这是我创建的数据库名及表 image
use my;
create table image(
id varchar(20) not null,
photo longlob,
primary key (id)
);下面是我仿照写的显示图片代码
liulan.php
<body>
<?php
//header("Content-type:image/jpeg");
$conn = @ mysql_connect("localhost","root","") or die("数据库链接错误");
mysql_select_db("my",$conn);
$result=mysql_query("select * from image")or die("can't perform query");while($row=mysql_fetch_object($result)){ ?><img src="<?php "secondtype.php"?>.jpg ">
<?php
}
?>
</body>secondtype.php
<body>
<?php
$result=mysql_query("select * from image")or die("can't perform query");
$row=mysql_fetch_object($result);
header("content-type: image/jpeg");
echo "$row[photo]";
?>
</body>
到底是什么问题,我刚学php 是个菜鸟,将图片以二进制存储到数据库费了不少劲,但是如果不显示在web页面也就没有意义了,望各位高手指教啊!先谢谢了!
use my;
create table image(
id varchar(20) not null,
photo longlob,
primary key (id)
);下面是我仿照写的显示图片代码
liulan.php
<body>
<?php
//header("Content-type:image/jpeg");
$conn = @ mysql_connect("localhost","root","") or die("数据库链接错误");
mysql_select_db("my",$conn);
$result=mysql_query("select * from image")or die("can't perform query");while($row=mysql_fetch_object($result)){ ?><img src="<?php "secondtype.php"?>.jpg ">
<?php
}
?>
</body>secondtype.php
<body>
<?php
$result=mysql_query("select * from image")or die("can't perform query");
$row=mysql_fetch_object($result);
header("content-type: image/jpeg");
echo "$row[photo]";
?>
</body>
到底是什么问题,我刚学php 是个菜鸟,将图片以二进制存储到数据库费了不少劲,但是如果不显示在web页面也就没有意义了,望各位高手指教啊!先谢谢了!
改成:
<img src="<?php echo "secondtype.php" ?>">
<body>
<?php
//header("Content-type:image/jpeg");
$conn = @ mysql_connect("localhost","root","") or die("数据库链接错误");
mysql_select_db("my",$conn);
$result=mysql_query("select * from image")or die("can't perform query");while($row=mysql_fetch_object($result)){ ?>
<img src="secondtype.php">
<?php
}
?>
</body>secondtype.php
<body>
<?php
$result=mysql_query("select * from image")or die("can't perform query");
$row=mysql_fetch_object($result);
header("content-type: image/jpeg");
echo "$row[photo]";
?>
</body>
中不能有其他字符输出
<img src="<?php "secondtype.php"?>.jpg ">
改成:
<img src="<?php echo "secondtype.php" ?>">
二:
secondtype.php中的 <body>和</body>要去掉
三:
echo "$row[photo]";
改成
echo "$row['photo']"; //加上引号
==========================================
另外,你应该先直接运行secondtype.php,看看能不能显示图片,排除一下其他代码问题。
<body>
<?php
//header("Content-type:image/jpeg");
$conn = @ mysql_connect("localhost","root","") or die("数据库链接错误");
mysql_select_db("my",$conn);
$result=mysql_query("select * from image")or die("can't perform query");while($row=mysql_fetch_object($result)){ ?><img src="<?php include "secondtype.php"?>" />
<?php
}
?>
</body>secondtype.php
<body>
<?php
$result=mysql_query("select * from image")or die("can't perform query");
$row=mysql_fetch_object($result);
//header("content-type: image/jpeg");//如果数据直接读出来是bytes可以直接显示不用发送头,但是1个IMG标签只能存放1个图片 ,这里的循环不合理
echo "$row[photo]";
?>
</body>
secondtype.php根据id参数取对应的数据
试了 不行 最后我把id换成了tinyint类型 谢了!