$sql="select * from T_goods";
$result=mysql_query($sql,$db) or die(mysql_error());
while($row=mysql_fetch_array($result)){
  echo $row["goodsname"].$row["goodspic"].substr($row["goodscontent"],0,50);
}

解决方案 »

  1.   

    <table>
    <?php
    $sql="select * from T_goods";
    $result=mysql_query($sql,$db) or die(mysql_error());
    while($row=mysql_fetch_array($result)){
    ?>
    <tr><td><?php echo $row["goodsname"];?></td><td><?php echo $row["goodspic"];?></td><td><?php echo substr($row["goodscontent"],0,50);?></td></tr>
    <?php
    }
    ?>
    </table>
      

  2.   

    <table>
    <?php
    $sql="select * from T_goods";
    $result=mysql_query($sql,$db) or die(mysql_error());
    while($row=mysql_fetch_array($result)){
    ?>
    <tr>
    <td><?php echo $row["goodsname"];?></td>
    <td><a href="XX.php"><img src="<?php echo $row["goodspic"];?>"/></a></td>
    <td><?php echo substr($row["goodscontent"],0,50);?></td></tr>
    <?php
    }
    ?>
    </table>
      

  3.   

    $sql="select * from T_goods";
    $result=mysql_query($sql,$db) or die(mysql_error());
    while($row=mysql_fetch_array($result)){
      echo $row["goodsname"].$row["goodspic"].substr($row["goodscontent"],0,50);
    }