
php从mysql读取图片路径并显示
<?Header("Content-type:image/gif");mysql_connect("localhost:/tmp/mysql/mysql.sock","r...
<?
Header( "Content-type: image/gif");
mysql_connect("localhost:/tmp/mysql/mysql.sock","root","mysql174") or die("unable to connect to SQL server");
mysql_select_db("members") or die("unable to select database");
$result=mysql_query("SELECT * FROM Images WHERE PicNum=5") or die("Cant perform Query");
$row=mysql_fetch_object($result);
echo "<IMG SRC=\"$row->Image\">";
?>
怎么显示出来图片呢,这样出来的是<IMG SRC="picture/AJ[9W]X7RO_{O$788E29IHF.jpg"> 展开
Header( "Content-type: image/gif");
mysql_connect("localhost:/tmp/mysql/mysql.sock","root","mysql174") or die("unable to connect to SQL server");
mysql_select_db("members") or die("unable to select database");
$result=mysql_query("SELECT * FROM Images WHERE PicNum=5") or die("Cant perform Query");
$row=mysql_fetch_object($result);
echo "<IMG SRC=\"$row->Image\">";
?>
怎么显示出来图片呢,这样出来的是<IMG SRC="picture/AJ[9W]X7RO_{O$788E29IHF.jpg"> 展开
展开全部
把picture/AJ[9W]X7RO_{O$788E29IHF.jpg恢复成完整文件名,比如基础目录是d:/uploads/,那么还原成d:/uploads/picture/AJ[9W]X7RO_{O$788E29IHF.jpg,然后读取文件内容,把内容echo就行了。
参考手册fopen的说明,里面会有例子。
参考手册fopen的说明,里面会有例子。
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