I have a Java client that is inserting (through a CSV and 'LOAD DATA INFILE') timestamps (type TIMESTAMP) in UTC format in a MySQL DB. My server's timezone is "America/New_York" and I need to keep daylight savings information. November 5th at 2:00am is when the time changes from EDT to EST and clocks are set back one hour to 1:00AM. So November 5th 1:30 AM, for instance,
我有一个Java客户端，它在MySQL数据库中以UTC格式插入（通过CSV和'LOAD DATA INFILE'）时间戳（键入TIMESTAMP）。 我的服务器的时区是“America / New_York”，我需要保留夏令时信息。 11月5日凌晨2点，时间从美国东部时间变为东部时间，时钟从早上1点恢复到凌晨1点。 例如，11月5日上午1:30，发生两次; 一次在美国东部时间，一次在东部时 我怎样才能让服务器区分两者？ 例如，如果我想插入'2017-11-05 01:30:0
I have to print the EST time in my Java application. I had set the time zone to EST using: Calendar cal = Calendar.getInstance(TimeZone.getTimeZone("EST")); But when the daylight savings is being followed in this timezone, my code does not print the correct time (it prints 1 hour less). How to make the code work to read the correct time always, irrespective of whether the daylight savings ar
我必须在我的Java应用程序中打印EST时间。 我使用以下方法将时区设置为EST： Calendar cal = Calendar.getInstance(TimeZone.getTimeZone("EST")); 但是，如果在此时区遵守夏令时，我的代码不会打印正确的时间（打印时间会少于1小时）。 如何使代码工作始终阅读正确的时间，无论是否遵守夏令时？ PS：我尝试将时区设置为EDT，但它不能解决问题。 首先是这个问题： Calendar cal = Calendar.getInstance(TimeZone.getTim
This question already has an answer here: Daylight saving time and time zone best practices [closed] 30 answers I would suggest storing all times in UTC format. Perform all your calculations and algorithms with the UTC time. You only care about timezones for display purposes. For a particular user, convert the UTC time to their preferred timezone.
这个问题在这里已经有了答案： 夏令时和时区最佳做法[已关闭] 30个答案 我建议以UTC格式存储所有时间。 用UTC时间执行所有计算和算法。 您只关心时区用于显示目的。 对于特定用户，将UTC时间转换为其首选时区。
I need to force any time related operations to GMT/UTC, regardless the timezone set on the machine. Any convenient way to so in code? To clarify, I'm using the DB server time for all operations, but it comes out formatted according to local timezone. Thanks! The OP answered this question to change the default timezone for a single instance of a running JVM, set the user.timezone system
无论机器上设置的时区如何，我都需要强制任何时间相关的操作到GMT / UTC。 任何方便的方式，以便在代码中？ 为了澄清，我使用数据库服务器时间进行所有操作，但它根据当地时区格式化。 谢谢！ OP回答了此问题以更改运行JVM的单个实例的默认时区，请设置user.timezone系统属性： java -Duser.timezone=GMT ... <main-class> 如果您需要在从数据库ResultSet检索Date / Time / Timestamp对象时设置特定时区，请使用
I'm using Intellij Idea 14 working with big java project where a lot of code is compiled and w/o sources. As far as I know Idea can't make a search in compiled code (as eclipse does), but it can decompile single file using build-in fernflower and this also gives a possibility to debug through compiled code. Is there a way to decompile all compiled code at once the same way Idea does a
我正在使用Intellij Idea 14与大型Java项目一起工作，其中大量代码被编译并且没有源代码。 据我所知， Idea无法在编译代码中进行搜索（就像eclipse一样），但它可以使用内置的fernflower反编译单个文件，这也可以通过编译代码进行调试。 有没有办法像IDE一样反编译所有编译后的代码，并将其附加到项目中，这样就可以在反编译的源代码中通过“项目和库”搜索不到调试功能的项目？ 我试过使用不同的fernflower选项来使用Byteco
I have a web service that had been working fine while we were on Java 7 version 75. We recently updated to Java 8 and now we are getting this error on the web service: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target This is the full dump: Charset [empty string] ErrorDetail I/O Exception:
我有一个在Java 7版本75上工作正常的Web服务。我们最近更新到Java 8，现在我们在Web服务上收到这个错误：PKIX路径构建失败：sun.security.provider.certpath。 SunCertPathBuilderException：无法找到要求的目标的有效证书路径 这是完整的转储： Charset [empty string] ErrorDetail I/O Exception: sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCer
This question already has an answer here: What does Serializable mean? 10 answers Serialization is the conversion of an object to a series of bytes, so that the object can be easily saved to persistent storage or streamed across a communication link. The byte stream can then be deserialized - converted into a replica of the original object. You can think of serialization as the process of
这个问题在这里已经有了答案： Serializable是什么意思？ 10个答案 序列化是将一个对象转换为一系列字节，以便该对象可以轻松地保存到永久性存储器中或通过通信链接进行流式处理。 字节流然后可以被反序列化 - 转换成原始对象的副本。 您可以将序列化看作将对象实例转换为字节序列（根据实现可能是二进制或非二进制）的过程。 当您想要通过网络传输一个对象数据时，例如从一个JVM传输到另一个JVM，这非常有用。 在Ja
I am looking for a java function that minimizes the number of edge crossings in a graph. The input is supposed to be a Graph G(E[],V[]) where V[] is an array of all nodes and E[] is an array of all the edges. The output should be a 2D array of edge elements (E[][]) which contains all pairs of edges crossing each other. Please see https://www.ads.tuwien.ac.at/research/graphDrawing.html sectio
我正在寻找一种Java函数，它可以最小化图形中边缘交叉点的数量。 输入应该是图G（E []，V []），其中V []是所有节点的数组，E []是所有边的数组。 输出应该是边缘元素的二维数组（E [] []），其中包含所有相互交叉的边。 例如，请参阅https://www.ads.tuwien.ac.at/research/graphDrawing.html“交叉最小化”部分。 图3a和3b显示了同一个图的不同表示。 但是图3b只有极少量的边缘交叉。 因此在这种特殊情况下，函数outputa
I have a problem by using Apache Camel in combination with Oracle Advanced Queues and JMS. It's about an application to distribute messages. Messages are received and enqueued in Oracle Advanced Queues with the help of Camel. Then they are consumed with Camel and forwarded to the target system. For the case that the message delivery fails, there is an retry count defined in the Advanced
通过将Apache Camel与Oracle Advanced Queues和JMS结合使用，我遇到了一个问题。 这是关于分发消息的应用程序。 在Camel的帮助下，收到消息并在Oracle高级队列中入队。 然后他们与骆驼消耗并转发到目标系统。 对于消息传递失败的情况，在高级队列中定义了重试计数，以便重复消息传递。 如果Camel现在将邮件出列并将其发送到不可用的目标系统，则会抛出HttpOperationFailedException或NoSuchEndpointException。 这些被捕
There is a simple iterative solution for these kind of problems. Node Insert(Node head,int data) { Node newNode = new Node(); newNode.data = data; if (head == null) { return newNode; } Node current = head; while (current.next != null) { current = current.next; } current.next = newNode; return head; } It works perfectly fine. But I want to le
对于这些问题有一个简单的迭代解决方案。 Node Insert(Node head,int data) { Node newNode = new Node(); newNode.data = data; if (head == null) { return newNode; } Node current = head; while (current.next != null) { current = current.next; } current.next = newNode; return head; } 它工作得很好。 但我想学习递归并从这个角度看待事物。 因此，我提出