Security :: Date With Timestamps In TRIGGER?
Jun 29, 2012
i m creating the dynamic table every month to maintain the particular month data seperately .when the records getting inserted in the table,trigger will automatically insert the records in the dynamic table. only date alone(without timestamp) getting inserted in the dynamic table from staging. so by default ,00:00:00 is getting appended with date instead of actual timestamp. tried select to_date(to_char(:new.ACTN_DATE,'YYYY-MM-DD HH24:MI:SS'),'YYYY-MM-DD HH24:MI:SS') INTO v_temp_actn_date from dual; but i am getting only date alone . in my table and dynmaic table datatype for date column is date
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Nov 7, 2013
How to write and fire trigger if any database object will created?
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Mar 10, 2011
We are trying to audit a view. We are currently seeing that view in production instance and trying to find out when exactly it got invalidated.
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Feb 1, 2013
In the attached PRE-INSERT Trigger Code i need to do a validation. Validatation: If the Date range is between FEB 6 2013 and MARCH 1 2013 then the code should work ELSE It should throw a message
Check_Product_Title;
Begin
if :prod2.internet_product_flag = 'Y' and :prod2.brnd_code is null then
soft_messages('E',TRUE,'Brand code must be specified for CCH online product');
end if;
[Code]....
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Mar 5, 2008
How to select the transactions out of the database that occurred within 70 seconds of each other. The toll_date field is a TIMESTAMP field.
Problem is, I seem to only get transactions that occurred within 70 minutes of each other. On the timestamp field I break the math down into the seconds in a day and I add 70. I then subtract that value and add that value to the timestamp and I should get anything between those values right?
SELECT Acct_ID, Ln, Tag_Rd, COUNT(*)
FROM (
SELECT T1.Account_ID Acct_ID, T1.Tag_Read Tag_Rd, T1.Revenue_Date Rev_Date, T1.Toll_Date, T1.Lane_ID, T1.Plaza, T1.Lane Ln, T2.Toll_Date, T2.Plaza, T2.Lane
FROM Toll T1
JOIN Toll T2
[code]......
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Apr 24, 2013
I ran this following query and somehow i feel the results are wrong.
SQL> select to_char(starttime,'dd-mm-YYYY hh24:mi:ss') from report where dateofmonth between to_timestamp_tz('22-Apr-2013 12:00:00','dd-mm-YYYY hh24:mi:ss') and to_timestamp_tz ('23-Apr-2013 14:00:00','dd-mm-YYYY hh24:mi:ss');
TO_CHAR(STARTTIME,'
-------------------
23-04-2013 22:43:59
23-04-2013 13:43:37
SQL> select to_timestamp_tz(starttime,'dd-mm-YYYY hh24:mi:ss') from report where dateofmonth between to_timestamp_tz('22-Apr-2013 12:00:00','dd-mm-YYYY hh24:mi:ss') and to_timestamp_tz ('23-Apr-2013 14:00:00','dd-mm-YYYY hh24:mi:ss');
TO_TIMESTAMP_TZ(STARTTIME,'DD-MM-YYYYHH24:MI:SS')
---------------------------------------------------------------------------
23-APR-13 10.43.59 PM -07:00
23-APR-13 01.43.37 PM -07:00
I am not sure why the 10 PM time is coming up in the result.
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Oct 5, 2012
My control file
LOAD DATA
Infile code_val_new.dat
BADFILE sample.bad
DISCARDFILE sample.dsc
[code]...
18 7BLM AGS ELEM BUILDING MATHEMATICAL EXTRACT Oct 4 2012 3:19:26:000AM
But all rows are getting rejected. What modification should i make?
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Jul 8, 2013
view the below select statement..why it's adding extra zero's...
select to_timestamp('2001-05-22 12:00:18.600','YYYY-MM-DD HH:MI:SS.ff3AM') from dual
output: 5/22/2001 12:00:18.600000000 PM ---why it's adding extra zeors's
my output should be as " 5/22/2001 12:00:18.600 PM"
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Jan 4, 2013
Have a table which has 3 columns id,name,time where time is of datatype timestamp and it stores the time when the row was inserted. Need an query which accepts 2 parameters as input Ex: Start_Time,End_Time and all the rows in between the above mentioned timestamps must be deleted.
