A configuration file is to be displayed with every empty line and every line that begins with # in the first column removed. Select the TWO commands that do this.

LPIC-1 Exam 101-500, objective 103. GNU and Unix commands hard

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The options

Choose 2.

Correct grep -Ev '^(#|$)' /etc/ssh/sshd_config

Correct. -E enables alternation without backslashes, ^(#|$) matches a line starting with # or a line with nothing in it, and -v inverts the selection.

Correct sed -E '/^(#|$)/d' /etc/ssh/sshd_config

Correct. The same pattern used as a sed address, with the d command deleting the matching lines from the output stream.

Not correct grep -E '^(#|$)' /etc/ssh/sshd_config

Wrong. Without -v this prints exactly the comments and blank lines that were to be discarded.

Not correct sed '/^#/d' /etc/ssh/sshd_config

Wrong, though half right. Comments go, but empty lines survive because nothing in the address matches them.

Not correct grep -v '^#|^$' /etc/ssh/sshd_config

Wrong. In a basic regular expression | is an ordinary character, so the pattern only matches a line consisting of the three characters #|^, and virtually everything is printed.

Why

Alternation is the clearest illustration of the basic and extended split: in an extended regular expression | separates alternatives and ( ) group them, while in a basic one those characters are literal unless escaped as \| and \( \). The other half of the item is that grep selects lines while sed transforms a stream, so an inverted match in grep and a d command in sed reach the same result. Note that both patterns anchor # to column one; leading whitespace before a comment would need ^[[:space:]]*#.

Where this comes from

Cited
manual page grep(1)

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