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Given a log file loga.log with timestamps of the format DD/MM/YYYY:hh:mm:ss, which command filters out all log entries in the time period between 8:00 am and 8:59 am?
Correct Answer: E
Explanation The command that filters out all log entries in the time period between 8:00 am and 8:59 am is grep -E ':08:[0-9]+:[0-9]+' loga.log. The grep command is used to search for a pattern in a file or standard input and print the matching lines. The -E or --extended-regexp option enables the use of extended regular expressions, which support more operators and syntax than the basic regular expressions. The pattern ':08:[0-9]+:[0-9]+' is an extended regular expression that matches a colon followed by 08, followed by another colon, followed by one or more digits, followed by another colon, followed by one or more digits. This pattern matches any timestamp that has 08 as the hour part, which corresponds to the time period between 8:00 am and 8:59 am. The loga.log file is the name of the log file that contains the timestamps of the format DD/MM/YYYY:hh:mm:ss. For example, running grep -E ':08:[0-9]+:[0-9]+' loga.log will produce an output like this: 01/01/2023:08:00:01 User logged in 01/01/2023:08:15:23 User performed an action 01/01/2023:08:30:45 User logged out 01/01/2023:08:45:12 User logged in again The other commands are either invalid or do not perform the desired task. The grep -E ':08:[09]+:[09]+' loga.log command will only match timestamps that have 0 or 9 as the minute and second parts, which is too restrictive. The grep -E ':08:[00]+' loga.log command will only match timestamps that have 0 as the minute and second parts, which is too specific. The grep -E loga.log ':08:[0-9]+:[0-9]+' command will not work, as the file name should come after the pattern, not before it. The grep loga.log ':08:[0-9]:[0-9]' command will not work, as it uses a basic regular expression without the -E option, and it will only match timestamps that have one digit as the minute and second parts, which is too narrow.