将数据插入外键时出错

Question

有人可以帮助我，我已经坚持了很长一段时间，对此我也很陌生。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。这就是我在失去理智

$query = "SELECT id FROM tbl_dealer_info WHERE account_name = '$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);
}

当我回显我的Insert语句然后将其粘贴到phpmyadmin上时，我收到错误

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

Answer 1

  

＆＃39;不正确的整数值：区域变化＆＃39;列＆＃39; type_of_call

type_of_call列中的integer，您尝试在其中插入'change in region'，string值。