php mysql解决方法
php mysql
这只是查询出表的数据,显示在li标签中,如何使得点击每条数据就跳转到另外页面查看详细信息?求助!
------解决方案--------------------
你是要把每天记录做个链接吗?那你把a标签放到循环里面来,比如:
<br /><li><br /> <ul class="topnews-ul"><br /> <li><br /> <?php<br /> $link=MySQL_connect('localhost','root','pqm4YAqGt7jx7jhn'); <br /> ?><br /> <?php<br /> mysql_select_db('php_db'); <br /> mysql_query('set names utf8');<br /> $query="select * from table1 where id<=6";<br /><br /> $result= mysql_query($query);//获取数据集<br /> while($rows=mysql_fetch_array($result))<br /> {<br /> ?><br /> <a href="case_cats.php?id=<?php echo $rows['id'];?>" title=""><p name="rowID" style="display:none;"><?php echo $rows['name'] ?></p></a><br /> <?php<br /> }<br /> ?><br /> </li><br /> </ul><br /></li> <br />
------解决方案--------------------
至于怎么显示html内容,自己调整下,以上只是示例

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