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Nov 28, 2010
I have a requirement to create the table monthly.The tablename will be the o/p of the following select query.
select 'TBR_AND_SECURITY_DEPOSIT'||TO_CHAR(sysdate, 'Monyy') from dual
I am trying to create the table for creating the backup of the existing table TBR_AND_SECURITY_DEPOSIT.
create table (select 'TBR_AND_SECURITY_DEPOSIT'||TO_CHAR(sysdate, 'Monyy') from dual) select * from TBR_AND_SECURITY_DEPOSIT
but it is throwing error as invalid table name.
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Jan 2, 2013
I have to create the following table. The fields Trend_Date, Price and Trend are already given. I have to calculate the field permanently and to insert the value in this permanent table.
Fields:
The field price belong to the value of a product during the trade.
The field trade_date belongs to the moment of the trade.
The field trend belongs to the future behavior of the the price. Here, the price of the present moment is compared to the following price (possible characteristics: 'UP', 'DOWN', 'STABLE').
The field permanently belongs to the time (in seconds) how long the value of the field Trend_Date (depending on the price) is still true.
For example:
Row 1: The trend in row 1 is 'UP' and it has a price of '11'. Until row 3 this remains true (the price is greater or equal to 11). In this case, the difference between row 1 and row 3 are 9801 (rounded) seconds.
Row 2: The trend in row 2 is 'DOWN' and it has a price of '12'. This remains true till to the end (the price is never greater than 12) In this case, the difference between row 2 and row 11 are 97346 (rounded) seconds. To calculate the 97346 seconds the field has to consider that between row 2 and row 11 are two days. There will be no trade between 18:00 and 07:00 o'clock. This belongs to 7 hours for each days, in seconds (2*46800) 93600.
-> 190945-93600 = 97346s
Row 6: The trend in row 6 is 'UP' and it has a price of '5'. This remains true till to the end (the price is never smaller than 5) In this case, the difference between row 6 and row 11 are 65729 (rounded) seconds. To calculate the 65729 seconds the field has to consider that between row 65729 and row 11 are one days. There will be no trade between 18:00 and 07:00 o'clock. This belongs to 7 hours for each days, in seconds (1*46800) 46800.
-> 112528-46800 = 65729s
Row 9: The trend in row 9 is 'STABLE' and it has a price of '8'. Until row 10 this remains true (the price is equal to 8 ). In this case, the difference between row 9 and row 10 is 14418 (rounded) seconds.
Row 11: Is empty because there are no values to compare.
Example Table
TRADE_DATE --PRICE --TREND --permanently
02.01.13 11:21:42,720000000--11--UP--9801
02.01.13 12:44:03,236000000--12--DOWN--97346
02.01.13 14:05:03,845000000--11--DOWN--92485
[Code]....
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May 26, 2010
Check the following
SQL> ALTER SESSION SET TIME_ZONE = '-01:00';
Session altered.
SQL> SELECT SysTimeStamp
2 FROM dual;
SYSTIMESTAMP
[code]...
My doubt is that why Extract(HOURFROMCURRENT_TIMESTAMP) is showing 11 rather than 10?
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Mar 8, 2013
When i run the below query the time zone changes from -06:00 to +06:00 How do i retain the same time zone -06:00? Im in Oracle 10GR2.
select TO_TIMESTAMP_TZ('2013-03-08'||'03:00:00.000-06:00','YYYY-MM-DD HH24:MI:SS.FF TZH:TZM') from dual;
Output: 3/8/2013 3:00:00.000000000 AM +06:00
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Mar 23, 2011
SQL statement:
SELECT user_id, jc_name, upd_time
FROM user_jc
WHERE user_id IN ( SELECT user_id
FROM user_jc
WHERE JC_NAME LIKE 'PPF\_S\_%' ESCAPE ''
[code]...
