I've got some Java code with SQL statements written as Java strings (please no OR/M flamewars, the embedded SQL is what it is - not my decision). I've broken the SQL statements semantically into several concatenated strings over several lines of code for ease of maintenance. So instead of something like: String query = "SELECT FOO, BAR, BAZ FROM ABC WHERE BAR > 4"; I have somethin
我有一些Java代码,用SQL语句写成Java字符串(请不要使用OR / M flamewars,嵌入式SQL就是这样 - 不是我的决定)。 为了便于维护,我已将SQL语句语义分解为多行代码中的几个连接字符串。 所以不是这样的: String query = "SELECT FOO, BAR, BAZ FROM ABC WHERE BAR > 4"; 我有这样的东西: String query = "SELECT FOO, BAR, BAZ" + " FROM ABC " + " WHERE BAR > 4 "; 这种风格使SQL
So, back again I have a JHipster generated project which uses an elasticsearch java client embedded in spring boot . I have recently done some major changes to the datasets since we've been migrating a whole new bunch of data from different repositories When deploying the application it all works fine, all SearchRepositories are loaded with no problem and all search capabilities roll sm
所以,再回来 我有一个JHipster它使用一个生成的项目elasticsearch java嵌入客户端spring boot 。 我最近对数据集进行了一些重大更改,因为我们已经从不同存储库中迁移了一大堆数据 部署应用程序时,它一切正常,所有SearchRepositories加载没有问题,所有的搜索功能平稳 从测试环境运行时出现问题。 目前还没有改变什么那么到application-test.yml文件,也不到elasticsearch Java的配置文件。 我们有一些更新索引的
This question already has an answer here: Big O, how do you calculate/approximate it? 22 answers Worst case is O(n^2) if you assume all blocks are "free" and they have a size of decreasing powers of 2 with last two being 1: 2^n, 2^(n-1) ... , 64, 32, 16, 8, 4, 2, 1, 1 First iteration merges the last two, triggers a recursive call that now operates on a list that looks like 2^n,
这个问题在这里已经有了答案: 大O,你如何计算/估计它? 22个答案 最坏的情况是O(n^2)如果你假设所有的块都是“空闲”的,并且它们的大小递减2,最后两个是1: 2^n, 2^(n-1) ... , 64, 32, 16, 8, 4, 2, 1, 1 第一次迭代合并最后两次,触发一个递归调用,该调用现在在看起来像的列表上运行 2^n, 2^(n-1) ... , 64, 32, 16, 8, 4, 2, 2 合并最后两个,递归调用,现在正在操作 2^n, 2^(n-1) ... , 64, 32, 16, 8, 4, 4
This question already has an answer here: Big O, how do you calculate/approximate it? 22 answers Well here goes my first answer... O() notation of an algorithm (though it could apply to any function) is somewhat like an upperbound on the algorithms growth rate. The case you are probably concerned with is the function that relates the number of operations your algorithm does for a given in
这个问题在这里已经有了答案: 大O,你如何计算/估计它? 22个答案 那么我的第一个答案是... 算法的O()表示法(虽然它可以应用于任何函数)有点像算法增长率的上限。 您可能担心的情况是与您的算法对给定输入大小执行的操作次数相关的函数。 你似乎要求一步一步的方法,所以我会尝试一个: 编写一个函数,用于描述算法针对n的输入大小(或n个元素的现有数据结构等)执行的最差操作次数。 使用Big-O符号的定义简化
This question already has an answer here: Big O, how do you calculate/approximate it? 22 answers How to find time complexity of an algorithm 10 answers Reference: How to Calculate Time complexity algorithm I found a good article related to How to calculate time complexity of any algorithm or program The most common metric for calculating time complexity is Big O notation. This removes
这个问题在这里已经有了答案: 大O,你如何计算/估计它? 22个答案 如何找到算法的时间复杂度10个答案 参考:如何计算时间复杂度算法 我发现了一篇与如何计算任何算法或程序的时间复杂度有关的好文章 计算时间复杂度的最常用指标是Big O符号。 这消除了所有常数因素,因此当N接近无穷大时,可以根据N估计运行时间。 一般来说,你可以这样想: statement; 是恒定的。 该陈述的运行时间不会相对于N而改变。 for (
