如何使用MySQL中的PHP中的按钮删除行

Question

I am having an issue deleting the row from the table. When I click on the 'delete' button it does take me to the next page and it says 'Removed 0 rows from player'. Basically, it is executing correctly, but I am unable to delete the selected row. I have been able to display and add to the table.

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

The problem lies in this line of code,

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

Here the value of $POST['id_player'] is Delete! because you are passing the name of the input type submit in your HTML code and I think that you don't have any id which equals to Delete! in your database.

The Solution

What you need to do is that you need to use a hidden input which will hold the id_Player value like this,

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

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