SQL：插入别名a以使列名具有2行

Question

我在OpenOffice Base中工作，类似于Access。 我有一个表（成本）和以下列（每个月有2个值（付费和待定）：

• 提供商

• Januarypay

• Januarypend

• Februarypay

• Februarypend

• Marchpay

• Marchpend

• Aprilpay

• Aprilpend

• Maypay

• Maypend

• Junepay

• Junepend

• Julypay

• Julypend

• Augustpay

• Augustpend

• Septemberpay

• Septemberpend

• Octoberpay

• Octoberpend

• Novemberpay

• Novemberpend

• Decemberpay

• Decemberpend

我正在尝试这样的事情

Select "provider",**"NEEDHELPHERE"**,(select (Januarypay + Januarypend + Februarypay + Februarypend + Marchpay + Marchpend + Aprilpay + Aprilpend + Maypay + Maypend + Junepay + Junepend + Julypay + Julypend + Augustpay + Augustpend + Septemberpay + Septemberpend + Octoberpay + Octoberpend + Novemberpay + Novemberpend + Decemberpay + Decemberpend) as TotalYear from "Costs")

替换“NEEDHELPHERE”以获取以下代码（我把它放了，否则这会太长）：

Januarypay as "January" || char(10)|| char(13) || "Paid",Januarypend as "January" || char(10)|| char(13) || "Pending",Februarypay as "February" || char(10)|| char(13) || "Paid",Februarypend as "February" || char(10)|| char(13) || "Pending",Marchpay as "March" || char(10)|| char(13) || "Paid",MarchpendMarchpay as "March" || char(10)|| char(13) || "Pending",Aprilpay as "April" || char(10)|| char(13) || "Paid",AprilpendAprilpay as "April" || char(10)|| char(13) || "Pending",Maypay as "May" || char(10)|| char(13) || "Paid",Maypend as "May" || char(10)|| char(13) || "Pending",Junepay as "June" || char(10)|| char(13) || "Paid",Junepend as "June" || char(10)|| char(13) || "Pending",Julypay as "July" || char(10)|| char(13) || "Paid",Julypend as "July" || char(10)|| char(13) || "Pending",Augustpay as "August" || char(10)|| char(13) || "Paid",Augustpend as "August" || char(10)|| char(13) || "Pending",Septemberpay as "September" || char(10)|| char(13) || "Paid",Septemberpend as "September" || char(10)|| char(13) || "Pending",Octoberpay as "October" || char(10)|| char(13) || "Paid",Octoberpend as "October" || char(10)|| char(13) || "Pending",Novemberpay as "November" || char(10)|| char(13) || "Paid",Novemberpend as "November" || char(10)|| char(13) || "Pending",Decemberpay as "December" || char(10)|| char(13) || "Paid",Decemberpend as "December" || char(10)|| char(13) || "Pending"

查询的结果应该是这样的：

表格信息

表格示例数据