Question

使用org.apache.wicket.util.resource.IResourceStream我在服务器和输出流中创建了zip文件，编写了相同的zip文件。但它引发了以下错误，我正在打破我的脑袋。有什么建议吗？

Timestamp: 7/25/2012 3:13:28 PM
Error: not well-formed
Source File: XXX
Line: 1, Column: 3
Source Code:
PK

Java代码

        AjaxButton one = new AjaxButton("one"){     
                @Override
                public void onSubmit(AjaxRequestTarget target,Form form) {
                    // TODO Auto-generated method stub
                    try {

                        {
                        IResourceStream iResourceStream  = null;
                        iResourceStream = new AbstractResourceStreamWriter(){

                            @Override
                            public String getContentType() {
                                // TODO Auto-generated method stub
                                return "application/zip";
                            }

                            @Override
                            public void write(OutputStream output) {
                                // TODO Auto-generated method stub
                                File tmpFile = null;
                                String batchFileName = "batch_"+dateFormat.format(new Date())".zip";
                                File zipFile = new File(batchFileName);
                                FileOutputStream zipFileOutputStream = null;
                                ZipOutputStream zipOutputStream = null;
                                try
                                {
                                    zipFileOutputStream = new FileOutputStream(zipFile);
                                    zipOutputStream = 
                                        new ZipOutputStream(zipFileOutputStream);    
                                    zipOutputStream.setLevel(Deflater.DEFAULT_COMPRESSION);
                                    for(XXX)
                                    {
// create tmpFile zip file here                                 
                                                    ZipEntry zipAdd = new ZipEntry(tmpFile.getName());  
                                                    System.out
                                                            .println(tmpFile.getName());
                                                    zipOutputStream.putNextEntry(zipAdd);
                                                    zipOutputStream.write(IOUtils.toByteArray(new FileInputStream(tmpFile)));
                                                    zipOutputStream.closeEntry();
                                                }
                                            }
                                        }
                                    }

                                }
                                catch (Exception e) {
                                    // TODO: handle exception
                                    e.printStackTrace();
                                }
                                finally
                                {
                                    if(zipOutputStream != null){
                                        try {
                                            zipOutputStream.flush();
                                            zipOutputStream.close();
                                        } catch (IOException e) {                   
                                            e.printStackTrace();
                                        }
                                    }

                                    if(zipFileOutputStream != null){
                                        try {
                                            zipFileOutputStream.flush();
                                            zipFileOutputStream.close();
                                        } catch (IOException e) {                   
                                            e.printStackTrace();
                                        }
                                    }
                                    try
                                    {
                                        {
                                            InputStream in = new FileInputStream(zipFile);
                                            byte[] buf = new byte[1024];
                                            int len;
                                            while ((len = in.read(buf)) > 0){
                                                output.write(buf, 0, len);
                                            }
                                            in.close();
                                            output.close();
                                        }
                                    }
                                    catch (Exception e) {
                                        // TODO: handle exception
                                        e.printStackTrace();
                                    }
                                }


                            }

                        };

                        getRequestCycle()
                        .setRequestTarget(new ResourceStreamRequestTarget(iResourceStream)
                        .setFileName("batch.zip"));


                    } else {

                    }

                } catch (Exception e)
{
}
}

Answer 1

您正在执行AJAX请求，但不是发回XML（浏览器期望的JS代码），而是发送二进制数据。这就是为什么你得到“格式不正确”的错误 - 它不是格式良好的xml。

有两种方法可以完成这项工作。一种是简单地不进行AJAX提交并使用常规Button而不是AjaxButton。我推荐这个。

如果您需要执行其他一些AJAX工作（更新面板或类似内容），然后想要提供下载，请查看：https://cwiki.apache.org/WICKET/ajax-update-and-file-download-in-one-blow.html

org.apache.wicket.util.resource.IResourceStream抛出javascript错误。不能下载文件

1 个答案: