How to pick system time of the file placed on UNIX?

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06-07-2016

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How to pick system time of the file placed on UNIX?

Hi ,I got one data file from external source and I have to load it in database through sql loader. I want to add 2 columns in database,one is file name and one is time of the file received on server.

Code:

-rwxr-x---    1 user user          238 Jun 03 16:32 90936264971326030616.ctr

From above case, I want one field as 'Jun 03 16:32' and one as "90936264971326030616" (file name without extension).

Thanks in advance

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Please use code tags as required by forum rules!

Last edited by RudiC; 06-07-2016 at 10:07 AM.. Reason: Added code tags.

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06-07-2016

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Not sure if your specifications requires the quotations or not but what about:

Code:

 awk '{print "\""$6,$7,$8"\"","\""substr($9,1,length($9)-4)"\""}' filename

Will produce:

"Jun 03 16:32" "90936264971326030616"

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06-07-2016

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Quite sparse a request, no? What tools would be preferred? What would the data be used for, and how? How or where to store the data? In a file? In stdout (to be piped into sth.)? In shell variables?

Try

Code:

awk '{sub (/\..*$/, _, $NF); print $NF; print $(NF-3) " " $(NF-2) " " $(NF-1)}' file
90936264971326030616
Jun 03 16:32

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06-08-2016

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Hi All,

Thanks for your response

The actual requirement I have is to take the timestamp (file name here can be ignored), append it to all rows in the file so that it can loaded in database through sqlldr.

Once I extract the timestamp of the file in UNIX and pass it in one variable like $1 then I can further use the below command to make a new data file which I will load.

Code:

sed -e 's/$/   May 04 09:02/' summary_file.sql > Data_file.sql

I hope I am clear now.

I tried your suggested command :

File content is:-
***************

Code:

-rw-r-----    1 owner owner        976 May 04 09:02 summary_file.sql

Code:

cat summary_file.sql
REPORT_ID       REPORT_NAME     FILE_NAME       FILE_DESCR      RPT_GEN_DT      PARTNER_NAME    HQ_ID   SOLD_TO SOLD_TO_NAME    RO_SEGMENT      REPORT_CATEGORY SUB_CATEGORY    LOB     SUB_LOB FISCAL  REGION  SALES_ORG       EXT_PUB_IND     DATA_LOAD_TS    REPORT_GEN_IND  REPORT_GEN_TS   CREATE_TS       SUB_REGION_CD   summary_file1

Approach 1:-
***********

Code:

awk '{print "\""$6,$7,$8"\"","\""substr($9,1,length($9)-4)"\""}'  summary_file.sql

Output
*****

Code:

"PARTNER_NAME HQ_ID SOLD_TO" "SOLD_TO_"

It gives me 6,7 and 8th field.

Approach 2:-
***********

Code:

awk '{sub (/\..*$/, _, $NF); print $NF; print $(NF-3) " " $(NF-2) " " $(NF-1)}' summary_file.sql

It gives me output as :-

Code:

summary_file1
REPORT_GEN_TS CREATE_TS SUB_REGION_CD

It is not giving me time. It just picks up the last field in one line and other last 3 columns in one line.

Please help.

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Mod Comment

Please use code tags as required by forum rules! It drastically improves readabilty of your post.

Last edited by RudiC; 06-08-2016 at 03:39 AM.. Reason: Added code tags

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06-08-2016

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An excellent example how incomplete and/or inaccurate specifications DO waste peoples' and requestor's time and effort!

To start from the beginning: What OS and shell version do you use? Do you have the stat or equivalent command at your disposal?

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Sorry for wasting your time ..its my first time to ask for a help like this..not sure what all info is required. Shell version is

Code:

/usr/bin/ksh

. It is an AIX server

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What be the stat equivalent on your system that can give the file's time stamp explicitly? (OK, if need be we can extract that from ls -l)

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