Derive date and its Volume number

Shell Programming and Scripting
1

Hi All,
I have a list of date number (YYYYMMDD) and its own number in a file.
Here is the example


20170320
D100001
D100002
D100003
20170321
D200001
D200002
D200004
D200005
20170322
D200005
D200006
D200007

If I grep a date ie. 20170321, I expecting to get the following list
D200001
D200002
D200004
D200005

Once it found next first integer, it will need to stop from proceeding.

Thank You everyone.

2

Hello ckwan123, Could you please try following and let me know if this helps you.

 awk '/^[0-9]/{val=""} /20170321/{val=1;next} val' Input_file

Thanks, R. Singh

3

Not as elegant as intended, but try

awk '$0+0 == $0 {P = 0} $0 == SRCH {P = 1; next} P' SRCH=20170321 file
D200001
D200002
D200004
D200005
4

Thanks Ravin and RudiC. However, I feel RudiC provided solution is easier to apply into my requirement as it is more flexible by replace the date with a variable. Ravin, yours is able to run with constant value, do you think it can change it to variable ?
RudiC, could you briefly your command ? Thanks so much for your given valuable solution.

5

Well, if it finds the search phrase, it set the boolean variable P to 1. P is evaluated, and if TRUE (or 1), the line is printed ( awk default behaviour). Later, if an integer is found ($0+0 casts the line to a number or (roughly speaking) to something different if a non-number ist encountered), P is reset to 0 and printing stops.

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