2010-11-17 69 views
1

我有一个处理多个上传的PHP上传脚本。该脚本效果很好,但我得到下面这个错误:未定义的索引错误

Notice: Undefined index: uploadedFile in C:\wamp\www\setapro\rmc_manager.php on line 52 

下面这条线出现上述错误:

我试图做这样的事情:

if (!isset($_FILES['uploadedFile']['name'])){ 
$_FILES['uploadedFile']['name'] = ''; 
} 

但这是行不通的。我希望对通知做些事情。任何建议,将不胜感激。由于

编辑:我的HTML的部分

form action="rmc_manager.php" method="post" enctype="multipart/form-data"><input type="hidden" name="MAX_FILE_SIZE" value="3000000" /> 

Please choose files to upload:<br /> 
<input name="uploadedFile[]" type="file" class="txt" /><br /> 
<input name="uploadedFile[]" type="file" class="txt" /><br /> 
<input type="submit" name="uploadfiles" value="Upload files" class="btn" /> 

PHP上传经理:

if (isset($_POST['uploadfiles'])) { 
$number_of_uploaded_files = 0; 
$number_of_moved_files = 0; 
$uploaded_files = array(); 
$upload_directory = dirname(__file__) . '/Uploads/'; 
for ($i = 0; $i < count($_FILES['uploadedFile']['name']); $i++) { 
    if ($_FILES['uploadedFile']['name'][$i] != '') { //check if file field empty or not 
     $number_of_uploaded_files++; 
     $uploaded_files[] = $_FILES['uploadedFile']['name'][$i]; 
     //if (is_uploaded_file($_FILES['uploadedFile']['name'])){ 
     if (move_uploaded_file($_FILES['uploadedFile']['tmp_name'][$i], $upload_directory . $_FILES['uploadedFile']['name'][$i])) { 
      $number_of_moved_files++; 
     } 

    } 

} 

} 
echo "Number of files submitted $number_of_uploaded_files . <br/>"; 
echo "Number of successfully moved files $number_of_moved_files . <br/>"; 
echo "File Names are <br/>" . implode("<br/>", $uploaded_files); 
echo "<br></br>"; 
echo "<p> Please find the processed GGAs with dates in the Setapro project folder</p>"; 

回答

0

测试$_FILES['uploadedFile']$_FILES['uploadedFile']['name']之前!

编辑:其实,只是测试$_FILES['uploadedFile'] ^^

+0

对不起,我该如何测试?这里是新的。我可以通过呼应来做到吗? – ibiangalex 2010-11-17 16:09:16

+0

您正在使用if()...继续! – MatTheCat 2010-11-17 16:15:02

0

你这样做

ü应该使用PHP函数isset(...)之前,

例如

if(isset($_FILES['uploadedFile']['name'])) 
+0

我做了类似的事情,但没有添加[名字]哪个不起作用。 – ibiangalex 2010-11-17 20:08:27

+0

嗨...我看到你有两个文件输入作为数组...如何访问和调用文件属性的索引,如果有两个文件输入,我认为你必须使用0和1作为索引从PHP端访问它们,因为$ _FILES ['uploadedFile'] ['name']指向无... – Alejandro 2010-11-18 02:08:02