无法在PHP中显示图像

这是我的图像显示代码-

$username = "xxxxxxxx";
$password = "xxxxxxxx";
$host = "000.001.000.000";
$database = "xxxxxxxx";

@mysql_connect($host, $username, $password) or die("Can not connect to database:         ".mysql_error());
@mysql_select_db($database) or die("Can not select the database: ".mysql_error());
$query = mysql_query("SELECT mimetype, Image FROM table ORDER BY id DESC LIMIT 0,1");
$row = mysql_fetch_array($query);
$content = $row['Image'];
header('Content-type: image/jpg');
echo $content;

这就是我得到的错误

“形象”http://www....“无法显示，因为它包含错误

---

怎么了？mysql中字段的数据类型是mediumblob

我建议在输出之前添加一个内容长度头：

header("Content-length: " . strlen($content))

---

我建议根据图像mime类型更改标题，使其成为动态的：

header('Content-type: '.$row['mimetype']);

---

好的，第一个测试，看看发生了什么：

$username = "xxxxxxxx";
$password = "xxxxxxxx";
$host = "000.001.000.000";
$database = "xxxxxxxx";

if( !mysql_connect($host, $username, $password) )
  die( 'Unable to connect to Server: '.mysql_error() );
if( !mysql_select_db($database) )
  die( 'Can not select the Database: '.mysql_error() );

$query = mysql_query( "SELECT mimetype, Image FROM table ORDER BY id DESC LIMIT 0,1" );

if( !$query )
  die( 'Query Failed: '.mysql_error() );
if( mysql_num_rows( $query )==0 )
  die( 'Query Returned No Records' );

$row = mysql_fetch_array($query);

echo '<pre>';
var_dump( $row );
echo '</pre>';

---

（假设

mimetype

字段类似于“image/jpg”）

图像文件有多大？在添加到数据库之前是否对其进行编码（例如mysql_escape_chars？）？是否确定显示的图像是jpeg？您正在获取mimetype和image，但没有使用mimetype-这告诉我您可能会存储多个文件类型。（gif/jpg/png等）注释掉

头（）

位，然后再次检查页面是否有错误。然后去掉那些错误抑制器。数据库、表和列都是utf8？？

$username = "xxxxxxxx";
$password = "xxxxxxxx";
$host = "000.001.000.000";
$database = "xxxxxxxx";

if( !mysql_connect($host, $username, $password) )
  die( 'Unable to connect to Server: '.mysql_error() );
if( !mysql_select_db($database) )
  die( 'Can not select the Database: '.mysql_error() );

$query = mysql_query( "SELECT mimetype, Image FROM table ORDER BY id DESC LIMIT 0,1" );

if( !$query )
  die( 'Query Failed: '.mysql_error() );
if( mysql_num_rows( $query )==0 )
  die( 'Query Returned No Records' );

$row = mysql_fetch_array($query);

header( 'Content-type: '.$row['mimetype'] );
echo $row['Image'];