创建具有特定扩展名的临时文件并在VS Code中打开

Question

I'm trying to pipe some output on bash stdout into a new temporary file and open this file using VS Code.

The file content might not always be plain text but also could be yaml and I want to create the temp file with yaml extension so VS Code uses the correct language mode.

I tried this but I end up with a file having the txt in VS Code.

cat yaml > $(mktemp $TMPDIR$(uuidgen).yaml) | code -

cat yaml is just an example for a command which might write yaml content to stdout .

mktemp --suffix=.yaml would be the way to go but this isn't supported on macOS.

Answer 1

I found a solution:

temp=$(mktemp $TMPDIR.XXXXXXXXXXXX.$(uuidgen).yaml) && ls -la > $temp && code $temp

Of course there might be better ones and I'm looking forward to it.

Answer 2

You can do it this way:

TMPDIR='/tmp' # Give a value to TMPDIR for the purpose of this answer
mktemp "${TMPDIR}/XXXXXXXXXXXX_$(uuidgen).yaml" |
  xargs -n 1 -I {} -- \
    code --file-uri "file://{}"

• mktemp "${TMPDIR}/XXXXXXXXXXXX_$(uuidgen).yaml" :
  Create the temporary file into $TMPDIR with name composed of:
  • 12 random alphanumeric mixed case letters
    replacing the upper_case X ,
  • followed by underscore _ ,
  • followed by a random UUID generated by the uuidgen command,
  • and suffixed by the .yaml extension.
• | xargs -n 1 -I {} | xargs -n 1 -I {} :
  Pass the newly created temporary file name to xargs , which will convert it into an argument:
  • -n 1 : One argument for each command
  • -I {} : Replace placeholder string {} with the argument
  • -- : End of the xargs options.
• code --file-uri "file://{}" :
  code is invoked by xargs .
  • --file-uri <uri> Opens a window with given file uri(s)
  • "file://{}" : The uri for the temporary file just created.
    xargs will replace {} by the temporary file path.