Searching by date range from filenames

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# 1  
Old 08-26-2010
Searching by date range from filenames

Hello all,

i have tons of files in folder named like this (yyyymmdd):



I need to find files with date range in there names lets say from 2010.07.30 - 2010.08.02

I tried with grep

grep -h --mmap 'horror' /usr/local/data/books/bookcollection201007[10-11]*

But it doesn't work, i can change just date, and not year and months.

The plan is that I will have a script that will ask from user date range input as 2010.07.30 - 2010.08.02 then I will add those years, months and days to variables that I can append to grep/sed command and display result

But I m stuck. Any ideas?

Last edited by Whit3H0rse; 08-26-2010 at 10:28 PM.. Reason: error
# 2  
Old 08-26-2010
ls bookcollection* |awk -v start=20100730 -v end=20100802 '{d=substr($0,15); if (d>=start&&d<=end) print}'



ls bookcollection* |awk -v start=$D1 -v end=$D2 '{d=substr($0,15); if (d>=start&&d<=end) print}'

Last edited by rdcwayx; 09-06-2010 at 03:47 AM..
# 3  
Old 08-27-2010
This doesn't work, I forgot to mention that there si time append so filenames are bookcollection20100728_001500.

Time doesn't matter to me, just yy, mm and dd. Any new suggestion or reformulation?

Thanks in advance


Tx a alot it works, I new at awk so I needed more time to figure it out Smilie

Last edited by Whit3H0rse; 08-28-2010 at 12:19 PM..
# 4  
Old 08-29-2010
Oke, this works great but how to include the "start" and "end" variables into results. Because when i enter range for today 2010.08.29 - 2010.08.29, it display nothing.

Now, it displays this for rage 2010.07.30 - 2010.08.02:

But I need that awk displays this:


# 5  
Old 08-29-2010
Do you mean something like this?

grep -h 'horror' $( ls -d bookcollection* | sed -n "/$D1/,/$D2/p" )

This is assuming that your ls sorts alphabetically, otherwise you could perhaps use:
grep -h 'horror' $( ls -d bookcollection* | sort | sed -n "/$D1/,/$D2/p" )

# 6  
Old 08-29-2010
ls bookcollection* |awk -F \_ -v start=20100730 -v end=20100802 '{d=substr($1,15); if (d>=start&&d<=end) print}'

# 7  
Old 09-05-2010
Ty people, this works great Smilie

---------- Post updated at 08:42 AM ---------- Previous update was at 03:05 AM ----------

Originally Posted by rdcwayx
ls bookcollection* |awk -F \_ -v start=20100730 -v end=20100802 '{d=substr($1,15); if (d>=start&&d<=end) print}'

SED version works but it is slow because of | sort|, so I prefer AWK, and this code above works great except it doesn't display last day =20100802 it stops on 20100801, also when I choose for example start=20100808
and end=20100808 it display nothing???

I tried modifying code buy my AWK skills are terrible:/, any help?
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