Howto have for every User_ID and the jc_name with the oldest upd_time? One record for each user_id with the oldest timestamp and the jc_name?
the output should be like this
UII00061PPF_S_SD_1st21.09.2010 19:23:46
UII00012PPF_S_Munich22.03.2011 15:44:20
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Jan 11, 2007
I'm trying to generate count of the number of entries in a table for each day.The problem is the date column is of datatype timestamp and looks like this "2006-12-30 18:42:03.0"
How would I generate a report of number of entries in the table for each date (I'm not intrested in the "time" only the "date" i.e YYYY-MM-DD)?
SELECT COUNT(*) FROM my_table_name
WHERE my_date_column LIKE '2006-12-30%'
GO
It returns zero rows ( and I kno there are rows in the table) I'm using Oracle 10g.
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Sep 25, 2012
how to caluclate days between two dates of single timestamp filed and with this
query
Select * from m_activity_transaction where actn_opp_id in (
Select actn_opp_id from m_activity_transaction where ACTN_ACTV_ID = 218
Group by actn_opp_id
[code]...
and i nedd to caluclate no.of days between two dates like 27-JAN-12 11.06.20.000000 AM and 08-FEB-12 05.32.54.000000 PM where actn_id is unique AND ACTN_OPP_ID IS NOT UNIQUE.
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Apr 28, 2008
Is there a way to query an oracle database in an automated fashion by a timestamp field based on current timestamp, like: 04/29/08 00:00:00 - 72 hours?
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Apr 12, 2011
calculating values of A & B
(both these fields a_std and a_time are coming as varchar from the parent table in a cursor.(basically they are time period and actual arrival time respectively)
i was juggling with the attempt to make varchar to timestamp or date..but caught with Round up /Round down)
Formula ->
A = Round down [A_TIME - A_STD]
B = Round up [A_TIME) - 10 minute + A_STD]
where
A_TIME VARCHAR2(8) N Time (Format" HH:MM AM/PM") eg "3:50 PM"
A_STD VARCHAR2(5) N Standard time (Format" HH:MM") eg "1:00"
Allowed values for A & B after round up/down = multiple of 10 ( 11:00,11:10,11:20 etc.)
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Jun 11, 2010
I have 3 tables, user_login_event, person and resource_viewed_event. What I want to do have a report for each month, users logged in our application and then show for each month, how many records were created in table person and how many resource views events were logged in resource_viewed_event.
Lets only worry about the timestamp fields in these tables now as I want to use them to join the tables together or at least build correlated subqueries along the months. I have tried several options, all not leading to a desired result:
Left outer join. Works but its incredibly slow:
SELECT
distinct to_char(ule.TIMESTAMP,'YYYY-MM') as "YYYY-MM",
count(distinct ule.id) as "User Logins",
count(distinct ule.user_id) as "Users logged on",
count(distinct p2.id) as "Existing Users",
count(distinct p1.id) as "New Users",
count(distinct r1.id) as "Resources created"
[code]....
Tried the same with left outer joins of temporary tables created through select statements:
select
distinct ule.month as "Month",
count(distinct p1.user_id) as "Users created",
count (ule.id) as "Logins",
count (distinct ule.user_id) as "Users logged in",
count(rv.id) as "Resource Views",
count(distinct rv.resource_id) as "Resources Viewed"
[code]....
Tried the same with left outer joins of temporary tables created through select statements:
select
distinct ule.month as "Month",
count(distinct p1.user_id) as "Users created",
count (ule.id) as "Logins",
count (distinct ule.user_id) as "Users logged in",
count(rv.id) as "Resource Views",
count(distinct rv.resource_id) as "Resources Viewed"
[code]....
another approach is to create my own temporary tables using select statements and create fixed Month values which I can use to directly link the sets together.
select
distinct ule.loginday as "Month",
count(distinct ule.id) as "Logins",
count(distinct ule.user_id) as "Users logged in",
count(distinct p1.user_id) as "Users created",
count(distinct p2.user_id) as "Existing users1"
[code]....
performance is OK with 2 tables but the example above takes forever to execute.
