我目前正在爲我的教員開發學生數據庫系統。我正在使用PHP和MySQL一起。我正在考慮爲學生上傳他們的個人資料照片創建一個選項,但我找不到任何適當的說明或教程。爲學生數據庫實施個人資料照片上傳
下面的代碼在處理文件上傳:
<?php
/* Script name: uploadFile.php
* Description: Uploads a file via HTTP with a POST form.
*/
if(!isset($_POST[‘Upload’]))
{
include(「form_upload.inc」);
}
else
{
if($_FILES[‘pix’][‘tmp_name’] == 「none」)
{
echo 「<p style=’font-weight: bold’>
File did not successfully upload. Check the
file size. File must be less than 500K.</p>」;
include(「form_upload.inc」);
exit();
}
if(!ereg(「image」,$_FILES[‘pix’][‘type’]))
{
echo 「<p style=’font-weight: bold’>
File is not a picture. Please try another
file.</p>」;
include(「form_upload.inc」);
exit();
}
else
{
$destination=’c:\data’.」\\」.$_FILES[‘pix’][‘name’];
$temp_file = $_FILES[‘pix’][‘tmp_name’];
move_uploaded_file($temp_file,$destination);
echo 「<p style=’font-weight: bold’>
The file has successfully uploaded:
{$_FILES[‘pix’][‘name’]}
({$_FILES[‘pix’][‘size’]})</p>」;
}
}
?>
代碼的文件上傳表單:
<!-- Program Name: form_upload.inc
Description: Displays a form to upload a file -->
<html>
<head><title>File Upload</title></head>
<body>
<ol><li>Enter the file name of the product picture you
want to upload or use the browse button
to navigate to the picture file.</li>
<li>When the path to the picture file shows in the
text field, click the Upload Picture
button.</li>
</ol>
<div align=」center」><hr />
<form enctype=」multipart/form-data」
action=」uploadFile.php」 method=」POST」>
<input type=」hidden」 name=」MAX_FILE_SIZE」
value=」500000」 />
<input type=」file」 name=」pix」 size=」60」 />
<p><input type=」submit」 name=」Upload」
value=」Upload Picture」 />
</form>
</div></body></html>
我,我無法找到該文件相同的結果是正在上傳,這是沒有被上傳到該位置,因爲它應該是。
如何手動將數據類型設置爲MySQL映像? – Steve87 2012-02-04 15:06:55