctype="multipart/form-data" action="upload.php" method="post">
Pilih Jumlah File
<select name ="jumfile" onChange="show()">
<option value=""> Silahkan Pilih </option>
<option value="1"> 1 </option>
<option value="2"> 2 </option>
<option value="3"> 3 </option>
<option value="4"> 4 </option>
<option value="5"> 5 </option>
<option value="6"> 6 </option>
<option value="7"> 7 </option>
<option value="8"> 8 </option>
<option value="9"> 9 </option>
<option value="10"> 10 </option>
</select><br><br>
<input type="hidden" name="MAX_FILE_SIZE" value="2000000" />
<div id="selectfile">
</div>
<br>
<input type="hidden" name="n"/>
<input type="submit" name="submit" value="Upload"/>
</form>

ini file yang ke dua

upload.php

<?php
$n= $_POST['n'];
//setting nama folder tempat upload
$uploaddir='data/';
for ($i=0; $i<=n-1; $i++) {
//membaca nama file yang akan diupload

$fileName=$_FILES ['userfile'.$i]['name'];

//membaca ukuran file yang akan di upload

$fileSIZE=$_FILES ['userfile'.$i]['size'];

//membaca file temporary yang akan disimpan di server

$tmpName=$_FILES['userfile'.$i]['tmp_name'];

//menggabungkan nama folder dan nama file

$uploadfile=$uploaddir.$fileName;

//proses upload file ke folder data

if ($fileSize>0){

if (move_uploaded_file($tmpName, $uploadfile)){

echo "File".$fileName."telah diupload<br>";

}
else {
echo "File".$fileName."gagal di upload <br>";
}

}
}

?>

mudah"n yang bantu jawab di tambah pahala nya n rezekinya :D

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