當用戶輸入他想要上傳的文件數量時,我正在創建一個頁面,然後在提交到另一個頁面上傳文件後,該頁面再次用表單呈現。
但是我收到一個錯誤[如果我上傳1個文件,否則問題是userfile的[極限-1]
Notice: Undefined index: userfile0 in E:\wamp\www\uploader\uploader.php on line 10
文件上傳的動態數量 - PHP
我在這裏貼我的代碼 的index.php
<html>
<?php
if (isset($_POST['num']) && !empty($_POST['num']))
{
?>
<form name="uploader" action="uploader.php" method="post">
<table>
<tr><td>Title</td><td>Select File</td><td>Description</td></tr>
<input type="text" name="number" value="<?php echo $_POST['num']; ?>"/>
<?php
for ($i=0;$i<$_POST['num'];$i++)
echo '<tr><td><input type="text" name="title'.$i.'"/></td>
<td><input type="file" id="userfile'.$i.'" name="userfile'.$i.'" size="30"></td>
<td><textarea name="desc'.$i.'" rows="4"></textarea></td></tr>';
?>
</table>
<input type="submit" name="b" value="Submit"/>
</form>
<?php
}
else
{
?>
<form name="form" method="post">
How many files to upload ? <input type="text" name="num" value="1"/>
<input type="submit" value="Submit"/>
</form>
<?php
}
?>
</html>
uploader.php
<?php
if (isset($_POST['number']))
{
for ($slot = 0; $slot < $_POST['number']; $slot++)
{
$title = $_POST["title$slot"];
$desc = $_POST["desc$slot"];
if (move_uploaded_file($_FILES["userfile$slot"]['tmp_name'],$_FILES["userfile$slot"]['name']))
echo $name.' file Uploaded !';
else
echo ' file not Uploaded ! ';
}
}
?>
Edit
刪除的SQL代碼
之前點擊_DOWN/CLOSE_ vote plz **提你的理由** !!! – Sourav 2011-06-04 08:33:25
你見過這個手冊頁嗎? [上傳多個文件](http://php.net/manual/en/features.file-upload.multiple.php) – hakre 2011-06-04 09:24:07
@hakre不,但是爲什麼? – Sourav 2011-06-04 09:24:51