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社区首页 >问答首页 >在网页上显示MySql记录

问在网页上显示MySql记录
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Stack Overflow用户
提问于 2015-11-03 18:31:22
回答 1查看 45关注 0票数 0

我试图从我的数据库中返回最后10条记录,并将它们显示在网页上。下面的代码只返回一条记录,有人知道我哪里出错了吗?我稍后会用一张桌子整理它,暂时我只需要最后10张唱片。

该表的名称为“devel”,列名为: id、STATION、REASON_CODE、ERROR、WEEK、注释、停机时间、日期。

代码语言:javascript
复制
$dbc = mysqli_connect(‘xxxx’, ‘xxxx’, 'xxxx', ‘xxxx’)
    or die('Error connecting to MySQL server.');

$query = "SELECT * FROM devel ORDER BY id DESC LIMIT 10";

$result = mysqli_query($dbc, $query)
    or die('Error querying database.');

while($record = mysqli_fetch_array($result)){
echo $record['id']. " " .$record['STATION'] . " " . $record['REASON_CODE'] . " " . $record['ERROR']. " " .$record['WEEK'] . " " .$record['COMMENT'] . " "    .$record['DOWNTIME'] . " " .$record['DATE'];
echo "<br /";
}
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2015-11-03 18:53:57

试试这个

代码语言:javascript
复制
$dbc = mysqli_connect(‘xxxx’, ‘xxxx’, 'xxxx', ‘xxxx’)
    or die('Error connecting to MySQL server.');
$query = "SELECT * FROM devel ORDER BY id DESC LIMIT 10"; 
$result = mysqli_query($dbc, $query)
    or die('Error querying database.');
echo "<table><tr>
                 <th>Id</th>
                 <th>STATION</th>
                 <th>REASON_CODE</th>
                 <th>ERROR</th>
                 <th>WEEK</th>
                 <th>COMMENT</th>
                 <th>DOWNTIME</th>
                 <th>DATE</th>
            </tr>";
while($record = mysqli_fetch_assoc($result)){
 echo "<tr>"; 
 echo "<td>".$record['id']."</td>";
 echo "<td>".$record['STATION']."</td>";
 echo "<td>".$record['REASON_CODE']."</td>";
 echo "<td>".$record['ERROR']."</td>";
 echo "<td>".$record['WEEK']."</td>";
 echo "<td> ".$record['COMMENT']."</td>";
 echo "<td> ".$record['DOWNTIME']."</td>";
 echo "<td> ".$record['DATE']."</td>";
 echo "</tr>";  
}
 echo "</table>";
票数 1
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/33506504

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