UNIX for Advanced & Expert Users

I have a file which contains the data like below.It conatins more data than i posted here.

BEGIN DSJOB
Identifier "TestPart"
DateModified "2008-10-08"
TimeModified "00.36.32"
Identifier "ROOT"
DateModified "1899-12-30"
TimeModified "00.00.01"
OLEType "CJobDefn"
Readonly "0"
Name "Test"
NextID "1"
Container "V0"
JobVersion "50.0.0"
ControlAfterSubr "0"
MetaBag "CMetaProperty"
BEGIN DSSUBRECORD
BEGIN DSJOB
Identifier "GetFile"
DateModified "2008-10-05"
Identifier "ROOT"
DateModified "1899-12-30"
TimeModified "00.00.01"
OLEType "CJobDefn"
Readonly "0"
Name "Test"
NextID "1"
Container "V0"
JobVersion "50.0.0"
ControlAfterSubr "0"
MetaBag "CMetaProperty"
BEGIN DSSUBRECORD
BEGIN DSJOB
Identifier "TrimFields"
DateModified "2008-10-05"
Identifier "ROOT"
DateModified "1899-12-30"
TimeModified "00.00.01"
OLEType "CJobDefn"
Readonly "0"
Name "Test"
NextID "1"
Container "V0"
JobVersion "50.0.0"
ControlAfterSubr "0"
MetaBag "CMetaProperty"
BEGIN DSSUBRECORD
BEGIN DSJOB
Identifier "ConvertDate"
DateModified "2008-10-05"
Identifier "ROOT"
DateModified "1899-12-30"
TimeModified "00.00.01"
OLEType "CJobDefn"
Readonly "0"
Name "Test"
NextID "1"
Container "V0"
JobVersion "50.0.0"
ControlAfterSubr "0"
MetaBag "CMetaProperty"
BEGIN DSSUBRECORD
BEGIN DSJOB
Identifier "LoadTable"
DateModified "2008-10-05"
Identifier "ROOT"
DateModified "1899-12-30"
TimeModified "00.00.01"
OLEType "CJobDefn"
Readonly "0"
Name "Test"
NextID "1"
Container "V0"
JobVersion "50.0.0"
ControlAfterSubr "0"
MetaBag "CMetaProperty"
BEGIN DSSUBRECORD
BEGIN DSJOB
Identifier "Test"
DateModified "2008-10-05"



In the above text I need to grep for "BEGIN DSJOB" and when ever i find this text i need the line after this text.

Out put :

Identifier "TestPart"
Identifier "GetFile"
Identifier "TrimFields"
Identifier "ConvertDate"
Identifier "LoadTable"
Identifier "Test"

Please give me some suggestion how can i acheive this.

Thanks

Code:

awk '{ if ( match($0, "^BEGIN DSJOB") ) { getline; print } }' filename

I have a question like if i want 3rd line from that record how can i get it.

Thanks

Sorry, your question is not clear.

Did you mean from the file

Code:

awk ' NR == 3 { print }' filename

or
from the output ?

From the same file I need to grep for "BEGIN DSJOB" and get the second line from that line

Ex output:

DateModified "2008-10-08"
DateModified "2008-10-05"
....

Thanks

You can do this simply enough with grep:

$ grep -A 2 '^BEGIN DSJOB' datafile | grep DateMod

you can simply do this..

Code:

awk 'c-->0;/BEGIN DSJOB/{c=2}' filename