我需要使用zip4j从zip中提取目录。我可以在目录中找到每个文件并将其解压缩。
如何列出目录中的文件?
或者,是否有一个实用程序将目录提取到路径?
答案 0 :(得分:1)
从Zip4j的ZipFile
,您可以获取此zip文件中所有文件头的列表。然后你可以从这个文件头检查,如果这个“文件”是一个目录。如果是,请提取它。
以下是仅从zip文件中提取目录的示例代码
import java.util.List;
import net.lingala.zip4j.core.ZipFile;
import net.lingala.zip4j.exception.ZipException;
import net.lingala.zip4j.model.FileHeader;
public class ExtractDirectory {
public static void main(String[] args) {
try {
ZipFile zipFile = new ZipFile("myZipWithDirectories.zip");
if (zipFile.isEncrypted()) {
zipFile.setPassword("test".toCharArray());
}
@SuppressWarnings("unchecked")
List<FileHeader> fileHeaders = zipFile.getFileHeaders();
for(FileHeader fileHeader : fileHeaders) {
if (fileHeader.isDirectory()) {
zipFile.extractFile(fileHeader, "anyValidPathToExtractTo");
}
//Alternatively, if you want to extract a directory by its name
//if (fileHeader.isDirectory() && fileHeader.getFileName().equals("myDirectoryName")) {
// zipFile.extractFile(fileHeader, "anyValidPathToExtractTo");
//}
}
} catch (ZipException e) {
e.printStackTrace();
}
}
}
答案 1 :(得分:0)
在zip4j中,您应该能够利用myZipFile.getFileHeaders()
或zipFile.getFileHeader("TargetFolder");
功能来提取目标文件夹。
//List<FileHeader> fHeaders = myZipFile.getFileHeaders(); for all file Header
FileHeader targetFileHeader = zipFile.getFileHeader("TargetFolder");
if (targetFileHeader.isDirectory()) {
File f = new File("anyGivenDirectory/" + targetFileHeader.getFileName());
f.mkdirs(); // mkdirs() is different from mkdir()
zipFile.extractFile(fileHeader, f.toString());
}
据我所知,另一个已知的好图书馆:zt-zip。但我不确定它支持解密。
答案 2 :(得分:0)
使用此库www.lingala.net/zip4j /
将此jar文件添加到app的lib文件夹中。
检查您的导入
import net.lingala.zip4j.core.ZipFile;
import net.lingala.zip4j.exception.ZipException;
import net.lingala.zip4j.model.FileHeader;
使用以下方法
解压缩( “/ SD卡/ file.zip”, “/ SD卡/ unzipFolder”)
public static void unzip(String Filepath, String DestinationFolderPath) {
try {
ZipFile zipFile = new ZipFile(Filepath);
List fileHeaders = zipFile.getFileHeaders();
for(FileHeader fileHeader : fileHeaders) {
String fileName = fileHeader.getFileName();
if (fileName.contains("\\")) {
fileName=fileName.replace("\\","\\\\");
String[] Folders=fileName.split("\\\\");
StringBuilder newFilepath=new StringBuilder();
newFilepath.append(DestinationFolderPath);
for (int i=0;i< Folders.length-1;i++){
newFilepath.append(File.separator);
newFilepath.append(Folders[i]);
}
zipFile.extractFile(fileHeader, newFilepath.toString(),null,Folders[Folders.length-1]);
}else {
zipFile.extractFile(fileHeader,DestinationFolderPath);
}
}
} catch (ZipException e) {
e.printStackTrace();
}
}
答案 3 :(得分:0)
使用 const MainNavigator = createStackNavigator(
{
Home: HomeScreen,
Categories: CategoriesScreen,
Recipe: RecipeScreen,
RecipesList: RecipesListScreen,
Ingredient: IngredientScreen,
Search: SearchScreen,
IngredientsDetails: IngredientsDetailsScreen,
Orders:OrdersScreen,
SingleOrder:SingleOrder,
OrderDetails:OrderDetailsScreen,
Register:RegisterScreen,
Edit:EditScreen,
EditScreenSingle:SingleProductEditScreen
},
{
initialRouteName: 'Home',
//headerMode: 'float',
defaulfNavigationOptions: ({ navigation }) => ({
headerTitleStyle: {
fontWeight: 'bold',
textAlign: 'center',
flex: 1,
},
})
}
);
const DrawerStack = createDrawerNavigator(
{
Main: MainNavigator
},
{
drawerPosition: 'left',
initialRouteName: 'Main',
drawerWidth: 250,
contentComponent: DrawerContainer
}
);
export const AppContainer = createAppContainer(DrawerStack);
方法提取文件和目录。如果要手动执行此操作,请尝试以下操作:
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.6.1/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.6.1/umd/react-dom.production.min.js"></script>