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I have a string and I need to add #, in the beginning,i.e. convert [ -n "$ID" -a "$ID" -le 200 ] && return to #[ -n "$ID" -a "$ID" -le 200 ] && return. I can use below command:

echo '[ -n "$ID" -a "$ID" -le 200 ] && return'| sed -n -e 's/\[ -n "$ID" -a "$ID" -le 200 \] && return/#&/p'

It works. Now I have two questions, regarding E, flag.

  1. For the string [ -n "$ID" -a "$ID" -le 200 ], if I escape brackets, it does not work; however, it works when I do not escape them i.e.

    sed -n -E 's/[ -n "$ID" -a "$ID" -le 200 ]/#&/p'

    works while

     sed -n -E 's/\[ -n "$ID" -a "$ID" -le 200 \]/#&/p'

    doesn't work.

  2. For, the full string [ -n "$ID" -a "$ID" -le 200 ] && return, it gives me wrong answer, when I do not escape them:

    echo '[ -n "$ID" -a "$ID" -le 200 ] && return'| sed -n -E 's/[ -n "$ID" -a "$ID" -le 200 ] && return/#&/p'

    It gives me output:

     [ -n "$ID" -a "$ID" -le 200 #] && return

I want to know how it is working.

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  • 6
    The [ and ] must be escaped in both BRE and ERE (otherwise they define a range expression). As to why you get no output when escaping them and using ERE, I think it's a difference in how $ is treated between BRE and ERE when it's not the final character of a pattern - see for example Bash sed replace double dollar sign $$ extended regular expressions Commented Sep 20, 2019 at 12:29
  • 2
    sed's -E and -e options are two completely different and unrelated things. they are not alternate versions of each other. -e tells sed that the next argument is a script to be run. -E tells sed to use extended regular expressions (ERE) instead of sed's default of basic regular expressions (BRE). Commented Sep 20, 2019 at 13:02
  • You can drop the "-e", it's only if you have separate expressions. The -E option would work if you escaped the "$"'s. Commented Sep 21, 2019 at 0:24

1 Answer 1

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sed -n -E 's/[ -n "$ID" -a "$ID" -le 200 ]/#&/p'

Does not work the way you expect it to work, it's more of a coincidence: [...] is a collection of characters that match. Inside that collection you have a range -n from whitespace (0x20) to n (0x6E), which can include [ (0x5B), depending on your locale settings. So the collection matches the first char. See what happens when you take #&___ as the replacement... not your intention, I guess?

And that's the reason for your full case to fail: The matching character for the collection is the ] because it is followed by the rest of the pattern, that's why the # is inserted there.

The problem with the extended regular expression is like @steeldriver assumed: The $ anchors the pattern:

A ( '$' ) outside a bracket expression shall anchor the expression or subexpression it ends to the end of a string; such an expression or subexpression can match only a sequence ending at the last character of a string. For example, the EREs "ef$" and "(ef$)" match "ef" in the string "abcdef", but fail to match in the string "cdefab", and the ERE "e$f" is valid, but can never match because the 'f' prevents the expression "e$" from matching ending at the last character.

So in ERE, a literal $ needs to be escaped, while in BRE it only needs to be escaped when being the last character of the pattern.

Also note that -e marks the next argument as a script and is optional if there is only one script, while -E is a switch. Dropping -e for '-E` only works if there is only one script because it was superfluous anyhow.

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