我有两个系列的帖子。每个帖子都有一定比例的进度（ // Creates the ArrayList
ArrayList<Integer> numberList = new ArrayList<Integer>();
// Loop creates an array of numbers starting at "from" ending at "to"
for (int i = from; i < to + 1; i++) {
numberList.add(i);
}
// Check the array to see whether number contains digit
// Code checks whether x contains 5, n == one digit
// IMPORTANT: Doesn't work on the first half of numbers i.e / will remove 3 but not 30
for (int j = 0; j < numberList.size(); j++) {
int number = (int) numberList.get(j); // This can be any integer
int thisNumber = number >= 0 ? number: -number; // if statement in case argument is negative
int thisDigit;
while (thisNumber != 0) {
thisDigit = thisNumber % 10; // Always equal to the last digit of thisNumber
thisNumber = thisNumber / 10; // Always equal to thisNumber with the last digit chopped off, or 0 if thisNumber is less than 10
if (thisDigit == omittedDigit) {
numberList.remove(j);
j--;
}
}
}
// Return the completed Array list
return numberList;
}
）和结束时间（project_percent）。我希望以后首先显示end_date的帖子（＆＃39; DESC＆＃39;）然后我想用过去的订单按百分比显示项目end_date。如何只使用一个查询来完成此操作？
（如果日期不是通过百分比不重要，只有日期很重要，但如果日期过去我想按百分比排序）
end_date
select * from posts ... *今天是--------------------------------------
| post | end_date | project_percent|
| P1 | 28-10-2016| 50 |
| P2 | 26-10-2016| 40 |
| P2 | 30-10-2016| 20 |
| P3 | 24-10-2016| 75 |
我希望输出如下：
27-10-2016*前两个按日期排序，后两个按persent排序
答案 0 :(得分：1)
尝试一下：
select *, end_date < now() as flag from posts
order by flag
, case when flag = 0 then end_date end desc
, case when flag = 1 then project_percent end desc;
使用
可以使其更简单select * from posts
order by
case when end_date < now() = 0 then end_date end desc
, case when end_date < now() = 1 then project_percent end desc;
但是第一个版本更适合调试。