I have a simple recursive solution as below: public int countPaths(int x, int y) { if(x == 0 && y == 0) { return 0; } else if(x == 0) { return 1; } else if(y == 0) { return 1; } else { int count = countPaths(x-1, y); count += countPaths(x, y-1); return count; } } This is to solve the following problem from the book: Cr
我有一个简单的递归解决方案,如下所示: public int countPaths(int x, int y) { if(x == 0 && y == 0) { return 0; } else if(x == 0) { return 1; } else if(y == 0) { return 1; } else { int count = countPaths(x-1, y); count += countPaths(x, y-1); return count; } } 这是解决以下问题的书:破解编码采访 想象一下,一个机器人坐在
Wildcard Matching Implement wildcard pattern matching with support for ' ? ' and ' * '. '?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence). The matching should cover the entire input string (not partial). The function prototype should be: bool isMatch(const char *s, const char *p) Some examples:
通配符匹配 实现通配符模式匹配,支持' ? '和' * '。 '?' 匹配任何单个字符。 '*'匹配任何字符序列(包括空序列)。 匹配应覆盖整个输入字符串(不是部分)。 函数原型应该是: bool isMatch(const char * s,const char * p) 一些例子: isMatch(“aa”,“a”)→false isMatch(“aa”,“aa”)→true isMatch(“aaa”,“aa”)→false isMatch(“aa”,“*”)→true isMatc
We want to quickly prototype widgets in Java. We overlay them on top of a display written in a proprietary 3rd party graphics package. We find that the Java GUI steals keyboard focus away from the window manager. The window manager is fvwm, I've tried configuring it so the Java app is setup not to get focus, and furthermore if it ever does get focus to take it away and give it to the othe
我们想要在Java中快速构建小部件的原型。 我们将它们覆盖在用专有第三方图形软件包编写的显示屏上。 我们发现Java GUI从窗口管理器中窃取键盘焦点。 窗口管理器是fvwm,我已经尝试过配置它,所以Java应用程序设置为不关注焦点,此外,如果焦点将其取出并将其提供给其他GUI。 如果我使用Java应用程序运行它,它不起作用(每当鼠标移到Java GUI上时它都有键盘焦点) - 如果我将一些标准X GUI小部件(XEyes)替换为Java GUI,
I encounter a very strange issue in my project using spring web flow 2.4.0. In the documentation of web-flow project we can read on chapter 2 the following statement: By default Web Flow does a client-side redirect upon entering every view state. My concern is when i submit a form web flow do not make redirection that implies a new form submission is made if user reload or refresh the page.
我在使用spring web flow 2.4.0的项目中遇到了一个非常奇怪的问题。 在web-flow项目的文档中,我们可以在第2章中阅读以下声明: 默认情况下,Web Flow在进入每个视图状态时执行客户端重定向。 我担心的是,当我提交表单Web流时,请勿重定向,这意味着如果用户重新加载或刷新页面,则会进行新的表单提交。 这很无聊。 我尝试了很多东西,并在网络上找到了一些解决方案,例如使用以下代码重定向程序: context.getExternalC
I am using Spring Web Flow 2 and am having a basic problem getting my transitions to appropriately fire. I have done a lot of searching on the internet and have been unable to find an in-depth explanation of how transitions are triggered from the view side. I have two states: enterBookingDetails and reviewBooking. EnterBookingDetails is working fine - the page is loading and on submit event, r
我正在使用Spring Web Flow 2,并且遇到了一个基本问题,让我的转换适当启用。 我在互联网上进行了大量搜索,并且一直无法从视角找到如何触发转换的深入解释。 我有两个状态:enterBookingDetails和reviewBooking。 EnterBookingDetails工作正常 - 页面正在加载并在提交事件,reviewBooking加载。 我的问题是,我无法从ReviewBooking转换到正常工作。 这是我的: 预订-flow.xml: <var name="bookingForm" class="com.