Tried an approach with union but this creates new rows for each table
SELECT DISTINCT p1.MONTH AS "Month",
COUNT(DISTINCT p1.user_id) AS "Users created",
NULL AS "Logins",
NULL AS "Users Logged in",
NULL AS "Resource views",
NULL AS "Resources viewed"
FROM (SELECT To_char(person.created_on_date, 'YYYY-MM') AS MONTH,
[code]....
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Dec 26, 2012
Any documentation supporting Oracle 11G and Advanced Security stating encryption at rest is FIPS 140-2 compliant?
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Nov 16, 2010
Lost Windows password? Forgot Windows password? Your PC was hacked? Therefore, it is a basic step for every Windows users to enhance the security of Windows password. In the networks, it is found that a number of user's passwords are easy to guess. Only the smallest groups are the most security conscious and select passwords that are mixed lowercase and uppercase letters, numbers and punctuation to create cryptic passwords. Adopting strong password is one of the most effective ways to ensure system security. Here are several methods for you to enhance the security of your passwords in Windows 7/2000/XP/Vista and so on. You'd better remember the methods below unless you want to reset Windows password from time to time.
1. Is random password a great password?
A common myth is that totally random passwords like Ht3&e#L%5d@$B are the best passwords. This is not true. While they may be strong passwords, they are usually difficult to remember, slow to type, and sometimes vulnerable to attacks against the password generating algorithm. It is easy to create passwords that are strong but much easier to remember by using a few simple techniques. For example, consider the password "Luck-73@Better?". This password utilizes uppercase and lowercase letters, two numbers, and three symbols. The password is 15 characters long and can be memorized with very little effort. Moreover, this password can be typed very fast. The portion"Luck" and "Better" alternate between left and right-handed keys on the keyboard, improving speed, decreasing typos, and decreasing the chances of someone being able to discover your password by watching you.
2. Create the long Windows password
Although a password may eventually be discovered through some means, it is possible to create a password that cannot be cracked in any reasonable time. If a password is long enough, it will take so long or require so much processing power to crack it. That is essentially the same as being unbreakable (at least for most hackers).
3. Create the Windows password constantly?
This may be good advice for some high-risk passwords, but it is not the best policy for every user. It is frustrating for a user to have to constantly think of and remember new passwords every 30 days. It may be better to focus on stronger passwords and better user awareness rather than limiting password age. A more realistic time for the common user may be 90-120 days.
4. Write down Windows password in a proper place
Sometimes it is necessary for some users losing and forgetting complex passwords easily to write down them somewhere proper. However, it is important to educate users on how to write down passwords properly. Obviously, a sticky note on the monitor is not a good idea, but storing passwords in a safe or even a locked cabinet may be sufficient.
5. 14 characters is the optimal password length
Each character that you add to your password increases the protection. Your passwords should be 8 or more characters in length; 14 characters or longer is the Optimal Password Length. Many systems also support use of the space bar in passwords, so you can create a phrase made of many words. It is not easier to forget and lose, as well as longer than a simple password, and harder to guess.
6. Try not to use the same Windows password for all accounts
Some users always make the same passwords for every account to make it easy to remember. In that case, when any one of them lost, your other information protected by that password will be in danger as well. It is serious to use different passwords for different systems and accounts.
7. Do not use some common words that other users maybe guess
Most of users prefer to use some common words to remember easily, for example, login name, birth date, driver's license, passport number, pets' name and other words contained their personal information someone knows. In that case, your Windows system will not be safe anymore. Moreover, do remember not to use some words spelled backwards, abbreviations, sequences or repeated characters and adjacent letters, such as, asdfgh, 123456, 888888, abcdef and so on.
