UNIX for Dummies Questions & Answers

Hi,

I have a file with rows of text like so :

E100005568374098100000015667
D100005568374032000000112682
H100005228374060800000002430


I need to grab just the last digits(bolded) of each line without the proceeding text/numbers.

Thanks

Code:

echo 'E100005568374098100000015667' | sed 's/.*\(.....\)/\1/'

To grab a value in a record without field seperators you have to know the positions of each field IMHO.
You can arbitrary begin at one of the position with a zero but you're never shure you have the right value.

Regards

actually now that I think about it - I don't know what's common among the values you return.
Is it preceeded with more than one '0'?

Code:

echo 'E100005568374098100000056607' | sed 's/.*00\(.*\)/\1/'

@vgersh99
You always cut the last 5 characters, no matter if he needs 4 or 6 like in the other rows.

@Franklin52
Yep, that's true. You have to know the possibilties of your file-to-be-parsed very well to limit the chances of unwanted behaviour.

So for the input presented we can say that when at least 5 zeros are in a row, that we can use them to separate the last digits we want:

Code:

sed 's/.*00000\([^0]*\)/\1/'

This will at least make sure, that even if you have 1 up to 4 zero in your last digits, the will be separated ie. parsed correctly. If you have at the end a "600000", you have a problem again. For this I don't have a good solution, as I just thought, that there at least a number of zeros following each other before the last digits come, you want.
If you know that the last digits are always not more than 6 digits, you can parse them like vgersh99 did, ie. add a dot or write it another way and after that cut off all leading zeros.


EDIT: @vgersh99
Oh you saw it already while I was writing

Thanks everybody, learnt a lot from this. The last piece of code posted has done the trick.