How to calculate the time to SELECT

I want to know how to calculate the time taken by the SELECT query.

For example: I run a select statement in TOAD (oracle version 9i) who takes 40 seconds to run, but in the output window, it displays few records. It says "list 1 of 500 records, more records exist". This table has total 20 million records.

Now my question is: have you my SELECT query took 40 seconds? or it took 40 seconds to SELECT 500 records.

If it took 40 seconds to SELECT 500 records then how to calculate the total amount?

Any help would be much appreciated.

To avoid the exit which is probably not really need you can make

begin
  for c in (select * from xy /*your query */)
  loop
    null;
  end loop;
end;
/

This gives an estimate of the time to recover all the lines.

Tags: Database

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How to calculate the time
I want to calculate the time. I want to subtract ' 01:00:00 ' AM currently, but I am getting error.

SQL > /.
Select to_char('01:00:00','HH24:MI:SS')-to_char(sysdate,'HH24:MI:SS') double
*
ERROR on line 1:
ORA-01722: invalid number

can someone help me

Thank you
Not sure what you mean. If you want to subtract an hour from a date, keep in date arithmetic spirit uses day as a unit. To subtract an hour from a date, you need to subtract 1/24:
```
SQL> alter session set nls_date_format = 'MM/DD/YYYY HH24:MI:SS'
  2  /

Session altered.

SQL> select sysdate,sysdate - 1/24 from dual
  2  /

SYSDATE             SYSDATE-1/24
------------------- -------------------
08/09/2009 09:55:28 08/09/2009 08:55:28

SQL> 
```
SY.

How to calculate the time difference

Hi friends,
I want to calculate and the sum of the time difference.
I have two fields in a table. Inside and OUTSIDE. for example.
OUT IN DIFFERENT
01:46 15:30?
01:47 15:45?
02:50 17:30?
I want less above figure and the result will appear in different column.
Help, please
Kind regards

Hello

Assuming that the two columns are the VARCHAR2 data type and "represent" the hours and minutes:

WITH sample_data AS (SELECT ' 01:46 ' in_col, 15:30 ' out_col FROM DUAL)

UNION ALL

SELECT ' 01:47 ' 15:45 ' OF THE DOUBLE

UNION ALL

SELECT ' 02:50 ', 17:30 ' OF THE DOUBLE

)

SELECT in_col,

out_col,

TO_CHAR (TO_DATE ((TO_DATE(out_col, 'HH24:MI')-TO_DATE(in_col, 'HH24:MI')) * 86400, «SSSSS»), "HH24") differ))

OF sample_data;

SQL > WITH sample_data AS (SELECT ' 01:46 ' in_col, 15:30 ' out_col FROM DUAL)

2. ANY TRADE UNION

3. SELECT ' 01:47 ' 15:45 ' OF THE DOUBLE

4 UNION ALL

5. SELECT ' 02:50 ', 17:30 ' OF THE DOUBLE

6 )

7. SELECT in_col,

out_col 8,.

9 to_char (TO_DATE ((TO_DATE(out_col, 'HH24:MI')-TO_DATE(in_col, 'HH24:MI')) * 86400, «SSSSS»), "HH24") differ))

10 FROM sample_data;

IN_COL OUT_COL DIFFER

------ ------- ------

01:46 15:30 13:44

01:47 15:45 13:58

02:50 17:30 14:40

SQL >

How to calculate the time interval
Dear Gururs,

I have table of details of the calls as below

Call_Detail DESC;
Name of Type Null
--------- ---------- --------------
Appeal of VARCHAR2 (20)
VARCHAR2 (20) called
Time DATE

in other words table contains each call on a day like this

The call | Called | Time
========================================
A | B | January 1, 2011 00:02:00
A | B | January 1, 2011 00:08:00
A | B | January 1, 2011 00:18:00
A | B | January 1, 2011 00:22:00
A | B | January 1, 2011 00:30:00
A | B | January 1, 2011 00:31:00

I want output like below

The call | Called | Time | Time Difference (sec)
==============================================================
A | B | January 1, 2011 00:02:00 | 360
A | B | January 1, 2011 00:08:00 | 600
A | B | January 1, 2011 00:18:00 | 240
A | B | January 1, 2011 00:22:00 | 480
A | B | January 1, 2011 00:30:00 | 60
A | B | January 1, 2011 00:31:00 | 0

Could someone help to above.

Concerning
Sanjeev
Hello

Use the analytical fucntion of LEAD to get a value on the next line.
For example:
```
SELECT       calling
,       called
,       NVL ( ( LEAD (time) OVER ( PARTITION BY calling
                                            ,               called
                         ORDER BY     time
                       )
             - time
             ) * 24 * 60 * 60
           , 0
           )     AS time_difference_sec
FROM      call_detail
ORDER BY  calling
,            called
,       time
;
```
If you would care to post CREATE TABLE and INSERT statements for your sample data, and then I could test this.

Published by: Frank Kulash, January 11, 2011 10:18

How to calculate the time difference in the two date columns time.
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I came back new requirement on obiee 10g.

I have two date columns in my table time, data as below.

StartDate: 18/03/2012-04:40:51

Closing date: 18/03/2012-04:50:55

I want to the output as below (Jet lag).

output
04:40:51 (less)-04:50:55 = 00:10:04

I hope you understand my requirement in OBIEE.

Note: I want to show above the output values in another State for OBIEE10g column.

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Satya
Hi Satya,

Here's another solution. You can try the formula below as it is:

Cast (floor (TIMESTAMPDIFF (sql_tsi_second, '-SNP_SESSION '.))) SESS_BEG, '-SNP_SESSION '. SESS_END) / 3600) as char). ':' ||
Cast (floor (mod (TIMESTAMPDIFF (sql_tsi_second, '-SNP_SESSION '.)))) SESS_BEG, '-SNP_SESSION '. ((SESS_END), 3600) / 60) as char). ':' ||
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I would like to know if that's what you're looking for.

Thank you

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How to calculate the time to SELECT

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