本文介绍了在控制器上注入构造函数java.lang.NullPointerException时出错的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我收到以下错误消息:

1)错误注入构造函数,java.lang.NullPointerException在
controllers.ProductsController。(ProductsController.java:20)

同时定位controllers.ProductsController
为router.Routes。(Routes.scala:34)的第4个参数,同时定位router.Routes同时定位
play.api.inject.RoutesProvider同时定位
play.api.routing.Router
为play.api.http.JavaCompatibleHttpRequestHandler的第一个参数。(HttpRequestHandler.scala:222)
同时定位play.api.http.JavaCompatibleHttpRequestHandler而
定位play.api.http.HttpRequestHandler
为play.api.DefaultApplication的第6个参数。(Application.scala:236)
play.api.DefaultApplication.class(Application.scala:235),而
定位play.api .DEF找到
时的aultApplication.api.Application

1) Error injecting constructor, java.lang.NullPointerException at controllers.ProductsController.(ProductsController.java:20)
while locating controllers.ProductsController for the 4th parameter of router.Routes.(Routes.scala:34) while locating router.Routes while locating play.api.inject.RoutesProvider while locating play.api.routing.Router for the 1st parameter of play.api.http.JavaCompatibleHttpRequestHandler.(HttpRequestHandler.scala:222) while locating play.api.http.JavaCompatibleHttpRequestHandler while locating play.api.http.HttpRequestHandler for the 6th parameter of play.api.DefaultApplication.(Application.scala:236) at play.api.DefaultApplication.class(Application.scala:235) while locating play.api.DefaultApplication while locating play.api.Application

1错误

这是堆栈跟踪:

com.google.inject.ProvisionException: Unable to provision, see the following errors:

1) Error injecting constructor, java.lang.NullPointerException
  at controllers.ProductsController.<init>(ProductsController.java:20)
  while locating controllers.ProductsController
    for the 4th parameter of router.Routes.<init>(Routes.scala:34)
  while locating router.Routes
  while locating play.api.inject.RoutesProvider
  while locating play.api.routing.Router
    for the 1st parameter of play.api.http.JavaCompatibleHttpRequestHandler.<init>(HttpRequestHandler.scala:222)
  while locating play.api.http.JavaCompatibleHttpRequestHandler
  while locating play.api.http.HttpRequestHandler
    for the 6th parameter of play.api.DefaultApplication.<init>(Application.scala:236)
  at play.api.DefaultApplication.class(Application.scala:235)
  while locating play.api.DefaultApplication
  while locating play.api.Application

1 error
     com.google.inject.internal.InjectorImpl$2.get(InjectorImpl.java:1028)
     com.google.inject.internal.InjectorImpl.getInstance(InjectorImpl.java:1054)
     play.api.inject.guice.GuiceInjector.instanceOf(GuiceInjectorBuilder.scala:409)
     play.api.inject.guice.GuiceInjector.instanceOf(GuiceInjectorBuilder.scala:404)

这是我的代码:

package controllers;

import java.util.List;

import javax.inject.Inject;
import javax.inject.Provider;
import javax.inject.Singleton;

import play.mvc.Controller;
import play.mvc.Result;

import play.data.Form;
import play.data.FormFactory;

import views.html.products.details;
import views.html.products.list;

import models.Product;

public class ProductsController extends Controller {

    @Inject FormFactory formFactory;
    private Form<Product> productForm = formFactory.form(Product.class);

    public Result list() {
        List<Product> products = Product.findAll();
        return ok(list.render(products));
    }

    public Result newProduct() {
        return ok(details.render(productForm));
    }

    public Result details(String ean) {
        return TODO;
    }

    public Result save() {
        return TODO;
    }
}

我真的不知道该怎么做。我一直在寻找各地,找不到任何有用的东西。想法?

I'm not really sure what should I do. I've been searching all over the place and couldn't find anything useful. Thoughts?

推荐答案

我想出来了,我需要将我的代码更改为如下所示:

I figured this out, I needed to change my code to be like the following:

package controllers;

import java.util.List;

import javax.inject.Inject;
import javax.inject.Provider;
import javax.inject.Singleton;

import play.mvc.Controller;
import play.mvc.Result;

import play.data.Form;
import play.data.FormFactory;

import views.html.products.details;
import views.html.products.list;

import models.Product;

public class ProductsController extends Controller {

    private Form<Product> productForm;

    @Inject
    public ProductsController(FormFactory formFactory) {
        this.productForm = formFactory.form(Product.class);
    }

    public Result list() {
        List<Product> products = Product.findAll();
        return ok(list.render(products));
    }

    public Result newProduct() {
        return ok(details.render(productForm));
    }

    public Result details(String ean) {
        return TODO;
    }

    public Result save() {
        return TODO;
    }
}

具体来说,我需要一个构造函数要注入的 FormFactory

To be specific, I needed to have a constructor with a FormFactory to be injected:

private Form<Product> productForm;

@Inject
public ProductsController(FormFactory formFactory) {
    this.productForm = formFactory.form(Product.class);
}

这篇关于在控制器上注入构造函数java.lang.NullPointerException时出错的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

07-29 12:46