如何获取剩余记录的特定记录？

Question

让我们说我有一些记录，而且我希望所有的汽车都能在 type =＆＃34; honda＆＃34;和门=＆＃34; 4＆＃34; 以及所有剩余记录

<Car id="1", type="honda", doors="4" created_at: "2015.01.01">
<Car id="2", type="jeep",  created_at: "2015.01.01">
<Car id="3", type="mazda", created_at: "2015.01.01">
<Car id="4", type="honda", doors="4" created_at: "2015.01.01">
<Car id="5", type="honda", doors="2" created_at: "2015.01.01">
<Car id="6", type="honda", doors="2" created_at: "2015.01.01">

我想得到这个：

<Car id="1", type="honda", doors="4" created_at: "2015.01.01">
<Car id="2", type="jeep",  created_at: "2015.01.01">
<Car id="3", type="mazda", created_at: "2015.01.01">
<Car id="4", type="honda", doors="4" created_at: "2015.01.01">

我有这个查询，但它只返回 type =＆＃34; honda＆＃34;和门=＆＃34; 4＆＃34; 主要是我不想列出IS NOT的其他类型，因为我不确定他们的价值观。

Car.where("type = ? AND doors = ?", "honda", "4")

Answer 1

使用OR运算符选择“honda”

以外的其他类型

Car.where("(type = ? AND doors = ?) OR type <> ?", "honda", "4", "honda")