Leap Year Date Issue (Example provided)
Please see link for document:
Dropbox - DateIssue.pdf
When the FROM dropdown has 02/28/2016 selected, the date for Monday throws an error.
The rest of the dates change but Monday does not change because it is supposed to be 02/29 (See Screenshot). Any help is appreciate! Thanks!!
First I would cleanup any errors that appear in the JavaScript console:
DeptData[cDeptName] is undefined
5:AcroForm:FromDate:Keystroke
TypeError: DeptData[cDeptName] is undefined
5:AcroForm:FromDate:Keystroke
I would next learn how to convert a date string to the date object.
Since you are formatting using the util.printd method, you can leave the days of the week field format to None.
The setFieldValues function:
function SetFieldValues(cDeptName)
this.getField("ToDate").value = event.target.value;
The custom calculation for the days of the week:
var f = this.getField("txtDate1");
// var d = new Date(f.value);
// convert the date value with format to date object;
var d = util.scand("mm/dd/yyyy", f.value);
// advance by 1 day;
d.setDate(d.getDate()+1);
event.value = util.printd("mm/dd", d);
Adjust the number of days to advance as needed.
Assuming that JavaScript is going always interpret the date string format may not always be a good approach.
Similar Messages
-
Loading leap year date using SQL*Loader
Hello,
I have a problem loading a date '29/02/2000' using SQL*Loader. This date is on a leap year. I'm getting an error message from SQL*Loader as 'ORA-01839: date not valid for month specified'. My colleague and I have tried using various date functions to convert the data into date, but no luck.
I would appreciate any helps,
BruceThanks for your help, I found the bug on my control file. I was using the RTRIM function to remove bad timestamp such as '29/02/2000 0:00:00'. So instead of using this statement:
LOG_DATE DATE "DD/MM/RRRR" "RTRIM(:LOG_DATE,'0:00:00')"
I was using the statement below with a space before the '0:00:00' string literal, with the intention to remove a space also:
LOG_DATE DATE "DD/MM/RRRR" "RTRIM(:LOG_DATE,' 0:00:00')"
Well, it turned out that if there was a space before the string literal, RTRIM function would trim the matching string plus any '0' characters from the right, including the '000' that belongs to '2000'. Thus, the error.
Thanks again,
Bruce -
IPhone Leap year calendar issues anyone?
Is anyone having a problem with scheduling meetings in outlook for February 29th and then they show up on the iPhone as March 1st instead? Is there something that is not configured right because of leap year?
I have a reoccurring appoint that happens every wednesday. But I noticed yesterday that it put that appointment on Tuesday - Feb 28th instead of Feb 29th. When you open up the event it says:
"Tuesday, 28 Feb, 2012
from 6:30 PM to 8:30 PM
repeats every week on Wednesday"
I'm quite sure I didn't add a single instance of this event on the Tuesday. I'm a software developer and took a guess that the weekly event falling on Feb 29th of a leap year was the most likely culprit of this bug.
Note that I used the iCloud control panel to push out my events from Outlook to iCloud a couple of weeks ago. So the logic error could be in that software. Or in Outlook, or in Google (as I used to synch my google calendar to outlook before going to iCloud). But I doubt it was caused by Outlook or Google calendar. No proof of that, just my gut feeling. -
Parser cannot handle leap years? simpler example
Here is a simpler example. If you change the <aDate> to 2001-02-29 the instance document still validates. However 2001-02-30 fails, as it should and 2001-02-28 validates, as it should.
The schema, d.xsd:
<?xml version = "1.0" encoding = "UTF-8"?>
<xsd:schema xmlns:xsd="http://www.w3.org/2000/10/XMLSchema"
targetNamespace="http://localhost/d"
elementFormDefault="qualified">
<element name="example">
<complexType>
<sequence>
<element name="aDate" type="xsd:date"></element>
</sequence>
</complexType>
</element>
</xsd:schema>
and the instance document d.xml with an incorrect date:
<?xml version = "1.0" encoding = "UTF-8"?>
<example xmlns="http://localhost/d"
xmlns:xsd="http://www.w3.org/2000/10/XMLSchema-instance"
xsd:schemaLocation="http://localhost/d
http://localhost.emv2.com/1914/d.xsd">
<aDate> 2001-02-29</aDate>
</example>
I used the XSDSample.java file provided with the xdk9 for validation.
Any ideas?
