我能够生成pdf并将其刷新到浏览器。但是,现在我的要求发生了变化。我需要生成多个pdf并将它们保存在一个zip文件中并将其刷新到浏览器。我遵循了http://www.avajava.com/tutorials/lessons/how-can-i-create-a-zip-file-from-a-set-of-files.html
但是找不到如何集成我的代码。这是我的代码。任何想法都将不胜感激。
for(int i = 0; i < 5 ; i++) {
byte[] documentBytes = TSService.generateDocument(dealKey, i);
String documentType = TSUtil.getDocumentType(i);
response.setHeader("Content-Disposition", "attachment;filename="+documentType);
response.setContentType("application/pdf");
response.setHeader("Expires", "0");
response.setHeader("Cache-Control", "must-revalidate, postcheck=0, pre-check=0");
response.setHeader("Pragma", "public");
response.setContentLength(documentBytes.length);
ServletOutputStream out = response.getOutputStream();
out.write(documentBytes);
out.flush();
out.close();
}
最初我只有循环代码。现在,我想基于i值生成5个报告。
更新了Alex
的代码String documentType = TSUtil.getDocumentType(Integer.valueOf(documentKey));
response.setHeader("Content-Disposition", "attachment;filename=dd.zip");
response.setContentType("application/zip");
response.setHeader("Expires", "0");
response.setHeader("Cache-Control", "must-revalidate, postcheck=0, pre-check=0");
response.setHeader("Pragma", "public");
ServletOutputStream out = response.getOutputStream();
ZipOutputStream zout = new ZipOutputStream(out);
for(int i = 1; i <= 5 ; i++) {
byte[] documentBytes = TSService.generateDocument(dealKey, i);
ZipEntry zip = new ZipEntry(i+".pdf");
zout.putNextEntry(zip);
zout.write(documentBytes);
zout.closeEntry();
}
zout.close();
答案 0 :(得分:2)
下面的代码应该可以工作,可以直接下载而无需创建任何临时文件。所有这些都是在运行中创建的,并且都在内存中。
response.setHeader("Content-Disposition", "attachment;filename="+***Your zip filename***);
response.setContentType("application/zip");
response.setHeader("Expires", "0");
response.setHeader("Cache-Control", "must-revalidate, postcheck=0, pre-check=0");
response.setHeader("Pragma", "public");
ServletOutputStream out = response.getOutputStream();
ZipOutputStream zout = new ZipOutputStream(out);
for(int i = 0; i < 5 ; i++) {
byte[] documentBytes = TSService.generateDocument(dealKey, i);
ZipEntry zip = new ZipEntry(***your pdf name***);
zout.putNextEntry(zip);
zout.write(documentBytes);
zout.closeEntry();
}
zout.close();
<强>已更新强>
我刚试过以下代码而没有问题。可以创建一个包含5个文本文件的新zip文件。所以我不知道你为什么会得到例外。
FileOutputStream out = new FileOutputStream("abc.zip");
ZipOutputStream zout = new ZipOutputStream(out);
for(int i = 0; i < 5 ; i++) {
byte[] documentBytes = "12345".getBytes();
ZipEntry zip = new ZipEntry(i+".txt");
zout.putNextEntry(zip);
zout.write(documentBytes);
zout.closeEntry();
}
zout.close();
答案 1 :(得分:1)
我用xml文件完成了相同的任务。我是我的代码
public String createZipFromXmlFile(List<String> filePath) {
String date = new SimpleDateFormat("yyyMMddHHmmssSS").format(new Date());
String fName = "zipDownload" + date + ".zip";
System.out.println(" File Path.................."+filePath);
String fileName = filePath.get(0).substring(0, filePath.get(0).indexOf("xml")) + "//" + fName;
try {
FileOutputStream fout = new FileOutputStream(fileName);
ZipOutputStream zout = new ZipOutputStream(fout);
for (String fnm : filePath) {
FileInputStream fin = new FileInputStream(new File(fnm));
ZipEntry zip = new ZipEntry(fnm.substring(fnm.indexOf("xml")));
zout.putNextEntry(zip);
byte[] bytes = new byte[1024];
int length;
while ((length = fin.read(bytes)) >= 0) {
zout.write(bytes, 0, length);
}
zout.closeEntry();
fin.close();
}
zout.close();
fout.close();
} catch (Exception e) {
}
return fileName;
}
其中filePath是包含xml文件路径的列表。