Question

我有一个页面，我使用PHP和HTML从Mysql渲染值表。

我想删除单击删除按钮时选中复选框的行。不幸的是，单击按钮时没有任何反应，我完全不知道在哪里看。

代码如下（数据库连接已被省略，但它有效）。

<body>

<?php
$delete = $_POST['checkbox'];
$query = "SELECT * FROM Cards";
$result = mysql_query($query);

echo "<table>";
echo "<tr><td>Name</td><td>Quantity</td><td>Color</td><td>Type</td></tr>";
while ($row = mysql_fetch_array($result)){
    echo "<tr><td><input type='checkbox' name='checkbox[]" . $row['id'] . "]'></td>";
    echo "<td>" . $row['Name'] . "</td>";
    echo "<td>" . $row['Quantity'] . "</td>";
    echo "<td>" . $row['Color'] . "</td>";
    echo "<td>" . $row['CardType'] . "</td></tr>";
}
mysql_free_result($result);
echo "</table>";
?>

</body>
<tr>
<td colspan="5" align="center"><input name="delete" type="SUBMIT" id="delete" value="delete" action="POST"></td>
</tr>

<?php
if (isset($_POST['delete'])) {
    $checkbox = $_POST['checkbox'];
    $count = count($checkbox);

    for($i = 0; $i < $count; $i++) {
        $id = (int) $checkbox[$i]; // Parse your value to integer

        if ($id > 0) { // and check if it's bigger then 0
            mysql_query("DELETE FROM Cards WHERE id = $id");
        }
    }
}
?>

<?php include "../footer.htm";?>

Answer 1

试试这个：

<body>

<?php

$connect=mysqli_connect("Host","Username","Password","Database"); /* REPLACE NECESSARY DATA */

if(mysqli_connect_errno()){

echo "Error".mysqli_connect_error();
}

if (isset($_POST['delete'])){
    $checkbox = $_POST['checkbox'];
    $count = count($checkbox);

    for($i=0;$i<$count;$i++){

        if(!empty($checkbox[$i])){ /* CHECK IF CHECKBOX IS CLICKED OR NOT */

        $id = mysqli_real_escape_string($connect,$checkbox[$i]); /* ESCAPE STRINGS */
        mysqli_query($connect,"DELETE FROM Cards WHERE id = '$id'"); /* EXECUTE QUERY AND USE ' ' (apostrophe) IN YOUR VARIABLE */

        } /* END OF IF NOT EMPTY CHECKBOX */

    } /* END OF FOR LOOP */

} /* END OF ISSET DELETE */

$query = "SELECT * FROM Cards"; /* SELECT FROM Cards TABLE */
$result = mysqli_query($connect,$query); /* EXECUTE QUERY */

echo "<form action='' method='POST'>"; /* SUBMIT PAGE ON ITSELF */

echo "<table>";
echo "<tr><td>Name</td><td>Quantity</td><td>Color</td><td>Type</td></tr>";
while ($row = mysqli_fetch_array($result)){ /* FETCH ARRAY */
    $id=mysqli_real_escape_string($connect,$row['id']);
    echo "<tr><td><input type='checkbox' name='checkbox[]' value='$id'></td>";
    echo "<td>" . $row['Name'] . "</td>";
    echo "<td>" . $row['Quantity'] . "</td>";
    echo "<td>" . $row['Color'] . "</td>";
    echo "<td>" . $row['CardType'] . "</td></tr>";
}
mysqli_free_result($result);
echo "</table>";
?>

<tr>
<td colspan="5" align="center"><input name="delete" type="SUBMIT" id="delete" value="delete" action="POST"></td>
</tr>
</form>

<?php include "../footer.htm";?>
</body>

我已将您的代码从已弃用的 MySQL 转换为 MySQLi 。
并使用 empty（）函数来确定选中的复选框。
您忘了为您的html添加<form>功能。
并且忘了为您的复选框使用值标记。

我在/ * / /中引用了我提供的代码中的解释。

Answer 2

试试这个：

<body>

<?php
$delete = $_POST['checkbox'];
$query = "SELECT * FROM Cards";
$result = mysql_query($query);

echo "<form method=post action=''>";
echo "<table>";
echo "<tr><td>Name</td>
          <td>Quantity</td>
          <td>Color</td>
          <td>Type</td></tr>";
while ($row = mysql_fetch_array($result)){

        $content .= <<< END
    <tr>
        <td><input type="checkbox" name="checkbox" value="{$row['id']}"></td>
        <td>{$row['Name']}</td>
        <td>{$row['Quantity']}</td>
        <td>{$row['Color']}</td>
        <td>{$row['CardType']}</td>
    </tr>
END;
}
echo $content;
mysql_free_result($result);
?>
<tr>
<td colspan="5" align="center"><input name="delete" type="SUBMIT" id="delete" value="delete"></td>
</tr>
</table></form>
</body>

<?php
if (isset($_POST['delete'])) {
    $checkbox = $_POST['checkbox'];

    foreach($checkbox as $id) {

        $id = (int) $id;

        mysql_query("DELETE FROM Cards WHERE id = $id");


    }

}
?>

<?php include "../footer.htm";?>

使用复选框删除Mysql行

2 个答案: