php - 向外键插入数据时出错

Question

有人可以帮助我吗，我已经被这个问题困扰了很长一段时间，而且对此也很陌生。我有一个表(tbl_call_log_detail)，其中包含2个外键(dealer_id引用tbl_dealer_info中的id和type_of_call引用tbl_call_types中的id)。我正在尝试更新我的表单，它可以正常工作。然后我希望更新中的相关字段自动插入到 tbl_call_log_detail 中。这就是我失去理智的原因

$query = "SELECT id FROM tbl_dealer_info WHERE account_name = '$account_name' INTO $dealer_id";
    $result = mysqli_query($conn, $query);
    $query2 = "SELECT id FROM tbl_call_types WHERE type_of_call = '$type_of_call' INTO $type_of_call";
    $result2 = mysqli_query($conn, $query2);

        $sql = "INSERT INTO tbl_call_log_detail ";
        $sql .= "(comment, time_stamp, type_of_call, dealer_id) ";
        $sql .= "VALUES ";
        $sql .= "('$_POST[comments]', '$_POST[time_stamp]', '$type_of_call', '$dealer_id') "; 
mysqli_query($conn, $sql);
}

当我回显插入语句然后将其复制粘贴到 phpmyadmin 时，出现错误

'Incorrect integer value: change in region' for column 'type_of_call'

Answer 1

'Incorrect integer value: change in region' for column 'type_of_call

此处type_of_call位于整数列中，并且您尝试插入'change in region'，一个字符串 其中的值(value)。