PHP超鏈接/彈出表

我創建從MySQL查詢PHP表如下：PHP超鏈接/彈出表

$resultkulim = mysql_query("SELECT customer_name, zone_name, segment_code, COUNT(segment_code) FROM complete_wk where zone_name = 'ZONE KULIM' and (segment_code = 's30' or segment_code='s40' or segment_code='s50'')"); 
    $totalrrkulim = mysql_query("SELECT zone_name, repeat_rc, COUNT(repeat_rc) FROM complete_wk where zone_name = 'ZONE KULIM' and repeat_rc>1 and (segment_code = 's30' or segment_code='s40' or segment_code='s50')"); 

       echo "<table class='table1'>"; 
       echo "<thead>"; 
        echo "<tr>"; 
         echo "<th>Zone</th>"; 
         echo "<th>Total TR</th>"; 
         echo "<th>Total RR (RR>1)</th>"; 
         echo "<th>%RR</th>"; 
        echo "</tr>"; 
       echo "</thead>"; 
       //kulim row 
       while($rowkulim = mysql_fetch_array($resultkulim)) 
        { 
         echo "<tbody>"; 
         echo "<tr>"; 
          echo "<td>Zone Kulim</td>"; 
          echo "<td >" . $rowkulim['COUNT(segment_code)'] . "</td>"; 
         while($rrkulim = mysql_fetch_array($totalrrkulim)) 
          { 
          $myresultkulim = $rrkulim['COUNT(repeat_rc)']/$rowkulim['COUNT(segment_code)'] * 100; 
            echo "<td>" . $rrkulim['COUNT(repeat_rc)'] . "</td>"; 
            echo "<td>" . number_format($myresultkulim) . "%" . "</td>"; 
           echo "</tr>"; 
          echo "</tbody>"; 
          } 
        } 
       echo "</table>";

而且表的例子結果如下：

Zone   Total TR Total RR %RR 
Zone Kulim  182   11  6%

我想要做的是，我希望能夠點擊Total TR的結果，即「182」，它會將我帶到包含該結果細節表的其他頁面。

我已經做了其他表，它位於「本地主機/ TMJ/index.php文件/ TR/Kulim的」

我希望有人能賜教如何要麼使一個彈出窗口或超鏈接到上面表格代碼中的其他頁面。

來源

2012-04-16 ketto

我不知道這是否是你想要的，嘗試

echo '<td ><a href="index.php/tr/kulim">' . $rowkulim['COUNT(segment_code)'] . "</a></td>";

或者這是一個動態鏈接？

來源

2012-04-16 02:14:23 Phphelp

PHP超鏈接/彈出表

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