It's well known how to pipe the standard ouput of a process into another processes standard input:
proc1 | proc2
But what if I want to send the standard error of proc1 to proc2 and leave the standard output going to its current location? You would think bash would have a command along the lines of:
proc1 2| proc2
But, alas, no. Is there any way to do this?
There is also process substitution. Which makes a process substitute for a file.
You can send stderr to a file as follows:
process1 2> file
But you can substitute a process for the file as follows:
process1 2> >(process2)
Here is a concrete example that sends stderr to both the screen and appends to a logfile
sh myscript 2> >(tee -a errlog)
You can use the following trick to swap stdout and stderr. Then you just use the regular pipe functionality.
( proc1 3>&1 1>&2- 2>&3- ) | proc2
Provided stdout and stderr both pointed to the same place at the start, this will give you what you need.
What the x>y bit does is to change file handle x so it now sends its information to where file handle y currently points. For our specific case:
3>&1 creates a new handle 3 which will output to the current handle 1 (original stdout), just to save it somewhere for the final bullet point below.1>&2 modifies handle 1 (stdout) to output to the current handle 2 (original stderr).2>&3- modifies handle 2 (stderr) to output to the current handle 3 (original stdout) then closes handle 3 (via the - at the end).It's effectively the swap command you see in sorting algorithms:
temp = value1;
value1 = value2;
value2 = temp;