Thanks,
-StephenI've tried 9.0.2 beta and I get the same behaviour.
I can also reproduce the same behaviour by taking the catalogue.xml and cat.xsd example in the xdk/demo/java/schema directory.
I changed the Date element type from "year" to "date" and modified the <Date> elements in catalogue.xml appropriately (e.g. 1954-02-28 instead of just 1954).
I get the same behaviour, Feb. 29 always passes validation no matter what year, Feb. 30 always fails:
Here's what it says when it fails on Feb. 30
file:/tmp/xdk92/xdk/demo/java/schema/catalogue.xml<Line 14, Column 40>: XSD-2025: (Error) Invalid text '1977-02-30' in element: 'Date'
Parser Exception: Invalid text '1977-02-30' in element: 'Date'
I'm using the Linux versions, if that makes a difference.
Also, I've tried referencing the xml instance two ways, via a file reference /home/ssimm...
and via a URL http://localhost/.... The behaviour is consistent. (I don't think this should make a difference, I'm just trying to get the behaviour to change)
Any ideas?
Thanks,
-Stephen -
How to handle leap year date in the query ?
Dear Exparts,
Hope you are fine.
Here is my banner,
Oracle Database 10g Enterprise Edition Release 10.2.0.3.0 - ProdI have a date field name birth_date where i store date of birth. I need to pick date of birth in a given range. For example
BIRTH_DATE
01-FEB-1980
12-FEB-1984
29-FEB-1992
01-MAR-1986
01-APR-1986
22-APR-1988
01-MAR-1992My searching parameter is
brith_date between 01/02/13 and 01/04/13I want the record
01-FEB-1980
12-FEB-1984
29-FEB-1992
01-MAR-1986
01-APR-1986Here is the script
create table
CREATE TABLE DATE_OF_BIRTH
BIRTH_DATE DATE
);data insert
insert into DATE_OF_BIRTH
values(to_date('01-FEB-1980','DD-MON-RRRR'));
insert into DATE_OF_BIRTH
values(to_date('12-FEB-1984','DD-MON-RRRR'));
insert into DATE_OF_BIRTH
values(to_date('29-FEB-1992','DD-MON-RRRR'));
insert into DATE_OF_BIRTH
values(to_date('01-MAR-1986','DD-MON-RRRR'));
insert into DATE_OF_BIRTH
values(to_date('01-APR-1986','DD-MON-RRRR'));Thanks in advance... (thanks to Frank Kulash)
ask2Learn
Edited by: Asked to Learn on Feb 24, 2013 7:40 PMHi,
Asked to Learn wrote:
Dear Exparts,
Hope you are fine.
Here is my banner,
Oracle Database 10g Enterprise Edition Release 10.2.0.3.0 - ProdI have a date field name birth_date where i store date of birth. I need to pick date of birth in a given range. For example
BIRTH_DATE
01-FEB-1980
12-FEB-1984
29-FEB-1992
01-MAR-1986
01-APR-1986
22-APR-1988
01-MAR-1992My searching parameter is
brith_date between 01/02/13 and 01/04/13I want the record
01-FEB-1980
12-FEB-1984
29-FEB-1992
01-MAR-1986
01-APR-1986Here is the script
create table
CREATE TABLE DATE_OF_BIRTH
BIRTH_DATE DATE
);data insert
insert into DATE_OF_BIRTH
values('01-FEB-1980');
insert into DATE_OF_BIRTH
values('12-FEB-1984');
insert into DATE_OF_BIRTH
values('29-FEB-1992');
insert into DATE_OF_BIRTH
values('01-MAR-1986');
insert into DATE_OF_BIRTH
values('01-APR-1986');
You defined the column as a DATE, which is correct, but you're trying to insert VARCHAR2 values, such as '01-FEB-1980' into that column.
Use TO_DATE to convert a VARCHAR2 into a DATE:
insert into DATE_OF_BIRTH (birth_date)
values ( TO_DATE ('01-FEB-1980', 'DD-MON'YYYY'));or use DATE literals:
insert into DATE_OF_BIRTH (birth_date)
values ( DATE '1980-02-01);The same goes for queries. -
Leap year bug adding days to a date?
Please try this xquery:
<dd>
for $i in 0 to 4
return
<d>
{xs:date('2008-12-29')+xs:dayTimeDuration(concat('P',$i,'D')) }
</d>
</dd>
If the starting date is a leap year date (eg: 2008-12-29), the result does not contain the 30rd of december!
