SQL & PL/SQL :: To Calculate Minutes From Difference Of To Number Fields

Apr 9, 2013

I need the query to calculate minutes from difference of to number fields. Test case is as below.

DROP TABLE tmp;

CREATE TABLE tmp
(
code NUMBER(4),
stime NUMBER(4,2),
otime NUMBER(4,2)
)
LOGGING
NOCOMPRESS
NOCACHE
NOPARALLEL
NOMONITORING;
[code]......

CODE STIME OTIME
---------- ---------- ----------
1065 20 19.49
1082 20 18.57
1279 19.3 18.59
2075 19.3 15.32

Required output is

CODE STIME OTIME HR_MIN MINUTES
---------- ---------- ---------- ------------- --------
1065 20 19.49 00 HR 21 MIN 21
1082 20 18.57 01 HR 03 MIN 63
1279 19.3 18.59 00 HR 31 MIN 31
2075 19.3 15.32 03 HR 58 MIN 238

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Forms :: Calculate Difference Time Between Two Fields

Mar 24, 2011

I have two different date in my payroll software,

1-Shift_date shift date *used to contain shift timings
2-Attendace_datedate *used to contain employee IN timings

As you all know that shift is a setup form, where user input data once in the starting of software so the shift_date can be "01/jan/2011 16:00 pm" but attendance loads daily and attendance field data can be in this form "24/mar/2011 16:15 pm"..Now I want to calculate difference time between these two fields therefore I used this statement

SQL> Select to_char(attendance_date,'HH24:MI') to_char(shift_date,'HH24:MI') from dual;

but it is showing error: ORA-01722: invalid number...I used hours/minutes format mask in my query because you can see there is a difference of dates between these fields and it will be increase in the coming future and I need late hours and minutes.

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Sep 27, 2010

setting up the query/correcting the syntax below so that it calculates the 'number of days difference' between whatever the 'Biggest Date' field value is and whatever the 'current date' is using the 'sysdate'. So far, I've only managed to get the query to calculate the number of days difference (days past due) between the 'need date' and 'estimated delivery date'.

CODESELECT
To_Date(need_date, 'YYYYMMDD') Need_Dt,
To_Date(Case when estimated_delivery > ' ' THEN estimated_delivery ELSE need_date END, 'YYYYMMDD') Biggest_Date,  
To_Date(need_date, 'YYYYMMDD') - To_Date(Case when estimated_delivery > ' ' THEN estimated_delivery ELSE need_date END, 'YYYYMMDD') Date_Diff
        
FROM tableT

WHERE
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ORDER BY Date_Diff ASC

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Feb 24, 2010

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The code gives an error, but how can I do this?

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Oct 25, 2012

I have a date in_sdate as In parameter defaulted to sysdate. Basing on this in_Sdate I calculate my start and end dates as:

v_sdate TRUNC (in_sdate, 'MI') - 15 / 1440 ;
v_edate := TRUNC (in_sdate, 'MI');

My procedure is run for every 15 minutes. Now suppose if I am running for old dat, then I should get the difference of dates by taking

v_old_Date := v_edate - in_Sdate;

Divide this by 15 , round that value and loop to run the procedure for that n times. My doubt is when I am saying

v_old_date := v_edate - in_sdate ; I am getting expression is of wrong type. How can I take the difference of the dates and get the minutes from that ?

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Oct 17, 2012

Detail table will look like below:

Product_id issue_date action_date Force_date
1 10/10/2012 10/10/2012 10/10/2012
2 10/10/2012 10/10/2012 10/10/2012
3 10/10/2012 13/10/2012 15/10/2012
[code]....

Need the data like

Issue_date count_action_date count_Force_date (diff(action_date,force_date) 1 2 3 4 5 6(days since over)

10/10/2012 3 4 1 4 2 1 0 0

How to get the data like this? automatically how to get 123.... and how to calculate the difference by which day the count of difference is going?

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Jun 14, 2012

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Aug 10, 2003

I want to calculate time difference b/w two time clocks, just like we calculate the date difference and answer is in days, In the same way i like to have answer in hrs,min and ss.

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Feb 28, 2013

I need to subtract multiple numbers to get the difference between each numbers, the amount of numbers to subtract between each others can vary between 2 and 10.

create table table_1 (
col1 varchar2(5),
col2 number(6,0),
col3 number(12,0)
)

insert into table_1 values ('AAA', 34379, 11111);
insert into table_1 values ('AAA', 39032, 11111);
insert into table_1 values ('AAA', 54337, 11111);
insert into table_1 values ('AAA', 78005, 11111);

insert into table_1 values ('AAA', 66793, 22222);
insert into table_1 values ('AAA', 74323, 22222);
insert into table_1 values ('AAA', 81426, 22222);

Expected Output:

col1 col2 col3
AAA -4653 11111
AAA -15305 11111
AAA -23668 11111
AAA -7530 22222
AAA -7103 22222

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Nov 29, 2011

during application migration, i got one table from MS Access, and have situation where two events are splited into 4 columns (start: date1 time1 and stop: dat2 and time2). How to properly calculate duration between these two events, and show it in format: hh:mi ?

CREATE TABLE ACCBTP_DCZASTOJ
(ID NUMBER(11,0),
NASTANAK_KVARA DATE,
VRIJ_NASTKVARA DATE,
VRIJEME_OPRAVKE DATE,
DATUM_OPRAVKE DATE);

[Code]....

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Jan 10, 2012

how to calculate the difference between multiple dates at the same time..

Select to_date('10/10/2011','mm/dd/yyyy')
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- to_date('08/10/2011','mm/dd/yyyy') from dual;

Giving me an error...

ORA 00932 : inconsisten data types:expected DATE Julkian got DATE..

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Sep 22, 2011

create or replace function getDate(p_joing_date Date,p_sysdate)
Return Date;
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v_compltd_mnths;
BEGIN
SELECT into v_compltd_mnths MONTHS_BETWEEN(TO_DATE('sysdate','MM-DD-YYYY'), TO_DATE('joing_date','MM-DD-YYYY') ) "Months"FROM DUAL;
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that i have worte..

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Aug 15, 2011

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e.g.

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Sep 28, 2011

SELECT emp.emp_id, emp.ename, TRUNC (attendancedatetime),
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)
/ 60
)
- 60

[Code]...

Following is the out put:

EMP_IDENAME Date Min Hrs
10013Javed Iqbal09/20/2011 00:00:0036.007.00
10013Javed Iqbal09/21/2011 00:00:0027.007.00
10013Javed Iqbal09/22/2011 00:00:0049.007.00
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i need the TOtal sum of Minutes and Hrs e.g 7+7+7=21 and also minutes but if minutes total increase from 60 minutes then it should add to hrs .how to get sum.

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Dec 15, 2011

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my tables are:

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empname varchar(20),
address varchar(20),
no_of_dependents number(5),
deptno number(5),
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I was reading the documentation for oracle 11gr2, with reference to URL>.....

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if that is the case, then number of hash keys should be 1900/55 = 34.55 which should have rounded up to 35.

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Mar 11, 2010

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custname custid custage
aaa 1 23
bbbbb 2 22
cccc 3 45
dddddddd 4 21

but i need to left align the custid and custage.

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insert into test values ('d10','d20','d30');

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[Code]....

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Feb 17, 2012

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why I got 1079 records.how to show the two extra columns in the out put whcich are not used in GROUP BY clause.

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GR
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3.>> Comparing string and number >>
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RESULT
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