[SOLVED] Explain what this bash script does

Question

bash

Good evening, hackers!

Some formatted text is passed by the daemon via STDIN to the sh script:

#!/bin/sh
     while read x y
     do  case "$x$y" in
             '') exec logger -t collectd${severity+" $severity" -p user."$severity"};;
             Severity:WARNING) severity=warning;;
             Severity:OKAY)    severity=notice;;
             Severity:FAILURE) severity=err;;
         esac
     done

Sample text:

Severity: FAILURE
  Time: 1200928930
  Host: myhost.mydomain.org
  \n
  This is a test notification to demonstrate the format

But I can't figure out what's going on.
I see a cycle ... and then it's not clear.

Please help with the following questions:
What happens (step by step) when executing the script?
How can I pass arbitrary text to this script via STDIN?
How to modify the script so that this arbitrary text is written to a file and not to syslog?

Thanks in advance.

UPD. Thanks for the help, but one point remained unclear - how to redirect all incoming text to the

UPD2 file. Gotta do it like this

#!/bin/sh
read severity
read time
echo $severity >> file
echo $time >> file

Answer 1

read xy - reads a line, divides it by spaces, as the shell divides command line arguments into separate arguments. The first and second blocks are assigned to the x and y variables, respectively.
Next, the read line is parsed. If the combination Severity:WARNING (OK, FAILURE) is read, then the corresponding value is written to the severity variable. If the string is empty, then logger is started with the required parameters. The rest of the lines are ignored.

Answer 2

Line by line reading. The variable $x gets the first word, $y the second.
Next, the $severity variable is set to a value depending on $x and $y.
When hitting an empty line, a log entry is made.
To write to a file, change line 4 to something like
'') echo "severity: $severity" > test.log;;