可能会多次询问，但我有一张有一些记录的表格。有些是明天的约会，有些是后天。我尝试以下
SELECT * FROM Games where GAMEID='".$id."' AND GAMEDATE >=CURDATE() + INTERVAL 1 DAY
我得到的日期为10/1/2017和11/1/2017。中午1天不能只获得10/1/2017的日期吗？
如果我使用
GAMEDATE >= DATE_SUB(CURDATE(), INTERVAL 1 DAY) AND GAMEDATE <= CURDATE()
我得到空查询
答案 0 :(得分：1)
答案 1 :(得分：0)
尝试使用var str = "hello world";
console.log(str.length); //11
str = "hello stack overflow";
console.log(str.length); //20
答案 2 :(得分：0)
你不能只使用这样的添加。您需要使用@Test
public void partitioning() throws InterruptedException {
final int N = 10;
Flux<Integer> source = Flux.range(1, 10000).share();
// partition source into publishers
Publisher<Integer>[] publishers = new Publisher[N];
for (int i = 0; i < N; i++) {
final int idx = i;
publishers[idx] = source.filter(v -> v % N == idx);
}
// create ParallelFlux each 'rail' containing single partition
ParallelFlux.from(publishers)
// schedule partitions into different threads
.runOn(Schedulers.newParallel("proc", N))
// process each partition in its own thread, i.e. in order
.map(it -> {
String threadName = Thread.currentThread().getName();
Assert.assertEquals("proc-" + (it % 10 + 1), threadName);
return it;
})
// collect results on single 'rail'
.sequential()
// and on single thread called 'subscriber-1'
.publishOn(Schedulers.newSingle("subscriber"))
.subscribe(it -> {
String threadName = Thread.currentThread().getName();
Assert.assertEquals("subscriber-1", threadName);
});
Thread.sleep(1000);
}
：
DATE_ADD()如果你想倒退，你将使用GAMEDATE >= DATE_ADD(CURDATE(), INTERVAL 1 DAY);
。