相关疑难解决方法(0)

我试图从我当前的JAR中提取ZIP文件:

InputStream resource = getClass().getClassLoader().getResourceAsStream(name);

这是正确的InputStream,但是当我尝试使用以下代码解压缩它时会出错(我将每个文件存储到a中Hashmap<file, filename>):

public static HashMap<String, String> readZip(InputStream inputStream) throws IOException {
    byte[] buffer = new byte[1024];
    HashMap<String, String> list = new HashMap<>();
    ZipInputStream zipInputStream = new ZipInputStream(inputStream);
    ZipEntry entry = zipInputStream.getNextEntry();
    while (entry != null) {
        if (!entry.isDirectory()) {
            StringBuilder stringBuilder = new StringBuilder();
            while (IOUtils.read(zipInputStream, buffer) > 0) {
                stringBuilder.append(new String(buffer, "UTF-8"));
            }
            list.put(stringBuilder.toString(), entry.getName());
        }
        zipInputStream.closeEntry();
        entry = zipInputStream.getNextEntry();
    }
    zipInputStream.closeEntry();
    zipInputStream.close();
    return list;
}

但是当我尝试这样做时,我得到了这个例外(上IOUtils.read)

java.util.zip.ZipException: …