使用 Java 解压密码保护的 zip 文件抛出 NullPointerException

Question

我需要解压缩受密码保护的 zip 文件。

到目前为止，我是在互联网的帮助下得到这个的，我正在努力理解代码。

但它抛出 nullpointerexception。我真的很困惑。是 "child" 或者是其他东西。请解释一下。

Exception in thread "main" java.lang.NullPointerException at unzip2.Unzip2.main(Unzip2.java:27) C:\Users\zxc\AppData\Local\NetBeans\Cache.2\executor-snippets\run.xml:53: Java returned: 1 BUILD FAILED (total time: 0 seconds)

它说它在第 27 行。

for (final File child : file. listFiles()) {

据我所知，nullpointerexception 意味着我的变量为 null，但我不知道这里的 null 在哪里以及如何修复它

这是我的完整代码。请赐教

 /*
     * To change this license header, choose License Headers in Project Properties.
     * To change this template file, choose Tools | Templates
     * and open the template in the editor.
     */
    package unzip2;
    /**
     *
     * @author zxc
     */
    import java.io.File;
    import java.util.List;

    import javax.swing.filechooser.FileNameExtensionFilter;

    import net.lingala.zip4j.core.ZipFile;
    import net.lingala.zip4j.exception.ZipException;
    import net.lingala.zip4j.model.FileHeader;

    public class Unzip2 {

        public static void main(String[] args) {

            final FileNameExtensionFilter extensionFilter = new FileNameExtensionFilter("N/A","zip");
            //Folder where zip file is present
            final File file = new File("C:/Users/zxc/Desktop/ziptest/ziptest2.zip");
            for (final File child : file.listFiles()) {
                try {
                    ZipFile zipFile;
                    zipFile = new ZipFile(child);
                    if (extensionFilter.accept(child)) {
                        if (zipFile.isEncrypted()) {
                            //Your ZIP password
                            zipFile.setPassword("password");
                        }
                        List fileHeaderList = zipFile.getFileHeaders();

                        for (int i = 0; i < fileHeaderList.size(); i++) {
                            FileHeader fileHeader = (FileHeader) fileHeaderList.get(i);
                            //Path where you want to Extract
                            zipFile.extractFile(fileHeader, "C:/Users/zxc/Desktop/zipfile");
                            System.out.println("Extracted");
                        }
                    }
                } catch (ZipException e) {
                    System.out.println("Please Try Again");
                }
            }

        }
    }

Answer 1

答案在您的 File 实例上方的一行

//Folder where zip file is present
final File file = new File("C:/Users/zxc/Desktop/ziptest/ziptest2.zip");

您需要定义 zip 文件所在的文件夹，而不是 zip 文件本身

final File file = new File("C:/Users/zxc/Desktop/ziptest");

Answer 2

您不需要迭代文件处理程序。只需在创建 ZipFile 对象时传递 file 对象即可。

public class Unzip2 {

public static void main(String[] args) {

    final FileNameExtensionFilter extensionFilter = new FileNameExtensionFilter("N/A","zip");
    //Folder where zip file is present
    final File file = new File("C:/Users/zxc/Desktop/ziptest/ziptest2.zip");
    //for (final File child : file.listFiles()) {
        try {
            ZipFile zipFile;
            zipFile = new ZipFile(file);
            if (extensionFilter.accept(file)) {
                if (zipFile.isEncrypted()) {
                    //Your ZIP password
                    zipFile.setPassword("password");
                }
                List fileHeaderList = zipFile.getFileHeaders();

                for (int i = 0; i < fileHeaderList.size(); i++) {
                    FileHeader fileHeader = (FileHeader) fileHeaderList.get(i);
                    //Path where you want to Extract
                    zipFile.extractFile(fileHeader, "C:/Users/zxc/Desktop/zipfile");
                    System.out.println("Extracted");
                }
            }
        } catch (ZipException e) {
            System.out.println("Please Try Again");
        }
    //}

}

Answer 3

简单，看一下javadoc:

public File[] listFiles()

Returns an array of abstract pathnames denoting the files in the directory denoted by this abstract pathname. If this abstract pathname does not denote a directory, then this method returns null.

您提供了一个文件名，并询问其路径。文件没有路径。因此返回 null，并且 for (whatever : null) 导致 NPE。

所以这里的重点是：阅读文档，而不是盲目地写下一些代码而不理解这些代码应该做什么。

使用 Java 解压密码保护的 zip 文件抛出 NullPointerException

Unzip password protected zip file using Java throws NullPointerException

java

zip4j