我想做一个简单的应用
自动显示出10个标题,点击标题显示出新闻,新闻中有图片
我在后台上传图片,我如何知道这个图片是对应的这个标题
的图片我现在是想知道这个方法,我图片上传到一个uploads的文件夹。如何保存,或者在数据库中怎么体现,求高手指点下
自动显示出10个标题,点击标题显示出新闻,新闻中有图片
我在后台上传图片,我如何知道这个图片是对应的这个标题
的图片我现在是想知道这个方法,我图片上传到一个uploads的文件夹。如何保存,或者在数据库中怎么体现,求高手指点下
新建个表
id auto_increment 主键
news_id
pic_url
将 news_id 与 news 表关联即可。
<head>
<title>Uploading...</title>
</head>
<body>
<h1>Uploading file...</h1>
<?php//Check to see if an error code was generated on the upload attempt
//判断错误
if ($_FILES['userfile']['error'] > 0)
{
echo 'Problem: ';
switch ($_FILES['userfile']['error'])
{
case 1: echo 'File exceeded upload_max_filesize';
break;
case 2: echo 'File exceeded max_file_size';
break;
case 3: echo 'File only partially uploaded';
break;
case 4: echo 'No file uploaded';
break;
case 6: echo 'Cannot upload file: No temp directory specified.';
break;
case 7: echo 'Upload failed: Cannot write to disk.';
break;
}
exit;
} //查看上传文件类型是否为图片
if ($_FILES['userfile']['type'] != 'image/gif')
{
echo 'Problem: file is not image';
exit;
} // 文件上传路径
$upfile = '/uploads/'.$_FILES['userfile']['name'];
//是否是由http上传
if (is_uploaded_file($_FILES['userfile']['tmp_name']))
{
//如果是那么则可以移动位置到$upfile
if (!move_uploaded_file($_FILES['userfile']['tmp_name'], $upfile))
{
echo 'Problem: Could not move file to destination directory';
exit;
}
}
else
{
echo 'Problem: Possible file upload attack. Filename: ';
echo $_FILES['userfile']['name'];
exit;
}
echo 'File uploaded successfully<br><br>'; // reformat the file contents
$fp = fopen($upfile, 'r');
$contents = fread ($fp, filesize ($upfile));
fclose ($fp);
$contents = strip_tags($contents);
$fp = fopen($upfile, 'w');
fwrite($fp, $contents);
fclose($fp); // show what was uploaded
echo 'Preview of uploaded file contents:<br><hr>';
echo $contents;
echo '<br><hr>';?>
</body>
</html>
这个是我的upload代码,上传没有成功,而且图片的名字我怎么保存到数据库呢
你都能把名字输出来了,不能存起来吗
通常放在最后,等文件上传成功了
把这个$_FILES['userfile']['name']值
插入数据库就行了