我搜索了这个答案但却找不到任何答案。希望得到一点帮助。
需要查询：每列中每个数字出现的次数。
tableName =＆＃34; Schedule＆＃34;
+-----+-------+-----------+-------+
| key | prime | secondary | third |
+-----+-------+-----------+-------+
| 1 | 6 | 6 | 7 |
| 2 | 7 | 9 | 9 |
| 3 | 6 | 9 | 6 |
| 4 | 9 | 9 | 9 |
+-----+-------+-----------+-------+
我希望我的结果是每列中每次出现的计数......
+--------+------ +-----------+-------+
| number | prime | secondary | third |
+--------+-------+-----------+-------+
| 6 | 2 | 1 | 1 |
| 7 | 1 | 0 | 1 |
| 9 | 1 | 3 | 2 |
+-------+-----------+--------+-------+
所以＃6在＆＃34; prime＆＃34;中出现2次。列，在＆＃34; secondary＆＃34;中出现1x列在＆＃34;第三个＆＃34;中出现1x列。
答案 0 :(得分：1)
使用import org.apache.camel.CamelContext;
import org.apache.camel.builder.RouteBuilder;
import org.apache.camel.impl.DefaultCamelContext;
import org.apache.camel.routepolicy.quartz.CronScheduledRoutePolicy;
public class App1 {
public static void main(final String[] arguments) {
final CamelContext camelContext = new DefaultCamelContext();
try {
camelContext.addRoutes(new RouteBuilder() {
@Override
public void configure() throws Exception {
CronScheduledRoutePolicy startPolicy = new CronScheduledRoutePolicy();
startPolicy.setRouteStartTime("0 0/5 * 1/1 * ? *");
from("file:E:\\TestingWatch1\\input")
.routeId("testRoute").routePolicy(startPolicy)
.to("file:E:\\TestingWatch1\\output");
}
});
camelContext.start();
Thread.sleep(10000);
//camelContext.stop();
} catch (Exception camelException) {
}
}
}
从三列中获取所有不同的数字。 union这些数字的每列的left join以获得最终结果。
count答案 1 :(得分：0)
select nall.id,
(select count(*) from Schedule where prime = nall.id) as prime,
(select count(*) from Schedule where secondary = nall.id) as secondary,
(select count(*) from Schedule where third = nall.id) as third
from
(select prime as id from Schedule
union
select secondary as id from Schedule
union
select third as id from Schedule ) nall;