Re: [Linux-programlama] upload ile ilgili

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From: Erdal YAZICIOGLU (erdal.yazicioglu@gmail.com)
Date: Tue 28 Dec 2004 - 15:35:39 EET


Asagidaki kod umarim isinize yarar. Yaptigi is 5 adet resim upload u yapiyor
ve yaptiginiz upload u resimler klasörü altinda folderName girerek
olustururdugunuz klasörün altina atiyor.

Kolay gelsin

<form name="form2" method="post" action="" enctype="multipart/form-data">
<input type="text" name="folderName"> <br>
<input type="file" name="imagefile1"> <br>
<input type="file" name="imagefile2"> <br>
<input type="file" name="imagefile3"> <br>
<input type="file" name="imagefile4"> <br>
<input type="file" name="imagefile5"> <br>
<input type="submit" name="Submit" value="Submit">

<?
if(isset( $Submit ))
{

if(!$_POST['folderName']) echo " Baslik girmediniz";
$newdir = $_POST['folderName'];
if ($_FILES['imagefile1']['type'] == "image/gif" ||
$_FILES['imagefile2']['type'] == "image/gif" ||
$_FILES['imagefile3']['type'] == "image/gif" ||
$_FILES['imagefile4']['type'] == "image/gif" ||
$_FILES['imagefile5']['type'] == "image/gif")
{
 chdir("resimler/");
 $dir= opendir(".");
 if(mkdir("$newdir",0775))
 {
 echo "<br>" . $newdir . " klasörü yaratildi";
 }
 else
 {
 "problem";
 }
    copy ($_FILES['imagefile1']['tmp_name'],
"$newdir/".$_FILES['imagefile1']['name']) or die ("1 nolu resim
girilmedi");
    copy ($_FILES['imagefile2']['tmp_name'],
"$newdir/".$_FILES['imagefile2']['name']) or die ("2 nolu resim
girilmedi");
    copy ($_FILES['imagefile3']['tmp_name'],
"$newdir/".$_FILES['imagefile3']['name']) or die ("3 nolu resim
girilmedi");
    copy ($_FILES['imagefile4']['tmp_name'],
"$newdir/".$_FILES['imagefile4']['name']) or die ("4 nolu resim
girilmedi");
    copy ($_FILES['imagefile5']['tmp_name'],
"$newdir/".$_FILES['imagefile5']['name']) or die ("5 nolu resim
girilmedi");

        echo "";
        echo "Upload tamam....";
        }
else {
            echo "";

        echo "Kopyalanamadi. Yanlis dosya tipi
(".$_FILES['imagefile']['name'].")";
        }
}

?> </form>

Erdal YAZICIOGLU

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