显示数据库中blob类型的图片，我图片显示不出来
showimage.php

<?php  
include_once dirname(dirname (dirname ( __FILE__ ))). '/yeqztsys1.0/utilityfunction/SqlHelp.class.php';
include_once dirname(dirname (dirname ( __FILE__ ))). '/yeqztsys1.0/Comm.class.php';

$SqlHelper = new SqlHelp ( 'read', 'qztoa' );
$sql = "select * from qzt_image";
$query = $SqlHelper->query($sql );
$query_num = $SqlHelper->num_rows ($query );

$rows = array ();  
$retarr = array (); 
$query =$SqlHelper->query ( $sql );
if($query)
{
$arr_teainfo2 = array ();
while ( $arr_teainfo = $SqlHelper->fetch_assoc($query) ) {
$arr_teainfo2 [] = $arr_teainfo;
$Id=$arr_teainfo["Id"];
echo $arr_teainfo["Id"];
echo "<IMG SRC=\"Second.php3?Id=$Id\">"; 

}
}



?>  

Second.php3

<?php  
include_once dirname(dirname (dirname ( __FILE__ ))). '/yeqztsys1.0/utilityfunction/SqlHelp.class.php';
include_once dirname(dirname (dirname ( __FILE__ ))). '/yeqztsys1.0/Comm.class.php';

$SqlHelper = new SqlHelp ( 'read', 'qztoa' );
$sql = "select * from qzt_image where Id=10";
$query = $SqlHelper->query($sql );
$query_num = $SqlHelper->num_rows ($query );

$rows = array ();  
$retarr = array (); 
$query =$SqlHelper->query ( $sql );
if($query)
{
$arr_teainfo2 = array ();
while ( $arr_teainfo = $SqlHelper->fetch_assoc($query) ) {
$arr_teainfo2 [] = $arr_teainfo;
$Id=$arr_teainfo["Id"];
$imagetype=$arr_teainfo["imagetype"];
//echo $arr_teainfo["images"];
//Header( "Content-type: $imagetype"); 
header('Content-type: $imagetype');
echo $arr_teainfo["images"];

}
}

?>   

请问我哪里有问题？PHP新人，不知道哪里错了 PHP
[解决办法]

引用:

Quote: 引用:

图片显示不出来，是程序报错？还是路径不对。
先打印下，看有数据没，如果有，再看下，显示的路径是否正确。

先运行php3把image打印出来了，正常的，没有报错

<?php  
include_once dirname(dirname (dirname ( __FILE__ ))). '/yeqztsys1.0/utilityfunction/SqlHelp.class.php';
include_once dirname(dirname (dirname ( __FILE__ ))). '/yeqztsys1.0/Comm.class.php';

$SqlHelper = new SqlHelp ( 'read', 'qztoa' );
$sql = "select * from qzt_image where Id=10";
$query = $SqlHelper->query($sql ); 
 
$query_num = $SqlHelper->num_rows ($query );

$rows = array ();  
$retarr = array (); 
$query =$SqlHelper->query ( $sql );
if($query)
{
$arr_teainfo2 = array ();
while ( $arr_teainfo = $SqlHelper->fetch_assoc($query) ) {
$arr_teainfo2 [] = $arr_teainfo;
$Id=$arr_teainfo["Id"];
$imagetype=$arr_teainfo["imagetype"];
$image= $arr_teainfo["images"];
}
//header('Content-type: image/jpeg ');
echo $image; 


}

程序没报错，把输出的数据复制到地址栏，有图片显示没？？
[解决办法]
你的图片格式有问题！

这样吧，你把 $image 写到文件里，比如 file_put_contents('test.jpg', $image);
然后把他发上来
[解决办法]

<?php
$query = "select * from qzt_image where id=10";
$result = mysql_query($query);
$row = @mysql_fetch_array($result);
Header( "Content-type: image/pjpeg");
echo "$row[uploadfile]";
?>

我的原始版代码。
是pjpeg哦。