尝试从 String 转换为 PNG 时由 ZipException 引起的 Java IIOException

Question

我的 Java Servlet 有一个来自客户端的字符串(我所知道的是，他实际上向我发送了一个 PNG 作为请求中的参数，其中 contentType = application/x-www-form-urlencoded)，我正在尝试将其转换回文件并保存。这是我的代码(基于 this ):

byte[] data = org.apache.commons.codec.binary.Base64.decodeBase64(request.getParameter("image").getBytes());
String filename = request.getParameter("filename");

try{
    java.awt.image.BufferedImage image = javax.imageio.ImageIO.read(new ByteArrayInputStream(data));
    javax.imageio.ImageIO.write(image, "png", new File(getUploadFolder() + filename + ".png"));
}catch(Exception e){
    System.out.println("Exception saving image: " + e.getMessage());
    e.printStackTrace();
}

这是我的错误堆栈跟踪:

Exception saving image: Error reading PNG image data
javax.imageio.IIOException: Error reading PNG image data
    at com.sun.imageio.plugins.png.PNGImageReader.readImage(PNGImageReader.java:1308)
    at com.sun.imageio.plugins.png.PNGImageReader.read(PNGImageReader.java:1577)
    at javax.imageio.ImageIO.read(ImageIO.java:1448)
    at javax.imageio.ImageIO.read(ImageIO.java:1352)
    at com.mypackage.servlet.UploadServlet.doPost(UploadServlet.java:113)
Caused by: java.util.zip.ZipException: invalid code lengths set
    at java.util.zip.InflaterInputStream.read(InflaterInputStream.java:164)
    at java.io.BufferedInputStream.fill(BufferedInputStream.java:235)
    at java.io.BufferedInputStream.read(BufferedInputStream.java:254)
    at java.io.FilterInputStream.read(FilterInputStream.java:83)
    at com.sun.imageio.plugins.png.PNGImageReader.decodePass(PNGImageReader.java:1085)
    at com.sun.imageio.plugins.png.PNGImageReader.decodeImage(PNGImageReader.java:1196)
    at com.sun.imageio.plugins.png.PNGImageReader.readImage(PNGImageReader.java:1301)
    ... 4 more

查看InflaterInputStream类代码，我发现它实际上是一个DataFormatException:

public int read(byte[] b, int off, int len) throws IOException {
    ensureOpen();
    if (b == null) {
        throw new NullPointerException();
    } else if (off < 0 || len < 0 || len > b.length - off) {
        throw new IndexOutOfBoundsException();
    } else if (len == 0) {
        return 0;
    }
    try {
        int n;
        while ((n = inf.inflate(b, off, len)) == 0) {
            if (inf.finished() || inf.needsDictionary()) {
                reachEOF = true;
                return -1;
            }
            if (inf.needsInput()) {
                fill();
            }
        }
        return n;
    } catch (DataFormatException e) {
        String s = e.getMessage();
        throw new ZipException(s != null ? s : "Invalid ZLIB data format");
    }
}

但这并不是真正那么有启发性。

Answer 1

正如 @leonbloy 所说，无需解码然后重新编码数据。

如果您确定 PNG 文件是好的*，您可以简单地将数据写入文件，如下所示，您应该完成。

byte[] data = org.apache.commons.codec.binary.Base64.decodeBase64(request.getParameter("image").getBytes());
String filename = request.getParameter("filename");

OutputStream output = new FileOutputStream(new File(getUploadFolder() + filename + ".png"));

try {
    output.write(data);
}
finally {
    output.close();
}

*) 到目前为止提供的证据表明情况并非如此，并且您使用的代码应该可以工作(即使这浪费了 CPU 周期)。看到产生您所看到的异常的数据样本，以找出异常的原因，这将是非常有趣的。