Question

我想弄清楚如何在保持行到位的同时链接表格行中的项目，我有以下代码：

echo "<br><br><table border=0 cellpadding=3>";
echo "<td><b>Player Name</b></td>";
echo "<td><b>Position</b></td>";
echo "<td><b>Height</b></td>";
echo "<td><b>Weight</b></td>";
echo "<td><b>Birthdate</b></td>"; 
echo "<td><b>NHL Rights</b></td>";
echo "<td><b>CNGHL Team</b></td>";
echo "<td><b>Current Team</b></td>";
echo "<td><b>Current League</b></td>";

while($row = mysql_fetch_array($oteaminfo)) 
{
echo "<tr>";
echo "<td>".$row['FullName']."</td> ";
echo "<td>".$row['Position']."</td> ";
echo "<td>".$row['Height']."</td> ";
echo "<td>".$row['Weight']."</td> ";
echo "<td>".$row['DOB']."</td> ";
echo "<td>".$row['Team']."</td> ";
echo "<td>".$row['CNGHLRights']."</td> ";
echo "<td>".$row['InternationalTeam']."</td> ";
echo "<td>".$row['InternationLeague']."</td> ";
echo "</tr>";
}
echo "</table>";

我尝试过使用

     echo "<a href=\"cnghlplayers.php?  PlayerID=".$row['PlayerID']."\">".$row['FullName']."<br>";

代替

      echo "<td>".$row['FullName']."</td> ";

在表格中但它将链接放在我当前表格上方的新行中。任何帮助将不胜感激，我尝试搜索这个主题，但我找不到任何有用的信息。

谢谢！

Answer 1

只需将<td>元素包裹起来：

echo '<td><a href="cnghlplayers.php?PlayerID='.$row['PlayerID'].'">'.$row['FullName'].'</a></td>';

Answer 2

用TD包裹您的链接：

echo "<td><a href=\"cnghlplayers.php?  PlayerID=".$row['PlayerID']."\">".$row['FullName']."</a></td>";

Answer 3

测试一下，表格标记存在一些问题。这应该更容易阅读。

?> <!-- this line ends your php code so that you can format the HTML sanely -->
<br />
<br />
<table border=0 cellpadding=3>
        <tr>
            <td><b>Player Name</b></td>
            <td><b>Position</b></td>
            <td><b>Height</b></td>
            <td><b>Weight</b></td>
            <td><b>Birthdate</b></td> 
            <td><b>NHL Rights</b></td>
            <td><b>CNGHL Team</b></td>
            <td><b>Current Team</b></td>
            <td><b>Current League</b></td>
        </tr>

    <?php while($row = mysql_fetch_array($oteaminfo)) { ?>
        <tr>
            <td><?php echo "<a href='cnghlplayers.php?PlayerID=".$row['PlayerID'].">".$row['FullName']."</a>"; ?></td> 
            <td><?php echo $row['Position']; ?></td> 
            <td><?php echo $row['Height']; ?></td> 
            <td><?php echo $row['Weight']; ?></td> 
            <td><?php echo $row['DOB']; ?></td> 
            <td><?php echo $row['Team']; ?></td> 
            <td><?php echo $row['CNGHLRights']; ?></td> 
            <td><?php echo $row['InternationalTeam']; ?></td> 
            <td><?php echo $row['InternationLeague']; ?></td> 
        </tr>
    <?php } ?>
    </table>
<?php //continues your php code below

使用php链接表中的项目

3 个答案: