Java 使用PlayFramework 2.4.2的ProvisionException
我正在将一个项目从Play 2.2.4迁移到Play 2.4.2,我遇到了一个我无法理解和解决的异常Java 使用PlayFramework 2.4.2的ProvisionException,java,playframework,dependency-injection,playframework-2.4,Java,Playframework,Dependency Injection,Playframework 2.4,我正在将一个项目从Play 2.2.4迁移到Play 2.4.2,我遇到了一个我无法理解和解决的异常 Unexpected exception ProvisionException: Unable to provision, see the following errors: 1) Error injecting constructor, java.lang.NullPointerException at controllers.Application.<init>(Appl
Unexpected exception
ProvisionException: Unable to provision, see the following errors:
1) Error injecting constructor, java.lang.NullPointerException
at controllers.Application.<init>(Application.java:33)
while locating controllers.Application
for parameter 1 at router.Routes.<init>(Routes.scala:36)
while locating router.Routes
while locating play.api.inject.RoutesProvider
while locating play.api.routing.Router
1 error
build.sbt
文件包括必要的配置
libraryDependencies ++= Seq(
javaJdbc,
cache,
javaWs
)
// Play provides two styles of routers, one expects its actions to be injected, the
// other, legacy style, accesses its actions statically.
routesGenerator := InjectedRoutesGenerator
哪些内容可能会丢失或出错?此行会导致空指针:
ws.url(“https://...");当Guice实例化应用程序
类时,code>ws为空。此外,将请求
作为控制器字段不是线程安全的。将代码更改为以下内容:
public class Application extends Controller {
private WSClient ws;
@Inject
public Application(WSClient ws) {
this.ws = ws;
WSRequest request = this.ws.url("https://...");
...
}
}
谢谢你,这解决了我的问题!我想知道,如果播放文档不起作用,为什么会这样声明:WSRequest-request=ws.url(“http://example.com");代码>此行应属于控制器方法。
public class Application extends Controller {
private WSClient ws;
@Inject
public Application(WSClient ws) {
this.ws = ws;
WSRequest request = this.ws.url("https://...");
...
}
}