java 无需在java中提取即可读取Zip文件内容
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Read Zip file content without extracting in java
提问by TNN
I have byte[] zipFileAsByteArray
我有字节 [] zipFileAsByteArray
This zip file has rootDir --|
| --- Folder1 - first.txt
| --- Folder2 - second.txt
| --- PictureFolder - image.png
What I need is to get two txt files and read them, without saving any files on disk. Just do it in memory.
我需要的是获取两个 txt 文件并读取它们，而不在磁盘上保存任何文件。只在记忆中做。
I tried something like this:
我试过这样的事情：
ByteArrayInputStream bis = new ByteArrayInputStream(processZip);
ZipInputStream zis = new ZipInputStream(bis);
Also I will need to have separate method go get picture. Something like this:
另外我需要有单独的方法去获取图片。像这样的东西：
public byte[]image getImage(byte[] zipContent);
Can someone help me with idea or good example how to do that ?
有人可以帮助我提出想法或很好的例子如何做到这一点吗？
采纳答案by dambros
Here is an example:
下面是一个例子：
public static void main(String[] args) throws IOException {
ZipFile zip = new ZipFile("C:\Users\mofh\Desktop\test.zip");
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry entry = (ZipEntry) e.nextElement();
if (!entry.isDirectory()) {
if (FilenameUtils.getExtension(entry.getName()).equals("png")) {
byte[] image = getImage(zip.getInputStream(entry));
//do your thing
} else if (FilenameUtils.getExtension(entry.getName()).equals("txt")) {
StringBuilder out = getTxtFiles(zip.getInputStream(entry));
//do your thing
}
}
}
}
private static StringBuilder getTxtFiles(InputStream in) {
StringBuilder out = new StringBuilder();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line;
try {
while ((line = reader.readLine()) != null) {
out.append(line);
}
} catch (IOException e) {
// do something, probably not a text file
e.printStackTrace();
}
return out;
}
private static byte[] getImage(InputStream in) {
try {
BufferedImage image = ImageIO.read(in); //just checking if the InputStream belongs in fact to an image
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ImageIO.write(image, "png", baos);
return baos.toByteArray();
} catch (IOException e) {
// do something, it is not a image
e.printStackTrace();
}
return null;
}
Keep in mind though I am checking a string to diferentiate the possible types and this is error prone. Nothing stops me from sending another type of file with an expected extension.
请记住，尽管我正在检查一个字符串以区分可能的类型，但这很容易出错。没有什么能阻止我发送具有预期扩展名的另一种类型的文件。
回答by dryairship
You can do something like:
您可以执行以下操作：
public static void main(String args[]) throws Exception
{
//bis, zis as you have
try{
ZipEntry file;
while((file = zis.getNextEntry())!=null) // get next file and continue only if file is not null
{
byte b[] = new byte[(int)file.getSize()]; // create array to read.
zis.read(b); // read bytes in b
if(file.getName().endsWith(".txt")){
// read files. You have data in `b`
}else if(file.getName().endsWith(".png")){
// process image
}
}
}
finally{
zis.close();
}
}
