How to enter data from excel sheet to mysql database using JSP? JSP + Servlet + MySQL Project

Question

So, basically what I want to do is, on a jsp page give option to browse and select an excel file. Then when the user clicks on Upload button, the data from excel is saved into database table.

I am able to insert into database table without the UI i.e., when there is only servlet and database. In that case, I am doing this:

    InputStream fis = new FileInputStream(new File("C:\\Users\\RAJYAWARDHAN\\Desktop\\Book1.xlsx"));

And after this I use fis to one by one enter data into Database table using Apache POI. But when JSP page is included I am using form:

<input type="file" name="file_uploaded" />

So, when not using JSP page, I have the address of file, that is on my machine only. But that's for test purpose. When user will upload, then I will not have the address of the file because file is on users machine. Hence, I cannot use new File("address_of_file") as file is not on my machine(server basically).

Also, I don't want to first upload file from users machine to my machine and then pick it up from that particular path. I just want to import data from the selected file by user and not actually upload and save that file.

What should I do? Please explain in simple terms. Thank You.

Please ask if question is not clear. English is not my first language.

Please be aware that JSP is a server side process (the same as a servlet). It sounds like you will have to POST, the file to a servlet and then in that servlet process the xls and insert into the DB — Scary Wombat
– Scary Wombat, Commented Mar 6, 2018 at 2:53
@ScaryWombat Yes I am doing it in POST only. Should I upload the whole code? — rsp
– rsp, Commented Mar 6, 2018 at 12:34

Lakmi · Accepted Answer · 2019-06-24 06:25:00Z

1

I think what you want to use is getPart method in servlets.

Just like request.getParameter you can use request.getPart("file_uploaded"). It gets file for the given name from front end(jsp/HTML).

This returns Part object which can be converted to InputStream. You don't need to save file to some location in this case.

Do not forget to forget to set enctype= multipart/form-data in your form.

edited Jun 24, 2019 at 6:25

Lakmi

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answered Jun 24, 2019 at 5:47

Vagisha Sinha

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