file upload with original uploaded file name

sumanadak · 2016-07-10T02:42:19+00:00

Hash fragments are never passed to a web server by a web browser, you will not be able to get the "#random=data" part of that url.I didn't know...

file upload with original uploaded file name

sumanadak

Dear all,

I tried to create a simple file upload api for my mobile client. Mobile client will post some zip file to server. server will save it disk( may be offload to other server for processing). I also wanted to keep the original uploaded file name while I am saving the file to disk.

I tried something like this. Not able to understand, how I would get the original uploaded file name.

Also would be interested to know other way to upload file. File size is very small like 5kb. But there might be one million mobile client request which will be sending data and accepted request rate will be nearly 500 to 1k/sec

content_by_lua_block {

local file

local upload_path = '/home/test/uploads/'

ngx.req.read_body()

local data = "">

local source_file_name = ngx.req.get_body_file() -- it gives me only temp file path with some 000001 name not original file name.

local file_name = ??????? [how to get the original uploaded file name]

if file_name then

file = io.open(file_name, 'w+')

if not file then

ngx.say("failed to open file ", file_name)

return

end

file:write(data)

file:close()

ngx.say("file uploaded")

}

rpaprocki

Hi,

On Jul 9, 2016, at 11:42, Suman Adak <suma...@gmail.com> wrote:

Dear all,

I tried to create a simple file upload api for my mobile client. Mobile client will post some zip file to server. server will save it disk( may be offload to other server for processing). I also wanted to keep the original uploaded file name while I am saving the file to disk.
I tried something like this. Not able to understand, how I would get the original uploaded file name.

Also would be interested to know other way to upload file. File size is very small like 5kb. But there might be one million mobile client request which will be sending data and accepted request rate will be nearly 500 to 1k/sec

If you're reading from the buffered content off the disk, you would have to examine the file contents and examine the multipart headers. Have a look at lua-resty-upload (https://github.com/openresty/lua-resty-upload), which lets you stream multipart uploads and examine each part of the request body, including the name.

sumanadak

Hi,

thanks for reply, I tried to parse header and got it but it seems bit complex or error prone to get filename from header. can we write more easier way or am i doing wrong?

local file_header = res[2]

local i, j = string.find(file_header,"filename=")

if j then

local file_name = string.sub(file_header,j+2,string.len(file_header)-1)

ngx.log(ngx.INFO,"file name is = ",file_name)

end

Along with that I also need to send uploaded file to rabbitMQ. I am not able to figure out how I would send file to rabbitMQ. Please help me.

On Sunday, July 10, 2016 at 1:09:19 AM UTC+5:30, rpaprocki wrote:

On Jul 9, 2016, at 11:42, Suman Adak <suma...@gmail.com> wrote:

Dear all,

I tried to create a simple file upload api for my mobile client. Mobile client will post some zip file to server. server will save it disk( may be offload to other server for processing). I also wanted to keep the original uploaded file name while I am saving the file to disk.
I tried something like this. Not able to understand, how I would get the original uploaded file name.

Also would be interested to know other way to upload file. File size is very small like 5kb. But there might be one million mobile client request which will be sending data and accepted request rate will be nearly 500 to 1k/sec

If you're reading from the buffered content off the disk, you would have to examine the file contents and examine the multipart headers. Have a look at lua-resty-upload (https://github.com/openresty/lua-resty-upload), which lets you stream multipart uploads and examine each part of the request body, including the name.

rpaprocki

Hi,

On Mon, Jul 11, 2016 at 9:11 AM, Suman Adak <suma...@gmail.com> wrote:

thanks for reply, I tried to parse header and got it but it seems bit complex or error prone to get filename from header. can we write more easier way or am i doing wrong?

local file_header = res[2]
local i, j = string.find(file_header,"filename=")
if j then
local file_name = string.sub(file_header,j+2,string.len(file_header)-1)
ngx.log(ngx.INFO,"file name is = ",file_name)
end

Can you please post a full, minimal configuration/code example so we can try to diagnose the problem? Small snippets like these are not useful to a mailing list who has no idea the context of what you're doing.

Along with that I also need to send uploaded file to rabbitMQ. I am not able to figure out how I would send file to rabbitMQ. Please help me.

Have a look at https://github.com/wingify/lua-resty-rabbitmqstomp. I suggest you read through the documentation and contact the author if you have trouble with this module.

sumanadak

dear rpaprocki,

Sorry for that, here is my full nginx conf file

worker_processes 1;

error_log logs/error.log info;

events {

worker_connections 1024;

}

http {

server {

listen 8083;

location /upload {

default_type multipart/form-data;

content_by_lua_block {

local resty_md5 = require "resty.md5"

local upload = require "resty.upload"

local cjson = require "cjson"

local chunk_size = 4096

local form = upload:new(chunk_size)

if not form then

ngx.log(ngx.ERR, "failed to new upload: ", err)

ngx.exit(500)

end

local md5 = resty_md5:new()

local file

local upload_path = '/home/suman/uploads/'

while true do

local typ, res, err = form:read()

if not typ then

ngx.say("failed to read: ", err)

return

end

if typ == "header" then

-- local file_name = upload_path .. ngx.time()

local file_header = res[2]

local i, j = string.find(file_header,"filename=")

if j then

local file_name = string.sub(file_header,j+2,string.len(file_header)-1)

ngx.log(ngx.INFO,"file name is = ",file_name)

file_name = upload_path..ngx.time()..'.'..file_name

if file_name then

file = io.open(file_name, 'w+')

if not file then

ngx.say("failed to open file ", file_name)

return

end

elseif typ == "body" then

if file then

file:write(res)

md5:update(res)

end

elseif typ == "part_end" then

file:close()

file = nil

startUpload = false

elseif typ == "eof" then

break

else

-- nothingman

end

local digest = md5:final()

md5:reset()

local str = require "resty.string"

ngx.say("md5: ", str.to_hex(digest))

}

thanks

suman

On Monday, July 11, 2016 at 9:45:06 PM UTC+5:30, rpaprocki wrote:

Hi,

On Mon, Jul 11, 2016 at 9:11 AM, Suman Adak <suma...@gmail.com> wrote:

thanks for reply, I tried to parse header and got it but it seems bit complex or error prone to get filename from header. can we write more easier way or am i doing wrong?

local file_header = res[2]
local i, j = string.find(file_header,"filename=")
if j then
local file_name = string.sub(file_header,j+2,string.len(file_header)-1)
ngx.log(ngx.INFO,"file name is = ",file_name)
end

Can you please post a full, minimal configuration/code example so we can try to diagnose the problem? Small snippets like these are not useful to a mailing list who has no idea the context of what you're doing.

Along with that I also need to send uploaded file to rabbitMQ. I am not able to figure out how I would send file to rabbitMQ. Please help me.

Have a look at https://github.com/wingify/lua-resty-rabbitmqstomp. I suggest you read through the documentation and contact the author if you have trouble with this module.