Is this a (known) bug or there is something wrong in my xquery?
Thanks!
Bye
MirkoHi Mirko,
There's nothing wrong with your query - that looks like a bug in our date displaying code. Thanks for your analysis of the problem - I'll take a look into fixing it.
John -
Is there a way of checking that an entered leap year date is valid?
Eg. User enters 2/29/2001 which is not a valid date because 2001 was not a leap year.You should be able to use the SimpleDateFormat class to do this by issuing a setLenient(false) call. However, it does not seem to detect a leap year problem with the date 2/29. If you run the code below, you'll see that in the first section it correctly detects that 2/30 is a bad date, but in the second section, it does not detect 2/29 as a bad date. In section 3, you can see how an added test can detect the problem date.
//Date validation
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
//Section 1 - correctly detects bad date
try {
String date = "2001-02-30";
System.out.println("Here's the date - " + sdf.parse(date));
sdf.setLenient(false);
System.out.println("Here's the date - " + sdf.parse(date));
} catch (ParseException pe) {
System.out.println("Bad date");
//Section 2 - does not detect bad date
try {
String date = "2001-02-29";
System.out.println("Here's the date - " + sdf.parse(date));
sdf.setLenient(false);
System.out.println("Here's the date - " + sdf.parse(date));
} catch (ParseException pe) {
System.out.println("Bad date");
//Section 3 - correctly detects bad date with additional check
try {
String date = "2001-02-29";
System.out.println("Here's the date - " + sdf.parse(date));
sdf.setLenient(false);
//Additional test to compensate for leap year failure
String dateString = sdf.format(sdf.parse(date));
if (!dateString.equals(date)) {
System.out.println("Bad date");
} else {
System.out.println("Here's the date - " + sdf.parse(date));
} catch (ParseException pe) {
System.out.println("Bad date");
} -
Hi All,
I just want to get the exactly one year previous date by using the below query: for the leap year...
update day
SET FLAG = 'Y'
WHERE PERIOD_DATE BETWEEN TRUNC(ADD_MONTHS(TO_DATE(20050228,'YYYYMMDD'),-12),'MON') and
ADD_MONTHS(TO_DATE(20050228,'YYYYMMDD'),-12)
In the same time while we use the same query for the leap year date:
select * from day
WHERE PERIOD_DATE BETWEEN TRUNC(ADD_MONTHS(TO_DATE(20080229,'YYYYMMDD'),-12),'MON')
AND TO_DATE(20080229,'YYYYMMDD') + NUMTOYMINTERVAL(-1,'YEAR')
I receive the following error - ORA-01839: date not valid for month specified.
Problem:FLAG was incorrectly set to 'Y' for Feb 29 in 2004 during the report processing of Feb 28 data. Because of this, the extra day of Feb 29 2004 of Sales data was incorrectly being added to a report, thus overstating MTD sales of Feb 28 2004 by one full day. This appears to be caused by the fact that there were 28 days in Feb 2005 and 29 days in Feb 2004 becuase of the leap year. This will not cause problems until the next leap year.
Can you please give me the resolution of the above.
Thanks & Regards,
Satheesh
Message was edited by:
user565141
Message was edited by:
user565141Hi Sateesh,
I love your question, I have been writting quite often about this.
ADD_MONTH is not the correct function to use, neither y2m interval.
First, you need to define what is 2004-02-29 + 1 year. In most laws I found on the net (like majority and retirement policies), a leap-year baby will have his birthday on March 1st in non-leap years. If you have a contract, you must define this too.
ex:
update day set flag='Y'
where period_date between
trunc(ADD_MONTHS(:d,-12),'MON') and
case to_char(:d,'MMDD')
when '0228' then add_months(trunc(:d,'Y'),-12)+58
else add_months(:d,-12) end;another approach would be
where to_number(to_char(period_date,'YYYYMMDD')) between
to_number(to_char(:d,'YYYYMM"00"'))-10000 and
to_number(to_char(:d,'YYYYMMDD'))-10000;but if you have an index on period_date, the first solution may be more efficient
Message was edited by:
Laurent Schneider
58 not 59
Message was edited by:
Laurent Schneider
depending on how you define 29Feb -1Y, you could add
when '0229' then add_months(trunc(:d,'Y'),-12)+59 -
Finding a function to get leap year
Hi everybody!
I need a function or code to find if a year is leap year or normal year.
