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从AS3和PHP中的Flash网站将图像上传到服务器

[英]Uploading images to server from a flash web site in AS3 and PHP

Ive been trying to create this code for some time now and just wondered if someone out there who may read code better than me may work out why the image doesnt end up on my server as i seem to have brain freeze on this :( 我一直在尝试创建此代码一段时间,只是想知道是否有人比我能读得更好的代码可能会弄清楚为什么图像无法最终显示在我的服务器上,因为我似乎对此不动了:(

var fileRef:FileReference = new FileReference(); 
fileRef.addEventListener(Event.SELECT, selectHandler); 
fileRef.addEventListener(Event.COMPLETE, completeHandler); 
try 
{ 
var success:Boolean = fileRef.browse(); 
} 
catch (error:Error) 
{ 
trace("Unable to browse for files."); 
} 
function selectHandler(event:Event):void 
{ 
var request:URLRequest = new URLRequest("http://localhost/upload.php") 
try 
{ 
fileRef.upload(request); 
} 
catch (error:Error) 
{ 
trace("Unable to upload file."); 
} 
} 
function completeHandler(event:Event):void 
{ 
trace("uploaded"); 
}
customerService.createClients(SlagsData);
}

And then here is the php on the server 然后这是服务器上的php

<?php
define('UPLOAD_DIR', 'c:/wamp/www/IMAGES/');
define('UPLOAD_DIR_default', 'c:/wamp/www/IMAGES/0/image01.jpg');
$hostname_thatexclients = "localhost";
$database_thatexclients = "myexclients";
$username_thatexclients = "root";
$password_thatexclients = "";
$thatexclients = mysql_connect($hostname_thatexclients, $username_thatexclients,   $password_thatexclients) or trigger_error(mysql_error(),E_USER_ERROR); 
set_time_limit ( 240 );




if($_FILES['yourpic']['size'] > 1) 
{
if($_FILES['yourpic']['size'] < 5000000) 
$newname = "image01.jpg";
$id = 0;
$query = ("SELECT * FROM clients WHERE  ID = (SELECT MAX(ID) FROM clients)") or die(mysql_error());
mysql_select_db($database_thatexclients, $thatexclients);
$Result2 = mysql_query($query, $thatexclients) or die(mysql_error());
While($row = mysql_fetch_array($Result2))
{
$id = $row["ID"];
$id = ($id + "1");
}  
mkdir(UPLOAD_DIR.$id, 0777, true) or die ("Could not make directory"); 
$idr = ($id.'/');  
move_uploaded_file($_FILES['yourpic']['tmp_name'],UPLOAD_DIR.$idr .$newname);// this one works  
}
else
{
$id = 0;
}

?>

Any help would be great :) 任何帮助将是巨大的:)

好吧,代码必须一直是狡猾的.jpeg图片,代码看起来一直在令人尴尬地工作着

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