如何从请求中获取excel文件？
如何从请求中获取excel文件？
提问于 2012-03-14 17:18:28
我从一个jsp页面上传一个excel文件。下面是代码。
<form action="Upload.jsp" enctype="MULTIPART/FORM-DATA" method=post >
<input type="file" name="filename" />
<input type="submit" value="Upload" />
</form> 但是如何在下一页(Upload.jsp)中获得excel文件呢？我正在使用，但在第二行得到了错误。
InputStream file = request.getInputStream();
POIFSFileSystem myFileSystem = new POIFSFileSystem(file );那么如何从请求中获取excel文件呢？
回答 3
Stack Overflow用户
发布于 2012-03-14 17:35:35
您将根据请求获得一个Multipart/form-data，您需要从中提取包含您的文件字节的部分。
最简单的方法是使用Apache Commons Fileupload
http://commons.apache.org/fileupload/
Stack Overflow用户
发布于 2012-03-14 18:31:04
创建FileUploader
import org.apache.commons.fileupload.disk.*;导入org.apache.commons.fileupload.servlet。；导入java.io。
公共类FileUploader {私有静态ServletFileUpload上传器；
private FileUploader()
{
}
public static synchronized ServletFileUpload getservletFileUploader(String tempDir, int maxSizeInMB)
{
if(uploader == null)
{
DiskFileItemFactory factory = new DiskFileItemFactory();
factory.setSizeThreshold(1024 * 1024);
factory.setRepository(new File(tempDir));
uploader = new ServletFileUpload(factory);
uploader.setFileSizeMax(maxSizeInMB * 1024 * 1024);
}
return uploader;
}}
然后在处理请求时使用它
protected MultiPartFormData handleMultiPartRequest(HttpServletRequest request)
throws FileSizeLimitExceededException
{
if(!isMultipartRequest(request))
return null;
ServletFileUpload upload = FileUploader.getservletFileUploader(tempDir, 50);
MultiPartFormData data = new MultiPartFormData();
try
{
List<FileItem> items = upload.parseRequest(request);
for (FileItem item : items)
{
if(item.isFormField())
{
data.getParameters().put(item.getFieldName(), item.getString());
}
else
{
String filename = item.getName();
//Internet explorer and firefox will send the file name differently
//Internet explorer will send the entire path to the file name including
//the backslash characters etc ... we should strip it down
//THIS IS HACKY
if(filename.indexOf("\\") != -1)
{
int index = filename.lastIndexOf("\\");
filename = filename.substring(index + 1);
}
if(filename == null || filename.equals(""))
{
//do nothing
}
else
{
String randomFileName = (new RandomGUID()).toString() + getFileExtension(filename);
File uploadFile = new File(uploadDir + File.separator + randomFileName);
item.write(uploadFile);
}
}
}
}
catch(FileSizeLimitExceededException e)
{
throw e;
}
catch(Exception e)
{
e.printStackTrace();
}
return data;
}供你参考。MultiPartForm数据如下所示
import java.util.Hashtable;导入java.util.ArrayList；
公共类MultiPartFormData {
private ArrayList<Integer> fids;
private Hashtable<String, String> parameters;
public MultiPartFormData()
{
this.fids = new ArrayList<Integer>();
this.parameters = new Hashtable<String, String>();
}
public ArrayList<Integer> getFids() {
return fids;
}
public void setFids(ArrayList<Integer> fids) {
this.fids = fids;
}
public Hashtable<String, String> getParameters() {
return parameters;
}
public void setParameters(Hashtable<String, String> parameters) {
this.parameters = parameters;
}}
Stack Overflow用户
发布于 2012-03-15 16:15:51
好了，伙计们，谢谢你们的回复。但是我已经用下面的流程解决了这个问题。
JSP内部：
<form action="/upload.do" enctype="MULTIPART/FORM-DATA" method=post >
<input type="file" name="file" id="file" size=25/>
<input type="submit" value="Upload" />
</form>创建了一个表单bean：
private FormFile file;
public void setFile(FormFile file) {
this.file = file;
}
public FormFile getFile() {
return file;
}在操作类上载中：
FileUploadForm uploadForm = (FileUploadForm) form;
FormFile file = uploadForm.getFile();
InputStream stream = file.getInputStream();
POIFSFileSystem fsFileSystem = new POIFSFileSystem(stream);
//
rest of code for reading the excel
//现在它工作得很好。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/9698928
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