我想禁用Sonar的规则,因此它不会在网页中显示结果.在我的情况下,我想隐藏(或不捕获)有关尾随注释的结果.
在某处配置它是否可行?
谢谢.
我正在尝试过滤我的jar中包含的资源.我正在使用shade-maven-plugin,这个将所有依赖项的所有资源添加到我生成的jar中,我只想包含我的项目资源.
这是我的pom定义:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>1.6</version>
<executions>
<execution>
<goals>
<goal>shade</goal>
</goals>
<configuration>
<minimizeJar>true</minimizeJar>
<transformers>
<transformer
implementation="org.apache.maven.plugins.shade.resource.ManifestResourceTransformer">
<mainClass>es.app.applet.MyApplet</mainClass>
</transformer>
</transformers>
</configuration>
</execution>
</executions>
<configuration>
<filters>
<filter>
<artifact>*:*</artifact>
<excludes>
<exclude>META-INF/*.SF</exclude>
<exclude>META-INF/*.DSA</exclude>
<exclude>META-INF/*.RSA</exclude>
<exclude>META-INF/*.RSA</exclude>
</excludes>
</filter>
</filters>
</configuration>
</plugin>
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我尝试添加下一个过滤器以删除所有资源,然后添加另一个过滤器,仅添加我的artifactID资源,但它不起作用.
<filter>
<artifact>*:*</artifact>
<excludes>
<exclude>resources/*.*</exclude>
</excludes>
</filter> <filter>
<artifact>my.groupId:my.artifactId</artifact>
<includes>
<include>resources/*.*</include>
</includes>
</filter>
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想法?
谢谢.
我想在Android中为我的应用程序添加一个按钮,并在按下按钮几秒钟后在此按钮(onclick)上捕获事件,因此在第一次触摸时它不会做出反应.这可以在Android上实现吗?
现在我有下一个代码,用于捕获主页按钮上的onclick(在ActionBar中).
public boolean onOptionsItemSelected(MenuItem item) {
if (item.getItemId() == android.R.id.home) {
showSendLogAlert();
}
return super.onOptionsItemSelected(item);
}
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当用户点击此按钮时,它会通过电子邮件发送一个小报告,我不想意外地启动此事件,这就是为什么我希望用户按下几秒钟以确保他想要这样做操作.
解:
基于以下评论,我得到了这个有效的解决方案:
@Override
protected void onCreate(final Bundle savedInstanceState) {
// stuff
// Set the home button clickable
getActionBar().setHomeButtonEnabled(true);
// Define a long click listener instead of normal one
View homeButton = findViewById(android.R.id.home);
homeButton.setOnLongClickListener(new View.OnLongClickListener() {
@Override
public boolean onLongClick(View v) {
showSendLogAlert();
return false;
}
});
// more stuff
}
Run Code Online (Sandbox Code Playgroud) 我正在尝试从wsdl文件生成一个带有maven和jaxb的客户端,里面有两个模式,一些具有相同名称的元素来自不同的模式
当我尝试执行编译时,我收到下一个错误:
Two declarations cause a collision in the ObjectFactory class.
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WSDL模式:
<wsdl:types>
<schema targetNamespace="http://ws.services" xmlns="http://www.w3.org/2001/XMLSchema">...</schema>
<schema targetNamespace="http://ws.models" xmlns="http://www.w3.org/2001/XMLSchema">...</schema>
</wsdl:types>
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我尝试重命名产生错误的元素,但然后我的spring客户端收到正确的SOAP消息,但它没有正确填充响应对象(它的所有属性都为null).我想问题可能来自重命名响应类,所以这就是为什么我试图生成不同的包保留所有类的原始名称.
