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Help need in awk command.

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# 1  
Old 03-10-2011
Help need in awk command.
Hi,

I need to get the line count of files through awk command which can be executed directly in command line without creating script.

Example.
Files:
File_X_20110305000000.dat
File_Y_20110306000000.dat
File_Z_20110307000000.dat
File_X_20110308000000.dat
File_Y_20110309000000.dat
File_Z_201103010000000.dat

Out Put:

File 20110305 2011038
X 10 11
Y 22 23

can we print in this way?

Thanks in advance SmilieSmilie
# 2  
Old 03-10-2011
Line count using AWK
Code:
awk 'END{print NR}' inputfile

Provide some more information on your desire o/p as cannot understand.
# 3  
Old 03-10-2011
If you want to get a count of files containing a patterns "X" ,"Y" and "Z" execute the following awk command. Otherwise pls clarify your question.
Code:
awk -F"_" '$2 ~ /X/ { x++  }; $2 ~ /Y/ { y++  }; $2 ~ /Z/ { z++  }; END { print "X:" x " Y:" y " Z:"z}' inputfile

where inputfile should have all the filenames in it.

Regards,
Rosh
Last edited by Franklin52; 03-10-2011 at 03:39 AM.. Reason: Please use code tags, thank you
# 4  
Old 03-10-2011
Quote:
Originally Posted by pravin27
Line count using AWK
Code:
awk 'END{print NR}' inputfile

Provide some more information on your desire o/p as cannot understand.
I will get set of files daily. The file name will be Type_DateTime.dat.

Type1_Date1Time1.dat
Type1_Date2Time2.dat
Type2_Date1Time1.dat
Type2_Date2Time2.dat

I want to print,

Type Date1 Date2
Type1 100 120
Type2 200 220
...

---------- Post updated at 04:06 AM ---------- Previous update was at 02:01 AM ----------

Any shell script also would be fine if it prints in the matrix format...
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