是否可以创建一个任何进程都可以使用的“文件”，而无需实际将文件保存在磁盘上？

Question

假设我做了一个grep给我很多结果的。我想使用 grep 的输出并将其传递给一个将文件名作为第一个参数的命令。例如 -

myCommand filename.txt

我想做这样的事情 -

grep "xyz" | myCommand

However, since myCommand doesn't read from stdin, this piping does not work. I don't want to create a temporary file on disk with the results of grep. Is it possible by any other means to create a "file", say in memory and let myCommand use that?

Answer 1

A simple solution is to use Bash's support for process substitution, like this:

myCommand <(grep "xyz" somefile)

This will connect the output of the grep command to a file descriptor and then pass /dev/fd/NNN (where NNN is the fd number) to your command as the input file (for example, your command might actually get run as myCommand /dev/fd/63). This solution does not create a temporary file.

You can also create a named pipe and use that to transfer data around:

$ mkfifo /tmp/myfifo
$ grep "xyz somefile > /tmp/myfifo &
$ myCommand /tmp/myfifo

But really, the first solution is simpler.

Stephane Chazelas comments:

zsh 还有 myCommand =(grep xyz somefile) 它使用一个普通的临时文件（在 myCommand 不能接受管道的情况下）。

这很重要，因为某些命令希望能够对其输入文件执行随机访问，这不适用于管道。使用此语法，您可以获得临时文件的好处，而无需担心自己清理它们。