Shell Programming and Scripting

Hi ,
I have .csv file with value separated by ";".
1. Using awk how to extract perticular colums and store in to array
2. For some columns I needs to extract last value of the column

How to do same please help me ASAP

Thanks and Regards,
Sushma

Please post a sample input & expected output

You could use something like this,


cat tst.csv | awk -F";" '{v1=$0;split(v1,names,";");for (idx=1;idx<NF+1;idx++) print idx, names[idx];}'

The input file is,
$ cat tst.csv
1;2;3;4;5;6;7;8
a;b;c;d;e;f;g;h

And the output is,
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
1 a
2 b
3 c
4 d
5 e
6 f
7 g
8 h

Hope it helps.

Hi Dennis,
Please find sample input and output

input:
StartTime;LastResetTime;CurrentTime;ElapsedTime(P);ElapsedTime(C);Rate(P);Rate(C);IncomingCall(P);In comingCall(C);OutgoingCall(P);OutgoingCall(C);MsgRecvPerS(P);MsgRecvPerS(C);MsgSendPerS(P);MsgSendPe rS(C);UnexpectedMsg(P);UnexpectedMsg(C);CurrentCall;InitSuccessfulCall(P);InitSuccessfulCall(C);Traf ficSuccessfulCall(P);TrafficSuccessfulCall(C);DefaultSuccessfulCall(P);DefaultSuccessfulCall(C);Abor tSuccessfulCall(P);AbortSuccessfulCall(C);FailedCall(P);FailedCall(C);FailedRefused(P);FailedRefused (C);FailedAborted(P);FailedAborted(C);FailedTimeout(P);FailedTimeout(C);ResponseTime(P);ResponseTime (C);ResponseTimeRepartition;<50;<75;<100;<150;<300;<5000;>=5000;
2009-10-23 16:55:14;2009-10-23 16:55:14;2009-10-23 16:55:14;00:00:00;00:00:00;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;00:00:00:000;00 :00:00:000;;0;0;0;0;0;0;0;
2009-10-23 16:55:14;2009-10-23 16:55:15;2009-10-23 16:55:15;00:00:01;00:00:01;1.998;1.99005;0;0;2;2;1.998;1.99005;2.997;2.98507;1;1;1;1;1;0;0;0;0;0;0;0 ;0;0;0;0;0;0;0;00:00:00:439;00:00:00:439;;0;0;0;0;0;1;0;

Output :
array_16 = column16
last_value = column_16_lastrecord

---------- Post updated at 12:28 AM ---------- Previous update was at 12:20 AM ----------

Hi Shan,
Thanks for ur reply.
Since .csv file containes 50 lacs of records, If I use for loop and store into array takes too much of time. And I don't want to store all coulmns data, only selected column I need to store. Please suggest if any method of doing without iterating to each row of the file.

first in your sample file, only 45 column, if split by ;

Code:

$ awk -F\; '{print NF}' input
45
45
45

If you need output selected columns, you can use below ways:

Code:

awk -F\; '{print $3,$10,$30}' input
cut -d\; -f3,10,20-30,42- <input

Still I don't get how you get this output from your input file, can you explain on it?

Code:

array_16 = column16
last_value = column_16_lastrecord

Hi,
I need last record(last row value) of column 16. Suppose column 16 has 2 records(2 rows) I want to grep and store 2nd record of column16 i.e 3 in my input file.

Sushma,
If I understand correctly then, for your 2nd query you can simply do a tail -1 and then cut the field whose value you require,

e.g. to get the last row value in the first field,

tail -1 file.csv | cut -d';' -f1

Regards,
Shantanu


With awk you can do,
awk -F";" 'END {print $1}' file.csv