Question

我正在尝试从我的清单文件中提取信息，以便在我的jar文件中的某个方法中显示，但似乎遇到了一些问题。任何帮助表示赞赏。

清单文件：

Manifest-Version: 1.0
Created-By: 1.8.0_60 (Oracle Corporation)
Main-Class: com.example.package1.myClass

Name: com/example/package1
Specification-Title: MyPackage
Specification-Version: v1.1
Specification-Vendor: MyCompanyName
Implementation-Title: MP
Implementation-Version: 2015-11-05-C
Implementation-Vendor: MyName

Name: com/example/package2
Specification-Title: MySecondaryPackage
Specification-Version: v2.0
Specification-Vendor: MyCompanyName
Implementation-Title: M2ndP
Implementation-Version: 2015-11-05-C
Implementation-Vendor: MyName

myClass.java：

package com.example.package1;
import com.example.package2;

class myClass {
  public static void main(String[] args) {
    try {
      myClass clz = new myClass();
      Thread.sleep(10000); //pause 10 seconnds so we can see what's spit out
    } catch (Exception e) {
    //excluded in example
    }
  }

  public myClass() {
    Package pkg = getClass().getPackage();
    if (pkg == null)
      System.out.println("No Package Found");
    else {
      System.out.println("specs: " + pkg.getSpecificationTitle() + " - " + pkg.getSpecificationVersion());
      System.out.println("imps:  " + pkg.getImplementationTitle() + " - " + pkg.getImplementationVersion());
      System.out.println("name:  " + pkg.getName());
    }
    //other code here excluded from example
  }
}

输出：

specs: null - null
imps:  null - null
name:  com.example.package1

那是什么给出的？看起来pkg对象的定义正确，但它不会读取任何规范或实现属性。

Answer 1

所以我终于想通了，并且认为我会分享任何其他人就像我一样对着众所周知的砖墙猛烈抨击。我永远无法让Package类中的方法返回null以外的任何内容。请参阅下面的修订代码，了解我是如何实现这一目标的。

package com.example.package1;
import java.util.*;
import java.util.jar.*;
import java.net.*;

class myClass {
  public static void main(String[] args) {
    try {
    new myClass();
    } catch (Exception e) {
    System.out.println(e.getMessage());
    } finally {
    System.out.println("Done");
    try{Thread.sleep(40000);}catch(Exception ee){}
    }
  }

public myClass() throws Exception {
  String clz = getClass().getSimpleName() + ".class";
  String pth = getClass().getResource(clz).toString();
  String mnf = pth.substring(0, pth.lastIndexOf("!") + 1) + "/META-INF/MANIFEST.MF";
  String pkg = getClass().getPackage().getName().replaceAll("\\.","/");
  URL url = new URL(mnf);
  Manifest manifest = new Manifest(url.openStream());

  Attributes attr = manifest.getAttributes(pkg);
  String value = attr.getValue("Specification-Title") + " - " + 
  attr.getValue("Implementation-Title") + " " + 
  attr.getValue("Specification-Version") + " build # " + 
  attr.getValue("Implementation-Version");
  System.out.println(value);
  }
}

<强>输出：

MyPackage - MP v1.1 build # 2015-11-05-C
Done

提取四个元数据的代码很多。

所以，如果你喜欢这里的一些线路，那就是我用的东西：

public myClass() throws Exception {
  Attributes attr = new Manifest(new URL(getClass().getResource(getClass().getSimpleName() + ".class").toString().substring(0, getClass().getResource(getClass().getSimpleName() + ".class").toString().lastIndexOf("!") + 1) + "/META-INF/MANIFEST.MF").openStream()).getAttributes(getClass().getPackage().getName().replaceAll("\\.","/"));
  String value = attr.getValue("Specification-Title") + " - " + attr.getValue("Implementation-Title") + " " + attr.getValue("Specification-Version") + " build # " + attr.getValue("Implementation-Version");
  System.out.println(value);
}

Answer 2

在包路径的末尾添加斜杠。即将com/example/package1更改为com/example/package1/。在com.example.package1包中要求一些课程（我们称之为Foo），一切都应该正常。

Package pkg = com.example.package1.class.getPackage();
String specVer = pkg.getSpecificationVersion();

尾随斜线似乎很重要。例如。这是Apache的ant.jar：

的清单

Name: org/apache/tools/ant/
Extension-name: org.apache.tools.ant
Specification-Title: Apache Ant
Specification-Version: 1.9.6
Specification-Vendor: Apache Software Foundation
Implementation-Title: org.apache.tools.ant
Implementation-Version: 1.9.6
Implementation-Vendor: Apache Software Foundation

如何从清单文件

2 个答案: