2010-01-10 46 views
0

我該如何做一個php和xhtml表單,允許用戶上傳圖片並提交所有數據發送到我的電子郵件?帶表格的PHP表格上傳

+0

給我TEH codez?請更具體地說明處理文件上傳的哪些部分存在問題。 – Gordon 2010-01-10 16:01:22

回答

0

主要PHP網站本身也包括對建議的做法和陷阱一些有用的信息,以及一些樣品代碼 - 請參閱'Handling file uploads'部分。

0

試試這個

<? 
//print_r($_POST); 

if($_POST["action"] == "Upload Image") 
{ 
unset($imagename); 

if(!isset($_FILES) && isset($HTTP_POST_FILES)) 
$_FILES = $HTTP_POST_FILES; 

if(!isset($_FILES['image_file'])) 
$error["image_file"] = "An image was not found."; 


$imagename = basename($_FILES['image_file']['name']); 
//echo $imagename; 

if(empty($imagename)) 
$error["imagename"] = "The name of the image was not found."; 

if(empty($error)) 
{ 
$newimage = "images/" . $imagename; 
//echo $newimage; 
$result = @move_uploaded_file($_FILES['image_file']['tmp_name'], $newimage); 
if(empty($result)) 
$error["result"] = "There was an error moving the uploaded file."; 
} 

} 

?> 


<form method="POST" enctype="multipart/form-data" name="image_upload_form" action="<?$_SERVER["PHP_SELF"];?>"> 
<p><input type="file" name="image_file" size="20"></p> 
<p><input type="submit" value="Upload Image" name="action"></p> 
</form> 

<? 
if(is_array($error)) 
{ 
while(list($key, $val) = each($error)) 
{ 
echo $val; 
echo "<br>\n"; 
} 
} 
?>