执行MySQL的查询语句，没有返回数据

Question

图一：数据库的表

 

图二：浏览器的echo出来的内容

 

如果换成

$sql = 'SELECT * FROM DcData WHERE TO_DAYS(NOW()) - TO_DAYS(uptime) <= 1';

又可以正确查询到数据。

 

 

//$sql = 'SELECT * FROM DcData WHERE TO_DAYS(NOW()) - TO_DAYS(uptime) <= 1';

$sql = 'SELECT * FROM DcData WHERE user_name = qwe';
echo "你输入的需求为：".$sql;
//执行查询语句，返回result为数据内容
if($result = mysqli_query($conn,$sql))
{

   //得到查询的记录的个数，
   $rowcount = mysqli_num_rows($result);
   //echo "return {$rowcount} records <br>";

   for($i = 0;$i <$rowcount;$i ++)
   {

       //按顺序一次读取一条记录，保存到sqldata中

       $sqldata = mysqli_fetch_assoc($result);


       echo

       "<center>

       <table>

       <tr>

           <td>".$sqldata['thname']."</td>

           <td>".$sqldata['conf']."</td>

           <td>".$sqldata['bndbox']. "</td>

           <td>".$sqldata['uptime']. "</td>


       </tr>

       </table>

       </center>";

   }

   mysqli_free_result($result);

}

Answer 1

$sql = "SELECT * FROM DcData WHERE user_name = 'qwe'";