php çoklu dosya yükleme komut

3 Cevap

Herkes iyi bir php biliyor veya bir web sunucusuna yüklemek için birden fazla dosya yükleme komut ajax mı?

Burada fark, hiçbir şey yok flaş yani istemci makinede gerekli olmasıdır!

Ben sadece tarayıcı ile çalışmak istiyorum.

3 Cevap

digitarald’s fancy upload

<table width="500" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<form action="multiple_upload_ac.php" method="post" enctype="multipart/form-data" name="form1" id="form1">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td><strong>multiple Files Upload </strong></td>
</tr>
<tr>
<td>Select file
<input name="ufile[]" type="file" id="ufile[]" size="50" /></td>
</tr>
<tr>
<td>Select file
<input name="ufile[]" type="file" id="ufile[]" size="50" /></td>
</tr>
<tr>
<td>Select file
<input name="ufile[]" type="file" id="ufile[]" size="50" /></td>
</tr>
<tr>
<td align="center"><input type="submit" name="Submit" value="Upload" /></td>
</tr>
</table>
</td>
</form>
</tr>
</table>

Adım 2: Dosya multiple_upload_ac.php Oluştur

<?php

//set where you want to store files
//in this example we keep file in folder upload
//$HTTP_POST_FILES['ufile']['name']; = upload file name
//for example upload file name cartoon.gif . $path will be upload/cartoon.gif

$path1= "upload/".$HTTP_POST_FILES['ufile']['name'][0];
$path2= "upload/".$HTTP_POST_FILES['ufile']['name'][1];
$path3= "upload/".$HTTP_POST_FILES['ufile']['name'][2];

//copy file to where you want to store file
copy($HTTP_POST_FILES['ufile']['tmp_name'][0], $path1);
copy($HTTP_POST_FILES['ufile']['tmp_name'][1], $path2);
copy($HTTP_POST_FILES['ufile']['tmp_name'][2], $path3);

//$HTTP_POST_FILES['ufile']['name'] = file name
//$HTTP_POST_FILES['ufile']['size'] = file size
//$HTTP_POST_FILES['ufile']['type'] = type of file
echo "File Name :".$HTTP_POST_FILES['ufile']['name'][0]."<BR/>";
echo "File Size :".$HTTP_POST_FILES['ufile']['size'][0]."<BR/>";
echo "File Type :".$HTTP_POST_FILES['ufile']['type'][0]."<BR/>";
echo "<img src=\"$path1\" width=\"150\" height=\"150\">";
echo "<P>";

echo "File Name :".$HTTP_POST_FILES['ufile']['name'][1]."<BR/>";
echo "File Size :".$HTTP_POST_FILES['ufile']['size'][1]."<BR/>";
echo "File Type :".$HTTP_POST_FILES['ufile']['type'][1]."<BR/>";
echo "<img src=\"$path2\" width=\"150\" height=\"150\">";
echo "<P>";

echo "File Name :".$HTTP_POST_FILES['ufile']['name'][2]."<BR/>";
echo "File Size :".$HTTP_POST_FILES['ufile']['size'][2]."<BR/>";
echo "File Type :".$HTTP_POST_FILES['ufile']['type'][2]."<BR/>";
echo "<img src=\"$path3\" width=\"150\" height=\"150\">";

///////////////////////////////////////////////////////

// Use this code to display the error or success.

$filesize1=$HTTP_POST_FILES['ufile']['size'][0];
$filesize2=$HTTP_POST_FILES['ufile']['size'][1];
$filesize3=$HTTP_POST_FILES['ufile']['size'][2];

if($filesize1 && $filesize2 && $filesize3 != 0)
{
echo "We have recieved your files";
}

else {
echo "ERROR.....";
}

//////////////////////////////////////////////

// What files that have a problem? (if found)

if($filesize1==0) {
echo "There're something error in your first file";
echo "<BR />";
}

if($filesize2==0) {
echo "There're something error in your second file";
echo "<BR />";
}

if($filesize3==0) {
echo "There're something error in your third file";
echo "<BR />";
}
?>

I found this code in http://www.phpeasystep.com/phptu/2.html
for more details check <a href="http://www.specialend.com/">Tutorial</a>

I plupload, temel tek dosya HTML4 yükleme ek istemci teknolojileri geniş bir yelpazede (html5, flash, silverlight, BrowserPlus, dişliler) destekleyen bir hayranı değilim, sadece flaş + HTML4 son çareleri yok SWFUpload aksine .

Ayrıca jQuery ile güzel bütünleşir.

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