Can we create FileItem
object from given file name and file location using Java?
If we use commons-fileupload jar we can upload file using jsp/html and in servlet on parsing request we will get List<FileItem>
. But I want to do file upload using plain java where I want to create FileItem
object manually (I don't want to use byte[]
array to store the file). So is there any way to create FileItem
object manually?
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icza
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CodeName007
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This guys answer was complete. https://stackoverflow.com/questions/4120635/java-lang-nullpointerexception-while-creating-diskfileitem – Mike Croteau Dec 21 '20 at 02:49
1 Answers
1
Yes, you can.
Use the DiskFileItemFactory like this:
DiskFileItemFactory factory = new DiskFileItemFactory();
FileItem fi = factory.createItem("formFieldName", "application/zip", false,
"/var/temp/somefile.zip");
Obviously use content type and other parameters appropriate to your case.

icza
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DiskFileItemFactory factory = new DiskFileItemFactory(); FileItem fi = factory.createItem("formFieldName", "application/zip", false,pathToLoad); File fileLoc=new File(pathToLoad); factory.setRepository(fileLoc); try { File file=new File file=new File(pathToLoad+File.separator+fileName); fi.write(file); – CodeName007 Jul 16 '14 at 10:20
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4java.lang.NullPointerException at org.apache.commons.fileupload.disk.DiskFileItem.isInMemory(DiskFileItem.java:275) at org.apache.commons.fileupload.disk.DiskFileItem.write(DiskFileItem.java:391) – CodeName007 Jul 16 '14 at 10:21
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