Extract part of file name with multiple sections

Question

I am trying to extract part of a file name to compare with other file names as it is the only part that does not change. here is the pattern and an example

clearinghouse.doctype.payer.transID.processID.date.time EMDEON.270.60054.1234567890123456789.70949996.20120925.014606403

all sections are the same length at all times with the exception of clearinghouse & doctype that can vary in character length.

The part of the filename that i need for comparison is the transID.

What would be the cleanest shortest way to do this in a shell script.

Thanks

jlunavtgrad · Accepted Answer · 2012-09-26 12:57:25Z

1

There are lots of ways to do this, the easiest tool for simple tasks is the cut command. Tell cut what character you want to use as a delemiter and which fields you want to print. Here is the command that does what you want.

file=EMDEON.270.60054.1234567890123456789.70949996.20120925.014606403
transitId=$(echo $file | cut -d. -f4)

Awk can do the same thing, and allows you do much more complicated logic as well.

file=EMDEON.270.60054.1234567890123456789.70949996.20120925.014606403
transitId=$(echo $file | awk -F. '{print $4}')

edited Sep 26, 2012 at 12:57

answered Sep 25, 2012 at 18:25

jlunavtgrad

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chepner · Accepted Answer · 2012-09-26 02:09:10Z

1

You can split the filename apart using the read command using an appropriate value for IFS.

filename="EMDEON.270.60054.1234567890123456789.70949996.20120925.014606403"
IFS="." read clHouse doctype payer transID procID dt tm <<< "$filename"
echo $transID

Since you only want the transaction ID, it's overkill to assign every part to a specific variable. Use a single dummy variable for the other fields:

# You only need one variable after transID to swallow the rest of the input without
# splitting it up.
IFS="." read _ _ _ transID _  <<< "$filename"

or just read each part into a single array and access the proper element:

IFS="." read -a parts <<< "$filename"
transID="${parts[3]}"

answered Sep 26, 2012 at 2:09

chepner

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jordanm · Accepted Answer · 2012-09-25 18:26:02Z

0

You can do this with a parameter expansion:

$ foo=EMDEON.270.60054.1234567890123456789.70949996.20120925.014606403
$ bar=${foo%.[0-9]*.[0-9]*.[0-9]*}
$ echo "${bar##*.}"
1234567890123456789

answered Sep 25, 2012 at 18:26

jordanm

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Vijay · Accepted Answer · 2012-09-26 02:15:13Z

0

tranid==`echo file_name|perl -F -ane 'print $F[3]'`

answered Sep 26, 2012 at 2:15

Vijay

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