在PHP中编辑表行，但它正在获取表的最后一行的详细信息

Question

我想编辑数据库中行的内容。为此，我使用编辑和删除按钮从数据库创建了一个表。删除按钮与ajax正常工作。单击编辑按钮并移至下一页时，它将显示表中最后一行的详细信息。我想获取当前行的详细信息，我点击了编辑按钮。

我的PHP代码是

<form method="post" action="edit_team.php"> 
<?php
error_reporting(0);
//session_start();
include 'database-config.php';

$selectquery = "SELECT * FROM `register_team`" ;

$group_result = mysql_query($selectquery, $conn);

$a=0;

while ($container_id_record = mysql_fetch_assoc($group_result)) {
$a++;

echo "<tr class='gradeA odd' role='row'><td class='sorting_1'>" . $a . "</td><td>" . $container_id_record['team_name'] . "</td><td>" .$container_id_record['team_coach']. "</td><td>" . $container_id_record['team_manager'] .  "</td><td>" . $container_id_record['team_contact'] .  "</td><td>". $container_id_record['team_email']. "</td><td><button type='submit'><i class='fa fa-edit edit' title='". $container_id_record['team_name_coach']. "'></i></button></td><td><i class='fa fa-trash delete' title='". $container_id_record['team_name_coach']. "'></i></td></tr>";
  $team = $container_id_record['team_name_coach'];
}

?>
<input type="hidden" name="team_name_coach" id="team_name_coach" value="<?php echo $team;?>"/>

我的edit_team.php是

$team_name_coach = $_POST['team_name_coach'];

从ajax做的时候，它没有显示下一页的当前值。

Answer 1

尝试这个..

{{1}}

在edit-team.php页面中使用$ _GET而不是$ _POST .. $ team_name_coach = $ _GET [＆＃39; team_name_coach＆＃39;];