为了做到这一点,我编写了下一个绑定文件,但我不知道我做错了它不能正常工作.
bindings.xml文件:
<?xml version="1.0" encoding="UTF-8"?>
<jaxb:bindings version="2.1"
xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/"
xmlns:xjc="http://java.sun.com/xml/ns/jaxb/xjc"
xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
xmlns:xs="http://www.w3.org/2001/XMLSchema" >
<jaxb:bindings schemaLocation="mywsdl.wsdl#types?schema1"
node="/xs:schema[@targetNamespace='http://ws.services']">
<jaxb:schemaBindings>
<jaxb:package name="package1" />
</jaxb:schemaBindings>
</jaxb:bindings>
<jaxb:bindings schemaLocation="mywsdl.wsdl#types?schema2"
node="/xs:schema[@targetNamespace='http://ws.models']">
<jaxb:schemaBindings>
<jaxb:package name="package2" />
</jaxb:schemaBindings>
</jaxb:bindings>
</jaxb:bindings>
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maven文件中的配置部分是下一个,以防它有用:
<groupId>org.jvnet.jax-ws-commons</groupId>
<artifactId>jaxws-maven-plugin</artifactId>
<version>2.3</version>
<executions>
<execution>
<goals>
<goal>wsimport</goal>
</goals>
</execution>
</executions>
<configuration>
<wsdlLocation>wsdl/mywsdl.wsdl</wsdlLocation>
<wsdlDirectory>src/main/resources/wsdl</wsdlDirectory>
<wsdlFiles>
<wsdlFile>mywsdl.wsdl</wsdlFile>
</wsdlFiles>
<bindingDirectory>src/main/resources/wsdl</bindingDirectory>
<bindingFiles>
<bindingFile>bindings.xml</bindingFile>
</bindingFiles>
<packageName>original.package</packageName>
<sourceDestDir>${basedir}/src/main/java</sourceDestDir>
</configuration> …
Run Code Online (Sandbox Code Playgroud) 我有下几个豆子:
Address {
String name;
String number;
String zipcode;
String town;
}
MyEntity {
Address address;
String value1;
String value2;
}
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我正在尝试执行下一个Hibernate查询:
private final List<String> propertiesDistinct = Arrays.asList("address.name");
private final List<String> properties = Arrays.asList("address.number",
"address.zipcode", "address.town")
ProjectionList projectionList = Projections.projectionList();
if (propertiesDistinct != null) {
ProjectionList projectionListDistinct = Projections.projectionList();
for (String propertyDistinct : propertiesDistinct)
projectionListDistinct.add(Projections.property(propertyDistinct).as(propertyDistinct));
projectionList.add(Projections.distinct(projectionListAgrupar));
}
if (properties != null)
for (String property : properties)
projectionList.add(Projections.property(property).as(property));
criterio.setProjection(projectionList);
// MORE FILTERS ON MyEntity FIELDS
//... criterio.add(Restrinctions...);
// I want to …
Run Code Online (Sandbox Code Playgroud) 假设我有下一个代码:
@Autowired
private IManager1 manager1;
@Autowired
private IManager2 manager2;
@Autowired
private IManager3 manager3;
@Transactional
public void run() {
manager1.doStuff();
manager2.registerStuffDone();
manager3.doStuff();
manager2.registerStuffDone();
manager1.doMoreStuff();
manager2.registerStuffDone();
}
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如果启动任何异常,我想回滚“doStuff()”方法完成的所有操作,但我不想回滚“registerStuffDone()”方法记录的数据。
我一直在阅读 @Transactional 注释的传播选项,但我不明白如何正确使用它们。
每个经理在内部都使用 hiberante 来提交更改:
@Autowired
private IManager1Dao manager1Dao;
@Transactional
public void doStuff() {
manager1Dao.doStuff();
}
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dao 看起来像这样:
@PersistenceContext
protected EntityManager entityManager;
public void doStuff() {
MyObject whatever = doThings();
entityManager.merge(whatever);
}
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这是我的 applicationContext 配置:
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSourcePool" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter"/>
</property>
</bean>
<bean id="entityManager" class="org.springframework.orm.jpa.support.SharedEntityManagerBean">
<property name="entityManagerFactory" …
Run Code Online (Sandbox Code Playgroud) 我是Lua的新手并且正在寻找互联网我找不到解决问题的方法,或者如果我真的可以做下一件事就给出答案.
我有下一张桌子.如你所见,钥匙有一个í:
DB = {
["Vigía"] = 112
}
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如果我尝试从表中检索该值,则返回nil.我尝试删除í字符然后我可以得到值112.
我可以在这种情况下使用拉丁字符作为键吗?
谢谢!