PHP 从表中退休错误

Question

我正在尝试从数据库中删除用户记录，我想从表中删除用户的所有附属信息，并且我有以下 php 代码用于从我的 mysql 数据库中删除记录。

function delete_records($array_refer)
    {
        global $link;
        foreach ($array_refer as $key => $array_value) {
        $table_name = substr($key, 0, -3);
            foreach ($array_value as $value) {
                $delete = "DELETE from $table_name where $key = $value";
                $exec_delete = @mysqli_query($link, $delete);
                if (!$exec_delete) {
                    rollback_die("Records from $table_name could not be deleted because of: ".mysqli_error($link));
                }
            }
        }
        return true;
    }

    @mysqli_query($link, "SET AUTOCOMMIT=0");
    $select_sui = "SELECT 
    shirt_users.*, 
    shirt_users_types.*, 
    shirt_orders.*, 
    shirt_shipping_addresses.*, 
    shirt_billing_addresses.*, 
    shirt_credit_cards.* 
    FROM shirt_users 
    LEFT JOIN shirt_users_types ON shirt_users.shirt_users_id = shirt_users_types.shirt_users_id
    LEFT JOIN shirt_orders ON shirt_users_types.shirt_orders_id = shirt_orders.shirt_orders_id
    LEFT JOIN shirt_shipping_addresses ON shirt_orders.shirt_shipping_addresses_id = shirt_shipping_addresses.shirt_shipping_addresses_id
    LEFT JOIN shirt_billing_addresses ON shirt_orders.shirt_billing_addresses_id = shirt_billing_addresses.shirt_billing_addresses_id
    LEFT JOIN shirt_credit_cards ON shirt_orders.shirt_credit_cards_id = shirt_credit_cards.shirt_credit_cards_id
    WHERE
    shirt_users.shirt_users_id = $shirt_users_id";

    $exec_select_sui = @mysqli_query($link, $select_sui);

当我运行此程序时，我收到以下错误消息:

The shirt_users_id is:4
Records from shirt_users_types could not be deleted because of: 
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

知道这意味着什么吗？我知道从衬衫_用户_类型检索数据时存在问题，但我没有看到它警告我的任何语法错误？

Answer 1

$key 和 $value 是否存在并且它们不是保留字吗？您还应该尝试'$value'。