PHP Mysql 获取字符串

Question

这个问题在这里已经有了答案:





mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource

(31 个回答)


8年前关闭。




我正在尝试制作新闻页面，但出现此错误

Warning: mysql_result() expects parameter 1 to be resource, boolean given in D:\News\Index.php on

第 37 行

这是代码

    $category = mysql_query("SELECT icon FROM website.catnews WHERE ncatid=\'".         mysql_real_escape_string($news['category']). '\'');
    echo '<td>
    <td><a href = "news.php?cat='. $news['category'].'"><img src="./img/'.mysql_result($category, 'icon').'" alt="" /></a></td>';

这是 mssql 中的全部源代码

    <br />
    <table style="width: 100%;">
        <tr>
        <td></td>
            <td id="key">Title</td>
            <td id="key" style="text-align: center;">Author</td>
            <td id="key" style="text-align: center;">Comments</td>
            <td id="key" style="text-align: center;">Views</td>
            <td id="key" style="text-align: center;">Date</td>
        </tr>
    <?php
        odbc_exec($mssql, 'USE [WEBSITE_DBF]');
        $count = odbc_exec($mssql, 'SELECT COUNT(*) as count FROM [web_news]');
        if(odbc_result($count, 'count') > 0) {
            $query = odbc_exec($mssql, 'SELECT TOP 10 * FROM [web_news] ORDER BY datetime DESC');
            while($news = odbc_fetch_array($query)) {
                $title = $news['title'];
                if(strlen($title) > 35) {
                    $title = substr($title, 0, 35).'...';
        }
                $comments = odbc_exec($mssql, 'SELECT COUNT(*) as count FROM [web_newscomments] WHERE nid=\''.mssql_escape_string($news['nid']).'\'');
                $category = odbc_exec($mssql, 'SELECT icon FROM [web_newscategories] WHERE ncatid=\''.mssql_escape_string($news['category']).'\'');
                echo '<tr>
                    <td><a href="news.php?cat='.$news['category'].'"><img src="./img/'.odbc_result($category, 'icon').'" alt="" /></a></td>
                    <td><a href="news.php?nid='.$news['nid'].'">'.$title.'</a></td>
                    <td style="text-align: center;">'.$news['author'].'</td>
                    <td style="text-align: center;">'.odbc_result($comments, 'count').'</td>
                    <td style="text-align: center;">'.$news['views'].'</td>
                    <td style="text-align: center;">'.date('Y-m-d', strtotime($news['datetime'])).'</td>';
            }
        } else {
            echo 'No news available!';
        }
    ?>
    </table>

Answer 1

解决了!

将图标表部分移到新闻表中
并使用:

<td><a href="news.php?cat='.$news['category'].'"><img src="./img/'. $news['icon'] .'" alt="" /></a></td>
instead of
<td><a href="news.php?cat='.$news['category'].'"><img src="./img/'.odbc_result($category, 'icon').'" alt="" /></a></td>