2011-06-23 45 views
-1

好吧,上传和删除图像php sql

我建立了一个插入图像页面(uploader.php),但我有2个问题。

的代码是:

<?php 

$target = "images/test/"; 
$target = $target . basename($_FILES['photo']['name']); 

$title=$_POST['title']; 
$desc=$_POST['desc']; 
$pic=($_FILES['photo']['name']); 

mysql_connect("dbhost", "dbuser", "dbpass") or die(mysql_error()) ; 
mysql_select_db("dbname") or die(mysql_error()) ; 
mysql_query("INSERT INTO `test` VALUES ('$title', '$desc', '$pic')") ; 

if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
{ 

echo "The file ". basename($_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; 
} 
else { 
echo "Sorry, there was a problem uploading your file."; 
} 
?> 

<form enctype="multipart/form-data" action="uploader.php" method="POST"> 
Title: <input type="text" name="title"><br> 
Description: <input type="text" name = "desc"><br> 
Photo: <input type="file" name="photo"><br> 
<input type="submit" value="Add"> 
</form> 

所以第一个问题是,信息不被输入到数据库 - 表中有4个领域 - ID(INT),标题(VARCHAR),递减(VARCHAR )和照片(varchar)。是否因为id字段没有被指定?这只是表格的自动递增主键。

第二个问题是,正在加载的图像中包含空格 - 例如,上传“test image.jpg”时 - 我想合并一个str_replace来创建“testimage.jpg”。你知道我将它插入代码的位置吗?再次

感谢您的帮助,

JD

回答