无法显示mysql里的图片

chaiyanlin 2010-04-04 08:27:13

一下是我的代码：
upf1.html
<html>
<head>
<title>上传图片</title>
</head>
<body>
<form method="post" action="upf2.php" enctype="multipart/form-data"><center><br><br><br><br>
<input type="hidden" value="204800" name="MAX_FILE_SIZE"/>
File: <input type="file" name="imgfile" /><br><br>
<input type="submit" value="OK" name="submitbtn" style="width:100px;height:23px"/></center>
</form>
</body>
</html>

upf2.php
<?php
//向数据库中插入图片
$imgfile=$_FILES['imgfile'];
$submitbtn=$_POST['submitbtn'];
if($submitbtn=='OK' && is_array($imgfile))
{
$name=$imgfile['name']; //取得图片名称
$type=$imgfile['type']; //取得图片类型
$size=$imgfile['size']; //取得图片长度
$tmpfile=$imgfile['tmp_name']; //图片上传上来到临时文件的路径
if($tmpfile && is_uploaded_file($tmpfile)) //判断上传文件是否为空，文件是不是上传的文件
{
//读取图片流
$file=fopen($tmpfile,"rb");
$imgdata=bin2hex(fread($file,$size)); //bin2hex()将二进制数据转换成十六进制表示
fclose($file);

$con=mysql_connect("localhost","root","199026"); //连接数据库函数
mysql_select_db("andy"); //选择数据库
//插入出数据库语句，图片数据前要加上0x，用于表示16进制数
if (mysql_query("insert into images(image) values('0x.$imgdata')") )
echo "<center>insert successful!<br><br><a href='showimage.php'>show image</a></center>";
else
echo "<center>insert failed!</center>";
mysql_close();
}
else
echo "<center>Choose image!<br><br><a href='upf1.html'>back</a></center>";
}
else
echo "<center>choose image！<br><br><a href='upf1.html'>back/a></center>";
?>

showimage.php
<?php
mysql_connect("localhost","root","199026");
mysql_select_db("test");
//显示最新插入的那张图片
$result=mysql_query("select image from images ");
$row=mysql_fetch_object($result);
header("Content-Type:image/pjpeg");
echo $row->image;
mysql_close();
?>
显示图片是报错，错误在showimage.php里
提示错误是：
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\test\showimage.php on line 6

Warning: Cannot modify header information - headers already sent by (output started at C:\AppServ\www\test\showimage.php:6) in C:\AppServ\www\test\showimage.php on line 7

菜鸟请求更正！

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