0

我面临的当前问题是，当我输入 url 时，我的 MySql 表没有显示：www.mydoman.com/search-results.php?town=Blackpool-venuetbl 中的结果没有显示并且打印错误正在说明Resource id #4

<?php
session_start();

include('dbconnect.php');

if(isset($_GET['town']))   {
$town = $_GET['town'];


            mysql_connect("$host", "$username", "$passwd")or die("cannot connect"); 
            mysql_select_db("$db")or die("cannot select DB");
    $res=mysql_query("SELECT * from venuetbl where town = '$town'") or die("error");        


}
?>
<?php while($row = mysql_fetch_array($res))    { ?>
        <div class="search-results">
            <div class="result">
                <img src="images/<? echo $rows['imageurl']; ?>.jpg" alt="" width="150" height="100" />
                <h2><? echo $rows['name']; ?></h2>
                <p><? echo $rows['description']; ?></p>

                <a class="button" href="http://www.mydomain.co.uk/result.php?id=<? echo $rows['venueid']; ?>">View More</a>
                <a class="button green" href="#">Add To My Venues</a>
            </div><!--END Result-->
        </div><!--END Search Results-->

                    <?php
        }
        ?>
4

1 回答 1

0

您正在尝试回显 $rows['SOME_KEY'] 而不是 $row['SOME_KEY']

于 2013-01-20T19:00:46.210 回答