超鏈接SQL查詢的第一個結果

Question

我希望添加一個鏈接到這個查詢輸出的結果之一。

<?php $tpw = "SELECT PLAYER, YEAR, WEEK, SUM(POINTS) "; 
    $tpw .= " FROM SCHEDULE"; 
    $tpw .= " WHERE OWNER = 'BRANDON' "; 
    $tpw .= " GROUP BY YEAR, WEEK, PLAYER "; 
    $tpw .= " ORDER BY SUM(POINTS) DESC "; 
    $tpw .= " LIMIT 5"; 
    $tpwresult = mysqli_query($con, $tpw); 
    //test for query error 
    if (!$tpwresult){ 
     die("Database Query Failed!"); 
    }?> 


    <table class="weeks"> 
    <th colspan="4">Most Points in a Week (Player)</th> 
    <tr><td>Year</td><td>Week</td><td>Player</td><td>Points</td></tr> 
     <?php while($tpwrow = mysqli_fetch_assoc($tpwresult)){?> 
    <tr><td><?php echo $tpwrow ["YEAR"];?></td> 
    <td><?php echo $tpwrow ["WEEK"];?></td> 
    <td><?php echo $tpwrow ["PLAYER"];?></td> 
    <td><?php echo $tpwrow ["SUM(POINTS)"];?></td></tr> 
<?php } ?></table> 

是否可以超鏈接此表的第一行/結果？

Answer 1

不是最有效的代碼，但保持它的簡單

<table class="weeks"> 
<th colspan="4">Most Points in a Week (Player)</th> 
<tr><td>Year</td><td>Week</td><td>Player</td><td>Points</td></tr> 
    <?php $first=1 
    while($tpwrow = mysqli_fetch_assoc($tpwresult)){?> 
<tr><td><?php echo $tpwrow ["YEAR"];?></td> 
<td><?php echo $tpwrow ["WEEK"];?></td> 

<td> 
<?php if ($first == 1) }?> 
<a href='http://www.youtube.com/watch?v=qiZXjcb77Qw'> 
<?php }?> 
<?php echo $tpwrow ["PLAYER"];?> 
<?php if ($first == 1) {?> 
</a> 
<?php $first = 0;}?> 
</td> 
<td><?php echo $tpwrow ["SUM(POINTS)"];?></td></tr> 
<?php } ?></table>