Question

最近我重建了我的网站，我使用这个简单的代码从mysql db中获取和获取数据为rss：

<?php



$connection = mysql_connect("localhost", "site_user", "1212121") or die (mysql_error()); // 'database_name' 
$db         = mysql_select_db("site_db", $connection) or die (mysql_error());

$rss_query = @mysql_query("SELECT * FROM jokes where valid = '1' order by id asc LIMIT 0,20");

echo $rss = "<?xml version='1.0' encoding='utf-8' ?>

<rss version='2.0'>

<channel>

<title>site</title>

<link>www.site.com</link>
<description>feed</description>
 ";

while ($q_rss = mysql_fetch_array($rss_query)) {

$id[]    =  $q_rss['id'];
$title[] =  $q_rss['joketitle'];
$des[]   =  $q_rss['preview'];
}    

$count = count($id); 
for ( $i = 0; $i <= $count-1; $i ++) {   
echo $r = " 
<item>
 <title>".$title[$i]."</title>

<link>www.site.com/view.php?byt=".$id[$i]."</link>

<description>...".substr($des[$i],0,70)."</description>

</item>
";
}
echo "
</channel>

</rss>";




?>

关于链接：www.site.com/view.php？byt =＆＃34; .$id [.i]。＆＃34;它改为www.site.com/byt/ [id]，其中id = numbers

出现的错误是：

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/site/public_html/rss.php on line 25

Notice: Undefined variable: id in /home/site/public_html/rss.php on line 32

我的php和mysql版本是最新的..

Answer 1

是否有特殊原因要将项添加到数组中，然后稍后遍历这些数组，而不是直接在记录集迭代中添加rss内容？

此外，不建议使用mysql_*函数套件，因此建议您迁移到mysqli - 尽管此处没有实际的SQL注入风险，因为您没有将用户输入作为参数sql。

也就是说，应该进行某种逻辑测试，看看查询是否产生了记录集。

<?php
    $connection = mysql_connect("localhost", "site_user", "1212121") or die (mysql_error());
    $db         = mysql_select_db("site_db", $connection) or die (mysql_error());
    $rss_query  = mysql_query("SELECT * FROM `jokes` where `valid` = '1' order by `id` asc LIMIT 0,20");

    echo $rss = "<?xml version='1.0' encoding='utf-8' ?>
    <rss version='2.0'>
        <channel>
            <title>site</title>
            <link>www.site.com</link>
            <description>feed</description>";

    if( $rss_query && mysql_num_rows( $rss_query ) > 0 ) {
        while( $q_rss = mysql_fetch_array($rss_query) ) {
            echo "
            <item>
                <title>{$q_rss['id']}</title>
                <link>{$q_rss['joketitle']}</link>
                <description>...{$q_rss['preview']}</description>
            </item>";
        }
    } else {
        echo "
            <item>
                <title>Error</title>
                <link></link>
                <description>The sql query has failed for some reason</description>
            </item>";   
    }

    echo "
        </channel>
    </rss>";

?>

mysql_fetch_array加上未定义变量的RSS提要问题

1 个答案: