我以前用bash做这个...
/bin/bash --rcfile /home/sindhu/bin/misc_scripts/shellrc/.bashrc_1
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我怎么能用zsh完成同样的事情?
谢谢.
我想在Scala中运行时创建一个类.现在,只考虑一个简单的例子,我想用相似的java bean来创建一些属性,我只在运行时知道这些属性.
如何创建scala类?我愿意从Scala的源文件创建,如果有编译它,并在运行时加载的方式,我可能要为我有时候有一些复杂的功能,我想添加到类.我该怎么做?
我担心我读到的scala解释器是沙盒化它加载的解释代码,以便托管解释器的一般应用程序无法使用它?如果是这种情况,那么我将无法使用动态加载的scala类.
无论如何,问题是,如何在运行时动态创建一个scala类并在我的应用程序中使用它,最好的情况是在运行时从scala源文件加载它,类似于interpreterSource("file.scala")
它并加载到我当前的运行时,第二最好的情况是通过调用方法来创建. createClass(...)
在运行时创建它.
谢谢,菲尔
我正在尝试将我的应用程序部署到Heroku.Heroku没有加载我的js和css文件
我跑了RAILS_ENV=production bundle exec rake assets:precompile
,它仍然无法正常工作.
然后我跑了
heroku rake assets:precompile
我在Heroku日志中收到此错误:
Error compiling asset application.css:
Sprockets::FileNotFound: couldn't find file 'jquery.ui.datepicker'
(in /app/app/assets/stylesheets/application.css.scss:13)
Served asset /application-989f5e5266d9b066eb316183d7db5c77.css - 500 Internal Server Error
Error compiling asset application.js:
Sprockets::FileNotFound: couldn't find file 'jquery.ui.datepicker'
(in /app/app/assets/javascripts/application.js:16)
Served asset /application-d81c946c6f47242e5e97de9bca4938be.js - 500 Internal Server Error
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config production.rb:
config.cache_classes = true
config.consider_all_requests_local = false
config.action_controller.perform_caching = true
config.serve_static_assets = true
config.assets.compress = true
config.assets.compile = true
config.assets.initialize_on_precompile = false …
Run Code Online (Sandbox Code Playgroud) 我正在使用Apache Maven3,因为两三天后一些依赖项无法解决,首先没有问题.更具体一点:
maven-findbgs-plugin:plugin:1.3.1
maven-cobertura-plugin:plugin:1.3
它们不应该被maven本身包括在内吗?
我的pom.xml
档案:
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<artifactId>xyz</artifactId>
<name>xyz</name>
<description>xyz</description>
<repositories>
<repository>
<id>prime-repo</id>
<name>PrimeFaces Maven Repository</name>
<url>http://repository.primefaces.org</url>
<layout>default</layout>
</repository>
</repositories>
<dependencies>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-webmvc</artifactId>
<version>${springframework-version}</version>
</dependency>
<dependency>
<groupId>org.springframework.webflow</groupId>
<artifactId>spring-faces</artifactId>
<version>${springwebflow-version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-core</artifactId>
<version>${springsecurity-version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-config</artifactId>
<version>${springsecurity-version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-web</artifactId>
<version>${springsecurity-version}</version>
</dependency>
<dependency>
<groupId>org.slf4j</groupId>
<artifactId>slf4j-api</artifactId>
<version>${org.slf4j-version}</version>
</dependency>
<dependency>
<groupId>org.slf4j</groupId>
<artifactId>jcl-over-slf4j</artifactId>
<version>${org.slf4j-version}</version>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>org.slf4j</groupId>
<artifactId>slf4j-log4j12</artifactId>
<version>${org.slf4j-version}</version>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>log4j</groupId>
<artifactId>log4j</artifactId>
<version>1.2.16</version>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>javax.inject</groupId> …
Run Code Online (Sandbox Code Playgroud) 我有一个像这样组织的项目:
core
-- /src/main/resources/company/config/spring-config.xml
webapp
-- /WEB-INF/applicationContext.xml
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webapp取决于我在部署时core.jar
正确包含的内容WEB-INF/lib
.
