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Java 将'init param'从GUI传递到servlet不起作用_Java_Google App Engine_Servlets_Guice_Jersey 2.0 - Fatal编程技术网

Java 将'init param'从GUI传递到servlet不起作用

Java 将'init param'从GUI传递到servlet不起作用,java,google-app-engine,servlets,guice,jersey-2.0,Java,Google App Engine,Servlets,Guice,Jersey 2.0,尝试使用Guice()将初始化参数传递给我的servlet: 我总是得到一个null。我做错了什么?我怀疑Guice和Jersey与GAE的集成有问题,但无法排除故障 档案: 配置: 适用于Java 1.9.26的Google应用程序引擎 Guice 4.0 泽西岛2.5.1 目标化5.1.7 将以下内容添加到servlet中有效: private String key1 = null; private String key2 = null; @Override publ

尝试使用Guice()将初始化参数传递给我的servlet:

我总是得到一个
null
。我做错了什么?我怀疑GuiceJersey与GAE的集成有问题,但无法排除故障

档案:

配置:

  • 适用于Java 1.9.26的Google应用程序引擎
  • Guice 4.0
  • 泽西岛2.5.1
  • 目标化5.1.7

将以下内容添加到servlet中有效:

private String key1 = null;

private String key2 = null;


@Override
public void init(ServletConfig config) throws ServletException {
    this.key1 = config.getInitParameter("key1");
    this.key2 = config.getInitParameter("key2");
}
然后在
doGet(…)
方法中:

    resp.getWriter().println("key1: " + this.key1);
    resp.getWriter().println("key2: " + this.key2);
private String key1 = null;

private String key2 = null;


@Override
public void init(ServletConfig config) throws ServletException {
    this.key1 = config.getInitParameter("key1");
    this.key2 = config.getInitParameter("key2");
}
    resp.getWriter().println("key1: " + this.key1);
    resp.getWriter().println("key2: " + this.key2);