0

我有 2 张桌子

1 个表 = 文件

      code                     | title

luLidwhSl8hmN0T6RsLaDmxAB09UZcX |这是Rar标题

4Xwvm1C3yTQJK7CnmxorUDI7sNSvcBK |这是JPG标题

...

2 表 = 命中

      page_name                                | hits

下载.php?code=luLidwhSl8hmN0T6RsLaDmxAB09UZcX |102

下载.php?code=4Xwvm1C3yTQJK7CnmxorUDI7sNSvcBK |87

...

我的查询是：

            include('db.inc.php');
            $query = mysql_query("SELECT t1.code, t1.title, RIGHT(t2.page_name, 31) as t2.page_name, t2.hits FROM files t1 INNER JOIN hits t2 ON t1.code= RIGHT(t2.page_name, 31) as t2.page_name ORDER by t2.hits DESC LIMIT 1, 7");

            while ($result = mysql_fetch_assoc($query)) {
                echo ' <div id="linkstyle"><strong><a href="http://localhost/edu/filesupload/download.php?code='. $result['t1.code'] . ' ">' , $result['t1.title'] , '</a></strong><br></div>';
                }

我收到这个错误

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Edu\filesupload\index.php on line 104

哪里有问题？

4

2 回答 2

1

改变这个：

$query = mysql_query("SELECT t1.code, t1.title, RIGHT(t2.page_name, 31) as t2.page_name, t2.hits FROM files t1 INNER JOIN hits t2 ON t1.code= RIGHT(t2.page_name, 31) as t2.page_name ORDER by t2.hits DESC LIMIT 1, 7");

对此：

$query = mysql_query("SELECT t1.code, t1.title, RIGHT(t2.page_name, 31) as t2.page_name, t2.hits FROM files t1 INNER JOIN hits t2 ON t1.code= RIGHT(t2.page_name, 31) as t2.page_name ORDER by t2.hits DESC LIMIT 1, 7") or die(mysql_error());

那么这应该告诉你错误是什么，因为如果它很好，它就不会返回 false。

于 2013-01-06T16:40:37.037 回答
1

试试这个查询：

SELECT t1.code, t1.title, RIGHT(t2.page_name, 31) as page_name, t2.hits FROM files t1 INNER JOIN hits t2 ON t1.code= RIGHT(t2.page_name, 31) ORDER by t2.hits DESC LIMIT 1, 7
于 2013-01-06T17:07:10.163 回答