I am not able to send mail. I have tried in multiple way. but still I am getting exception. I am pasting stacktrace and my code over here. Thank you in advance. I am sending test mail from rest ful api. This is my mail code. @Controller public class TestSendMail { @RequestMapping(value = "Customer/mailcheck.htm") @ResponseBody public String testMail(HttpServletRequest request)
我无法发送邮件。 我尝试过多种方式。 但我仍然得到例外。 我在这里粘贴stacktrace和我的代码。 先谢谢你。 我正在发送来自rest ful api的测试邮件。 这是我的邮件代码。 @Controller public class TestSendMail { @RequestMapping(value = "Customer/mailcheck.htm") @ResponseBody public String testMail(HttpServletRequest request) throws Exception, JSONException { String to = "shah20199
I am struggling to create a form where I can insert multiple readings of electric meters. I have constructed a wrapper that holds all the values. All looks good until the moment I submit the readings as I get java.lang.NullPointerException. Here is part of a controller: @RequestMapping("/registerEnergyReadings") public ModelAndView registerEnergyReadings( @ModelAttribute("energyReadi
我正在努力创造一种可以插入多个电表的读数的表格。 我构建了一个包含所有值的包装器。 所有看起来都不错，直到我提交读数，因为我得到java.lang.NullPointerException。 这是控制器的一部分： @RequestMapping("/registerEnergyReadings") public ModelAndView registerEnergyReadings( @ModelAttribute("energyReadingWrapper") EnergyReadingWrapper, energyReadingWrapper, BindingResult result) { // load
I'm getting the NullPointerException and I can't figure out where are my mistakes. The error is in selectByExample function. The code is: package com.service.test; @Service public class TestService { [...] @Autowired private TestMapper TestMapper; [...] public List<Short> test(){ List<Short> lista = new ArrayList<Short>(); testExample me = new testExample
我得到NullPointerException ，我无法弄清楚我的错误在哪里。 该错误在selectByExample函数中。 代码是： package com.service.test; @Service public class TestService { [...] @Autowired private TestMapper TestMapper; [...] public List<Short> test(){ List<Short> lista = new ArrayList<Short>(); testExample me = new testExample(); me.or().andTestEqualTo(1); List<
I dont know what is wrong in my listContacts in controller map.put("contactList", contactService.listContact()); Can somebody help me? SEVERE: Servlet.service() for servlet [dispatcher] in context with path [/test] threw exception [Request processing failed; nested exception is java.lang.NullPointerException] with root cause java.lang.NullPointerException at pl.ivmx.contact.controll
我不知道我的listContacts在控制器map.put("contactList", contactService.listContact()); 有人能帮助我吗？ SEVERE：Servlet.service（）for servlet [dispatcher]与path [/ test]的上下文抛出异常[Request processing failed; 嵌套异常是java.lang.NullPointerException]，其根本原因是java.lang.NullPointerException在pl.ivmx.contact.controller.ContactController.listContacts（ContactController.java:26
I am trying to insert values into Oracle 10g database table. I want to insert value for column FirstName and value is having two single quotes like this: O'Rielly's Edition . I am using the prepared statement of JDBC to pass the value for parameter FirstName . It is working fine with one single quote(O'Rielly). But when I use two single quotes it's giving SQL error as ORA-00
我正在尝试将值插入到Oracle 10g数据库表中。 我想为FirstName列插入值，并且值有两个单引号，如： O'Rielly's Edition 。 我正在使用JDBC的准备好的语句来传递参数FirstName的值。 它用单引号（O'Rielly）正常工作。 但是当我使用两个单引号时，它会给ORA-00907: missing right parenthesis提供SQL错误ORA-00907: missing right parenthesis 。 有关如何处理PreparedStatement中的两个单引号的任何建议？
I have a general Java method with the following method signature: private static ResultSet runSQLResultSet(String sql, Object... queryParams) It opens a connection, builds a PreparedStatement using the sql statement and the parameters in the queryParams variable length array, runs it, caches the ResultSet (in a CachedRowSetImpl ), closes the connection, and returns the cached result set. I ha
我有一个具有以下方法签名的普通Java方法： private static ResultSet runSQLResultSet(String sql, Object... queryParams) 它打开一个连接，使用sql语句和queryParams变量长度数组中的参数构建PreparedStatement ，运行它，缓存ResultSet （在CachedRowSetImpl ），关闭连接并返回缓存的结果集。 我在记录错误的方法中有异常处理。 我将sql语句记录为日志的一部分，因为它对调试非常有帮助。 我的问题是，日志记录的字符
I'm pretty new to Spring in general, as well as Amazon AWS, so I apologize in advance if I'm doing something horribly wrong. I can not seem to get my Spring Boot application to communicate with AWS services in any way when deployed onto Elastic Beanstalk. I've been following the guide here to try to set my project up correctly. I'm essentially taking the code they provide and
我对Spring以及Amazon AWS都很陌生，所以如果我正在做一些可怕的错误，我会提前道歉。 我似乎无法让Spring Boot应用程序在部署到Elastic Beanstalk时以任何方式与AWS服务进行通信。 我一直在这里遵循指南，试图正确设置我的项目。 我基本上采取他们提供的代码并试图从我的项目运行： @Component public class Storage { @Autowired private ResourceLoader resourceLoader; Logger logger = LoggerFactory.
If i create a keystore and sign a jar with this commands: keytool -genkey -alias myalias -keyalg RSA -validity 365 -keystore keys/my.keystore jarsigner -verbose -keystore keys/my.keystore -signedjar my_signed.jar my_unsigned.jar myalias Then i would generate a certificate which is 365 days valid. What does happen to the jar file after 365 days? Is it still runnable? Can someone still get th
如果我创建一个密钥库并用这个命令签署一个jar文件： keytool -genkey -alias myalias -keyalg RSA -validity 365 -keystore keys/my.keystore jarsigner -verbose -keystore keys/my.keystore -signedjar my_signed.jar my_unsigned.jar myalias 然后我会生成365天有效的证书。 365天后的jar文件会发生什么？ 它仍然可以运行吗？ 当他使用“jarsigner verify”时，有人还能得到它的签名者吗？ 证书如何更新？ jarsigner
I am using apache POI to change cells in an excel sheet. After I change the values, the cells with formulas corresponding to cells that have been changed are not updating. When I go into excel and click on the cell with the formula, and then click in the function bar, the formula updates. I have calculation options set to automatically update. Quick example. Row : [A2 + A3] [1] [2] A
我正在使用Apache POI来更改Excel表格中的单元格。 在更改值后，具有与已更改的单元格对应的公式的单元格不会更新。 当我进入excel并用公式单击单元格，然后单击功能栏时，公式更新。 我将计算选项设置为自动更新。 快速示例。 行： [A2 + A3] [1] [2] 这里A1等于3 当我使用POI更改它时： [A2 + A3] [2] [5] A1仍然等于3，直到我点击该单元格。 刷新工作簿或工作表也不起作用。 这是Excel或POI的问题吗？
Shape.java public interface Shape { String draw(); void calcSomething(); } Square.java public class Square implements Shape { @Override public String draw() { return "SQUARE"; } @Override public void calcSomething() { } } When I implement an interface , the method that I implemented must be public . I want to make draw() public method but calcSomet
Shape.java public interface Shape { String draw(); void calcSomething(); } Square.java public class Square implements Shape { @Override public String draw() { return "SQUARE"; } @Override public void calcSomething() { } } 当我实现一个interface ，我实现的方法必须是public 。 我想使draw()公共方法，但calcSomething() 私有或受保护 。 我已经通过了Java