Question

Echo'Hello Programmers';

我正在处理记录删除功能。但是，我现在对重复发生的未定义索引问题感到有点迷失。

这是前端代码。

<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
Select member to <b> DELETE! </b>: <select name="mid">
<?php
while($row = mysqli_fetch_assoc($result))
echo "<option value='{$row['mid']}'>{$row['mid']} </option>";
?>
</select>
<input type="submit" value=">!DELETE!<" />
</form>

<?php
}

else
{
$mid = $_POST['mid'];
$name = $_POST['name'];
$address = $_POST['address'];
$postcode = $_POST['postcode'];
$photo = $_POST['photo'];
$db1 = new dbmember();
$db1->openDB();
$numofrows = $db1->delete_member($mid, $name, $address, $postcode, $photo);
echo "Success. Number of rows affected:
<strong>{$numofrows}<strong>";


$db1->closeDB();
}

所以我们正在从下拉菜单中选择一个ID。然后，当按下删除按钮以执行删除时，我们将传递此ID。

function delete_member($mid, $name, $address, $postcode, $photo) {
        $esc_name = mysqli_real_escape_string($this->conn, $name);
        $esc_address = mysqli_real_escape_string($this->conn, $address);
        $esc_postcode = mysqli_real_escape_string($this->conn,$photo);
        $esc_photo = mysql_reali_escape_string($this->conn, $photo);

        $sql = "DELETE FROM member WHERE mid = $mid";
        $result = mysqli_query($this->conn, $sql);

       if ($result) {
             $numofrows = mysqli_affected_rows($this->conn);
            return $numofrows;
            }
            else
                $this->error_msg = "could not connect for some wierd reason";
                        return false ;
     }

Notice: Undefined index: name in C:\xampp\htdocs\dbm\deletemember.php on line 120

也许我需要更换'Else'操作符，或者将它们全部一起删掉？谢谢。

Answer 1

变化：

$db1->delete_member($mid, $name, $address, $postcode, $photo);

要：

$db1->delete_member($mid);

并改变：

function delete_member($mid, $name, $address, $postcode, $photo) {

要：

function delete_member($mid) {

删除delete_member函数中的第1行到第4行

PHP删除未定义的索引？

1 个答案: