Question

我在从表中删除行时遇到问题。当我单击“删除”按钮时，它确实将我带到下一页，并显示“从播放器中删除了0行”。基本上，它可以正确执行，但是我无法删除所选的行。我已经能够显示并添加到表格中。

Player.php

<table id="table table-bordered">
<tr>
    <th>Id#</th>
    <th>Player(s)</th>
    <th>Position</th>
</tr>

if(!($stmt = $mysqli->prepare("SELECT id_Player, name_Player, position_Player FROM player s ORDER BY position_Player ASC"))){
echo "Prepare failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
}

if(!$stmt->execute())
{
echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
}
if(!$stmt->bind_result($id_Player, $name_Player, $position_Player))
{
 echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;   
}

    while($stmt->fetch()){
    echo "<tr><td>  $id_Player </td> <td>  $name_Player </td><td>  $position_Player  </td>";

    ?>
        <td>
            <form id="delete" method="post" action="deletePlayers.php">

                <input type="submit" name="id_Player" value="Delete!"/>    

            </form>
        </td>
    </tr>

deletePlayers.php

if(!($stmt = $mysqli->prepare("DELETE FROM player WHERE id_Player  = ?"))){
    echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;}
if(!($stmt->bind_param("s",$_POST['id_Player']))){
    echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;}
if(!$stmt->execute()){
    echo "Execute failed: "  . $stmt->errno . " " . $stmt->error;}     
else {
    echo "Removed " . $stmt->affected_rows . " row from player. <br/><br/><strong> Returning to 'Add Players'</strong>";}

Answer 1

问题出在这行代码中，

if(!($stmt->bind_param("s",$_POST['id_Player']))){}

此处$POST['id_player']的值为Delete!，因为您正在HTML代码中传递输入类型Submit的名称，并且我认为您没有任何等于{{1 }}。

解决方案

您需要做的是，您需要使用一个隐藏的输入，该输入将保持Delete!的值，

id_Player

如何使用MySQL中的PHP中的按钮删除行

1 个答案: