stdin into zip command, how do I specify a file name? [duplicate]

Answer 1

Write echo output to the desired filename, then use the -m flag to zip and remove the original file in one step.

echo 'foo bar' > filename.txt && zip -m filename.txt.zip filename.txt

Answer 2

Use a fifo (a named pipe) instead of piping stdin directly!

mkfifo text.txt               # create fifo
echo 'foo bar' > text.txt &   # pipe data to fifo, in background
zip --fifo file.zip text.txt  # note the `--fifo` argument to zip
rm text.txt                   # cleanup

Result:

$ unzip -l file.zip
Archive:  file.zip
Length      Date    Time    Name
---------  ---------- -----   ----
8          2020-11-29 12:30   text.txt
---------                     -------
8                             1 file

Note that the fifo is a pipe, no data is stored on your hard drive, it is streamed just like "|" (an anonymous pipe).

Answer 3

What you can do is pipe the file in normally, then rename it in the zip

echo 'foo bar' | zip  > file.zip
printf "@ -\n@=filename.txt\n" | zipnote -w file.zip

Answer 4

Whenever stdout needed to be used as a command argument, xargs is the answer. In this sample the output comes as last argument to xargs.

cat filelist.txt | xargs zip files.zip

Supposing filelist.txt a source of file names:

echo dir1/file1.cfg         > filelist.txt
echo dir2/file2.log        >> filelist.txt
echo dir2/dir3/file3.txt   >> filelist.txt

Answer 5

echo 'foo bar' | zip -@ /path/to/zipfile.zip should do it while keeping filename integrity