php获取<a>标签内容并匹配MySQL数据库显示详细信息?
$dbc=mysqli_connect('localhost','root','root','jyx')or die('数据库连接失败');
$query="SELECT * FROM `xinwen` WHERE `leibie` = '站内新闻'";
$result=mysqli_query($dbc,$query) or die('数据库查询失败');

echo "<table>
    <tr><td align='center'>序号</td><td align='center'>标题</td><td align='center'>时间</td></tr>";
    while($row = mysqli_fetch_array($result)){
        echo "<tr><td align='center'>".$row[0]."</td><td><a href='xxnr.php' id='title'>".mb_strimwidth($row[1],0,55,'...','utf-8')."</a></td><td align='center'>".$row[4]."</td></tr>";
    }
echo "</table>";
mysqli_free_result($result);
mysqli_close($dbc);

这是获取的数据库内容,我给标题加了a标签,但是我怎么获取echo里面的a标签标题内容匹配数据库并在另一个页面显示标题的文章内容?
数据库字段是 id,title, leibie, content, addtime

1个回答

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