在web.xml
我有:
<param-value>
/WEB-INF/applicationContext.xml
</param-value>
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在applicationContext.xml
我有:
<import resource="classpath:/company/config/spring-config.xml" />
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但是当我跑步时,我收到了这个错误:
2012-10-04 20:03:39,156 [localhost-startStop-1] springframework.web.context.ContextLoader ERROR: Context initialization failed
org.springframework.beans.factory.parsing.BeanDefinitionParsingException: Configuration problem: Failed to import bean definitions from URL location [classpath:/company/config/spring-config.xml]
Offending resource: ServletContext resource [/WEB-INF/applicationContext.xml]; nested exception is org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from class path resource [company/config/spring-config.xml]; nested exception is java.io.FileNotFoundException: class path resource [company/config/spring-config.xml] cannot be opened because it does not exist
at …
Run Code Online (Sandbox Code Playgroud) 我正在尝试为进入Hapi处理程序的JSON对象编写Joi验证.到目前为止代码看起来像这样:
server.route({
method: 'POST',
path: '/converge',
handler: function (request, reply) {
consociator.consociate(request.payload)
.then (function (result) {
reply (200, result);
});
},
config: {
validate: {
payload: {
value: Joi.object().required().keys({ knownid: Joi.object() })
}
}
}
});
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到目前为止,您可以在上面的config:validate:code部分中看到Joi对象验证.进入的JSON看起来像这样.
"key": '06e5140d-fa4e-4758-8d9d-e707bd19880d-testA',
"value": {
"ids_lot_args": {
"this_id": "stuff",
"otherThign": "more data"
},
"peripheral_data": 'Sample peripheral data of any sort'
}
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在上面的JSON中,需要对象根目录的键和值,并且ids_lot_args
需要调用的部分.以peripheral_data开头的部分可以在那里,也可以是任何其他JSON有效负载.没关系,只ids_lot_args
需要根级别和值内部的键和值.
到目前为止,我一直在努力让Joi验证工作. 有关如何设置的任何想法?Joi的代码仓库位于https://github.com/hapijs/joi,如果想要查看它.到目前为止,我一直在尝试允许对象上的所有函数无效.
org-element
是一个新的模块org-mode
,我认为这对解析org文件很好,并且不想用旧方法用match-string解析它.但玩了几个小时之后,我不得不承认我没有足够的org-mode经验,所以主人能给我一些线索,非常感谢!
我的要求很简单,我希望得到所有标题和内容.
* headline
:PROPERTIES
** subheadline
content1
** subheadline
content2
Run Code Online (Sandbox Code Playgroud) 是否有(built'in/easy)方法以递归方式显示names
作为树的互连列表?(可能输出类似于tree
shell命令.)
例如,对于列表X,具有两个列A和B,A包含在两个子列a1和a2中
nametree(x)
X
??? A
? ??? a1
? ??? a2
??? B
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names(X)
只会显示 [1] "A" "B"
我是使用Java Config配置Spring Security的新手.我试图关注这个帖子.但是,当我运行我的应用程序时,我会对所有URL 进行基本Auth挑战,包括.输入下面的任一userid/pass组合似乎不起作用./
我的控制器:
package com.xxx.web;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
@Controller
@RequestMapping("/")
/**
* Controller to handle basic "root" URLs
*
* @author xxx
* @version 0.1.0
*/
public class RootController {
/**
* Handles '/'
* @param model
* @return
*/
@RequestMapping
public String index(Model model) {
return "index";
}
/**
* Handles '/signup'
* @param model
* @return
*/
@RequestMapping("/signup")
public String signup(Model model) {
return …
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有谁知道Curator锁配方中的哪一个会产生短暂的节点?
我测试了InterProcessMutex
锁,但据我所知(使用zkClient
),它不会在释放后删除节点或关闭会话.
这是我用于锁定密钥的代码.
public void lock(final LockKey lockkey, final LockAcquiredListener listener) throws Exception {
final String lockKeyPath = lockkey.toString();
LOGGER.info("Trying to acquire the lock {}", lockKeyPath);
final InterProcessMutex lock = new InterProcessMutex(client, LOCKS_PREFIX + lockKeyPath);
if (!lock.acquire(LOCK_MS_TIME_OUT, TimeUnit.MILLISECONDS)) {
LOGGER.info("Could not acquire the lock {}", lockkey.toString());
throw new LockAcquisitionTimeOutException("Could not acquire the lock: " + lockKeyPath);
}
try {
if (listener != null) {
LOGGER.info("Lock acquired for key {}", lockKeyPath);
listener.lockAcquired(lockkey);
}
} finally {
LOGGER.info("Release …
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