php multiple file upload script

Question

Кто-нибудь знает о хорошем сценарии загрузки файлов php или ajax для загрузки на веб-сервер?

Разница заключается в том, что на клиентской машине ничего не требуется, т.е. нет вспышки!

Я бы хотел, чтобы он работал только с браузером.

Answer 1

digitarald’s fancy upload

Answer 2

• FancyUpload
• SWFUpload

Answer 3

Я фанат plupload, которые поддерживают широкий спектр клиентских технологий (html5, Flash, SilverLight, browserplus, шестерни) в дополнение к базовому одиночному файлу html4 upload, в отличие от SWFUpload, который только делает flash + html4.

Он также прекрасно сочетается с jQuery.

Answer 4

<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> 

STEP2: Создать файл multiple_upload_ac.php

<?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>