根据表(单元格)中的位置执行查询

时间:2018-11-16 23:55:59

标签: php mysql

我用以下代码编写了一个表,并且单击表上的任何结果(单元格)时都需要进行第二次查询:

 while($test = mysql_fetch_array($result))
 {
 $id = $test['id']; 
 echo"<td>".$test['name']."</td>";
 echo"<td><a href='#' data-toggle='modal' data-target='#add_new_record_modal' class=text-dark>".$test['open']."</a></td>";
 echo"<td><a href='#' data-toggle='modal' data-target='#add_new_record_modal' class=text-dark>".$test['close']."</a></td>";
 echo "</tr>";
 }
 mysql_close($conn);
 ?>

结果是以下表格:

+--------+----------+----------+
|  Name  |   Open   |   Close  |
+--------+----------+----------+
| user1  |     0    |     1    |
| user2  |     3    |     0    |
| user3  |     2    |     1    |
| user4  |     0    |     3    |
| user5  |     1    |     1    |
+--------+----------+----------+

实际上,我以此打开了一个新窗口: 例如,在第2行( user2 )和第1列( Open )上单击数字 3 ,然后需要获取确切包含以下名称的变量:第2行和第1列:

<div class="modal fade" id="add_new_record_modal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">

<?php
include("conn.php");

//$result = mysql_query("Select * from table where stat = '$column_1' and name = '$row_2'
$result = mysql_query("Select * from table where stat = 'Open' and Name 'user2'");**

while($test = mysql_fetch_array($result))
{
    $id = $test['id']; 
    echo"<div class='form-group'>";
    echo"<label for='s'>".$column_1." ".$row_2."</label>";
    echo"</div>";
    }
    mysql_close($conn);
?>
</div>

如何为正确的变量替换“ 打开”和“ User2 ”? 谢谢!

0 个答案:

没有答案