php实现上传图片保存到数据库的方法

### 这篇文章主要介绍了php实现上传图片保存到数据库的方法,可通过将图片保存在数据库实现多台服务器共享文件的功能,非常具有实用价值,需要的朋友可以参考下

php实现上传图片保存到数据库的方法。分享给大家供大家参考。具体分析如下：

php 上传图片，一般都使用move_uploaded_file方法保存在服务器上。但如果一个网站有多台服务器，就需要把图片发布到所有的服务器上才能正常使用（使用图片服务器的除外）
如果把图片数据保存到数据库中，多台服务器间可以实现文件共享，节省空间。

首先图片文件是二进制数据，所以需要把二进制数据保存在mysql数据库。
mysql数据库提供了BLOB类型用于存储大量数据，BLOB是一个二进制对象，能容纳不同大小的数据。

BLOB类型有以下四种，除存储的最大信息量不同外，其他都是一样的。可根据需要使用不同的类型。

| 类型  | 限制  |
| :------------: | :------------: |
| TinyBlob  | 最大 255B  |
| Blob   |  最大 65K |
| MediumBlob  | 最大 16M  |
|LongBlob|最大 4G|

数据表photo，用于保存图片数据，结构如下：
```sql
CREATE TABLE `photo` ( 
  `id` int(10) unsigned NOT NULL auto_increment, 
  `type` varchar(100) NOT NULL, 
  `binarydata` mediumblob NOT NULL, 
  PRIMARY KEY  (`id`) 
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
```
upload_image_todb.php：
```php
<?php 
// 连接数据库 
$conn=@mysql_connect("localhost","root","")  or die(mysql_error()); 
@mysql_select_db('demo',$conn) or die(mysql_error()); 
// 判断action 
$action = isset($_REQUEST['action'])? $_REQUEST['action'] : '';

// 上传图片 
if($action=='add'){ 
    $image = mysql_escape_string(file_get_contents($_FILES['photo']['tmp_name'])); 
    $type = $_FILES['photo']['type']; 
    $sqlstr = "insert into photo(type,binarydata) values('".$type."','".$image."')"; 
    @mysql_query($sqlstr) or die(mysql_error()); 
    header('location:upload_image_todb.php'); 
    exit(); 
// 显示图片 
}elseif($action=='show'){ 
    $id = isset($_GET['id'])? intval($_GET['id']) : 0; 
    $sqlstr = "select * from photo where id=$id"; 
    $query = mysql_query($sqlstr) or die(mysql_error()); 
    $thread = mysql_fetch_assoc($query); 
    if($thread){ 
        header('content-type:'.$thread['type']); 
        echo $thread['binarydata']; 
        exit(); 
    } 
}else{ 
// 显示图片列表及上传表单 
?> 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> 
<html> 
 <head> 
  <meta http-equiv="content-type" content="text/html; charset=utf-8"> 
  <title> upload image to db demo </title> 
 </head> 
 
 <body> 
  <form name="form1" method="post" action="upload_image_todb.php" enctype="multipart/form-data"> 
  <p>图片：<input type="file" name="photo"></p> 
  <p><input type="hidden" name="action" value="add"><input type="submit" name="b1" value="提交"></p> 
  </form> 
 
<?php 
    $sqlstr = "select * from photo order by id desc"; 
    $query = mysql_query($sqlstr) or die(mysql_error()); 
    $result = array(); 
    while($thread=mysql_fetch_assoc($query)){ 
        $result[] = $thread; 
    } 
    foreach($result as $val){ 
        echo '<p><img src="upload_image_todb.php?action=show&id='.$val['id'].'&t='.time().'" width="150"></p>'; 
    } 
?> 
</body> 
</html> 
<?php 
} 
?>
```
![](http://blog.berfen.com/usr/uploads/2020/01/2521406621.png)
![](http://blog.berfen.com/usr/uploads/2020/01/3068961089.png)
希望本文所述对大家的php程序设计有所帮助。

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