PHP超链接/弹出表
我正在从MySQL查询创建一个PHP表,如下所示:PHP超链接/弹出表,php,mysql,hyperlink,html-table,popup,Php,Mysql,Hyperlink,Html Table,Popup,我正在从MySQL查询创建一个PHP表,如下所示: $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'')");
$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>";
我想做的是,我希望能够点击Total TR的结果,即182,它将带我进入另一个页面,该页面包含182结果的详细信息表
我已经完成了另一个表,它位于localhost/tmj/index.php/tr/kulim
我希望有人能告诉我如何从上面的表代码中创建一个弹出窗口或链接到其他页面 我不确定这是否是你想要的, 试一试 或者这是一个动态链接
Zone Total TR Total RR %RR
Zone Kulim 182 11 6%
echo '<td ><a href="index.php/tr/kulim">' . $rowkulim['COUNT(segment_code)'] . "</a></td>";