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Justin
Asked: 2014-10-18 15:54:38 +0800 CST

Sed to change a birth date using a regular expression

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I'm interested in changing a birth date in a text file to a different date of 11/14/46. I want to do this without having to know the previous birth date and instead be able to search generically. I know the person's name that the birth date is associated with, so I can use that to make the substitution

I've tried this:

sed "s/^PersonsName\/[1-12]\/[0-31]\/[0-99]/11\/14\/46/g" txt.file

But it doesn't work and I'm not sure how to proceed

The suggested answers below haven't worked for me. I'm going to give some example data and see if this helps to clarify. 

Popeye Sailor:156-454-3322:945 Bluto Street, Anywhere, USA 29358:3/19/35:22350

Let's pretend I'm trying to change Popeye's birth date (which is at the end), but I want to search without having to know his current one. I know using a regular expression is key here, but so far the examples I've seen haven't worked. Maybe providing the above example data will help

[Expected Output]

Popeye Sailor:156-454-3322:945 Bluto Street, Anywhere, USA 29358:11/14/46:22350

All I'm interested in changing is the birth date at the end to 11/14/46

sed

3 Answers

  1. [1-12]\/[0-31] and [0-99] do not mean 1 to 12, 0 to 31 and 0 to 99.

    • [1-12] means a range of characters starting from 1 till 1, and a 2. In essence: [12]. (Compare with [0-1a].)
    • [0-31] means a range of characters starting from 0 till 3, and a 1: [0123].

    Similarly for [0-99]. Expressing a range of numbers with regular expressions isn't easy.

    If the numbers don't use double digits (01/06/33), you'll have to do something like: [0-9]{1,2} for day and month, and [0-9]{2} for the year, or the regex will quickly grow unwieldy. This, of course, risks the selection of something like 99/99/99.

    For an example:

    $ sed "s/^\(muru\/\)[0-9]\{1,2\}\/[0-9]\{1,2\}\/[0-9]\{2\}/\111\/14\/46/g" <<<'muru/1/4/85'
muru/11/14/46
$ sed "s/^\(muru\/\)[0-9]\{1,2\}\/[0-9]\{1,2\}\/[0-9]\{2\}/\111\/14\/46/g" <<<'muru/10/4/85'
muru/11/14/46
$ sed "s/^\(muru\/\)[0-9]\{1,2\}\/[0-9]\{1,2\}\/[0-9]\{2\}/\111\/14\/46/g" <<<'muru/10/28/85'
muru/11/14/46

    I assume you want to retain the name, so I used backreferences to refer to it in the substitution (the \1).

    • 0

  2. Dates are tricky, but simplifying to a similar extent to your attempt…

    sed -r 's,^PersonsName1?[0-9]/[1-3]?[0-9]/[0-9]{2},11/14/46,' txt.file

    Explanation

    • Use -r for extended regular expressions. This allows (for example) ?.
    • Generally, I prefer single quotes ' rather than double ", so there's no shell expansion.
    • Instead of using / as sed delimiters, you can use whatever you want. Here I've used ,, so that we don't need to escape literal /s.
    • 1?[0-9] match an optional 1 and required 0-9, for the month.
    • [1-3]?[0-9] match and optional 1-3, and a required 0-9 for the day.
    • [0-9]{2} match two digits for the year.
    • N.B. that this will potentially have some false matches. e.g. 0/0/00 and 19/39/99 will match, but you can tweak the regex more if you like. See the link above for more information.

    More comments

    Note that this matches something like PersonsName11/11/11, and replaces the whole thing, similar to what your attempt tried to do. If you want to only replace the date, you can use the following instead.

    sed -r 's,^(PersonsName)1?[0-9]/[1-3]?[0-9]/[0-9]{2},\111/14/46,' txt.file

    Here, we capture PersonsName, using parens. Then we can refer to it again in the replace section with \1.

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  3. Best Answer

    You could use the below sed command,

    $ sed '/^Popeye Sailor:/s/:[^:]*\(:[^:\/]*\)$/:11\/14\/46\1/' file
Popeye Sailor:156-454-3322:945 Bluto Street, Anywhere, USA 29358:11/14/46:22350

    It changes the date only on the lines which starts with Popeye Sailor:

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