You can smoothly use your Windows now because the strong and powerful Windows password is created successfully, Certainly, I believe that many users lost Windows password and forgot Windows password, then you need have to reset Windows password or recover Windows password. It is a big problem for plenty of Windows users that how to reset Windows password. how to recover Windows password and they are puzzled by resetting windows password, for instance, reset Windows 7 password, recover password Windows XP, remove Windows Vista password and other operating systems after they create the password with complex letters, numbers and symbols. However, it is unnecessary to worry and it is said that things will eventually sort themselves out. There are many ways to reset forgotten Windows password, including use windows password reset disk and windows password reset software, like Super Windows Password Reset, a professional windows password reset software which could enable you to logon to Windows smoothly without reinstalling system.
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Aug 18, 2010
In Sybase, my application was using system tables to perform application login security. Those tables obviously don't exist in Oracle. I am looking for ways to provide the following functionality in an Oracle world:
1. How to determine 'x' days of inactivity based on "last login date"?
2. How to determine when a new user logs in for the first time and force them to change their password?
3. If we need to reset a users password, how can we require the user to change their password?
4. Is there any other option other than storing a user-id/password in the application code for locking a user's account if their account needs to be locked due to inactivity?
5. In the USER_USERS view there is a status column. What the different status's can be?
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Jun 1, 2010
I'm trying to work out how to take a table like this:
IDDate
12502-Feb-07
12516-Mar-07
12523-May-07
12524-May-07
12525-May-07
33302-Jan-09
33303-Jan-09
33304-Jan-09
33317-Mar-09
And display the data like this:
IDPeriodPeriod StartPeriod End
125102-Feb-0702-Feb-07
125216-Mar-0716-Mar-07
125323-May-0725-May-07
333102-Jan-0904-Jan-09
333217-Mar-0917-Mar-09
As you can see, it's split the entries into date ranges. If there is a 'lone' date, the 'period start' and the 'period end' are the same date.
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Apr 12, 2010
I have a two date fields in my form; valid from date and expiry date.
Currently my valid from date has an inital value property of $$date$$ which automaitcally brings up todays date.
I need my expiry date to automatically show a date 15 years after this date?
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Jul 27, 2010
Provide me the script which would track all the users security violations like ... say for example i want to find which users logged in and what he did in database prospective.
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Nov 29, 2012
how can we mask value of some columns in table? For example: user A is supervisor, he can query salary column in employee table, but for user B, he is staff member, he can query salary column but system just shows ***** or something like that for salary value.
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Jul 1, 2011
How can "call one trigger of item in trigger of form"
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Dec 3, 2012
Version : 4.1.1, I have a tabular form on a DB table. One of the columns is a date field. When the user hits the "add Row" button on the tabular form, I want the Date field to be defaulted to sysdate. Here is what I have tried so far,
1. Created a "hidden" item P1_SYSDATE and populated the default value with sysdate. After this, under the DB tabular report date field, I used default type - Item/application on this page and entered P1_sYSDATE
2. Instead of populating the default value of the P1_SYSDATE hidden item, I created a before regions process and added
:P1_SYSDATE := sysdate
and added P1_SYSDATE to default type of the tabular date field with default type as "ITem/application on this page.
I get the error
ORA-01790: expression must have same datatype as corresponding expression
I tried to_Char(sysdate,'dd-mon-yyyy') and then converting it back to to_date. still no luck.
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Oct 24, 2013
When I run a query form the the Query Window in Visuial Studios 2012 all the date fields truncated to 'mm/dd/yyyy', but i need the full date returned. I am able to get full date from TO_char(MyDateField, 'yyyy-mm-dd hh24:mi:ss'), but if I do TO_DATE(MyDateField, 'yyyy-mm-dd hh24:mi:ss') it only returns 'mm/dd/yyyy'. I'm sure this is a simple setting in Visual studios but I cant find it to save my life. Is there there a way to have the full date returned by default?
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Sep 27, 2007
I am a novice in oracle
I have 2 columns in my table
->Col1 with experience in years entered as an integer
->Col2 with current date
I need to add another column as a date value adn for that i need to subtract Currentdate-Col1 when i tried currentdate-Col1 it just subtracted the days i need the formula to subtract years and give a date
I have worked in DB2 and all u need to do there was add the keyword years at the end but in oracle the same does not work
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