//myphoto.php<?php $mylink=mysql_connect("localhost","root","only");mysql_select_db("test",$mylink);$sql1="setnamesgb2312";$result1=mysql_query($sql1);$result=mysql_query(...
//myphoto.php
$mylink = mysql_connect("localhost","root","only");
mysql_select_db("test",$mylink);
$sql1 = "set names gb2312";
$result1 = mysql_query($sql1);
$result=mysql_query("SELECT * FROM myimage;");
?>
我的图片
While($row=mysql_fetch_object($result))
{ ?>
echo "$row->name
";
echo "name\" >";
?>
//myphoto2.php
$mylink = mysql_connect("localhost","root","only");
mysql_select_db("test",$mylink);
$sql1 = "set names gb2312";
$result1 = mysql_query($sql1);
?>
$result=mysql_query("SELECT * FROM myimage where name=$name") ;
$row=mysql_fetch_object($result);
Header( "Content-type:image/jpg");
echo $row->image;
?>
我已经用phpmyadmin建好了数据库,name varchar(20),image BLOB;并且存储了图片,用mysql-front也能看到图片了。上面的程序是网上最多的程序,我不知道哪里错了,老是提示supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache2\htdocs\myphoto2.php on line 10
也就是这句:$row=mysql_fetch_object($result);。不知道怎么改?
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