Thanks a lot of!
DMHere is some code that I had in my bag of tricks.
report zrich_0001 .
parameters: p_datum type sy-datum.
* check if it is a leap year.
data: march1 type sy-datum.
data: last_day_feb type sy-datum.
march1 = p_datum.
march1+4(4) = '0301'.
last_day_feb = march1 - 1.
if last_day_feb+6(2) = '29'.
write:/ 'This is a leap year'.
endif.
Regards,
Rich Heilman -
How to do a date validation with leap years
I'm doing a date validation program in my Java class, and well it's pretty hard (for me that is). I have to be able to type in a date, have it say whether it's a leap year or not, and print out the number of days in the month. It seems pretty straight forward, but I get confused on trying to do the 'if else' statements and even the simplest things like getting the day prompting to work. >< The years I'm doing only goes through 1000 to 1999, so that's why those numbers are there. The program isn't complete, so if anyone could help show me what I'm doing wrong in the areas I'm working on then I'd appreciate it...and I'm still kind of in the basics of Java so if you do hint me with some code then I'd appreciate it if it was stuff that's not too advanced so yea.
// Dates.java
// Determine whether a 2nd-millenium date entered by the user
// is valid
import java.util.Scanner;
public class Dates
public static void main(String[] args)
int month, day, year; //date read in from user
int daysInMonth; //number of days in month read in
boolean monthValid, yearValid, dayValid; //true if input from user is valid
boolean leapYear; //true if user's year is a leap year
Scanner scan = new Scanner(System.in);
//Get integer month, day, and year from user
System.out.print("Type in the month: " );
month = scan.nextInt();
System.out.print("Type in the day: " );
day = scan.nextInt();
System.out.print("Type in the year: " );
year = scan.nextInt();
//Check to see if month is valid
if (month >= 1)
month = month;
else
if (month <= 12)
month = month;
else;
//Check to see if year is valid
if (year >= 1000)
year = year;
else
if (year <= 1999)
year = year;
else;
//Determine whether it's a leap year
//Determine number of days in month
if (year == 1 || 3 || 5 || 7 || 8 || 10 || 12)
System.out.println (Number of days in month is 31);
else (year == 4 || 6 || 9 || 11)
System.out.println (Number of days in month is 30);
//User number of days in month to check to see if day is valid
//Determine whether date is valid and print appropriate message
// Dates.java
// Determine whether a 2nd-millenium date entered by the user
// is valid
import java.util.Scanner;
public class Dates
public static void main(String[] args)
int month, day, year; //date read in from user
int daysInMonth; //number of days in month read in
boolean monthValid, yearValid, dayValid; //true if input from user is valid
boolean leapYear; //true if user's year is a leap year
Scanner scan = new Scanner(System.in);
//Get integer month, day, and year from user
System.out.print("Type in the month: " );
month = scan.nextInt();
System.out.print("Type in the day: " );
day = scan.nextInt();
System.out.print("Type in the year: " );
year = scan.nextInt();
//Check to see if month is valid
if (month >= 1)
month = month;
else
if (month <= 12)
month = month;
else;
//Check to see if year is valid
if (year >= 1000)
year = year;
else
if (year <= 1999)
year = year;
else;
//Determine whether it's a leap year
//Determine number of days in month
if (year == 1 || 3 || 5 || 7 || 8 || 10 || 12)
System.out.println (Number of days in month is 31);
else (year == 4 || 6 || 9 || 11)
System.out.println (Number of days in month is 30);
//User number of days in month to check to see if day is valid
//Determine whether date is valid and print appropriate message
}Here are some helpfull hints for you:
1. Your code is really hard to read, there are two main reasons for this. First, your indentation sucks. Second, you seem to be fascinated with saving two (ok four if you count the shift key) keypresses to avoid using { and }.
2. Not using the brackets (you know { and } which you like to avoid) also is causing your code to do some stuff you don't realize or want to happen, or at least it would be if your code compiled.
3. If statements require arguements, "year == 1" is an arguement, "3" is not an arguement. Each operator like the or operator ("||") is essentially a new if and requires a complete arguement. So the following code peice:
if (year == 1 || 3 || 5 || 7 || 8 || 10 || 12)Literally translates to if year equals 1 or if 3 or if 5 or if 7 or if 8 or if 10 or if 12. Doesn't make much sense in english, and it doesn't make much sense in Java either.
4. I am pretty sure "year" is not the variable you want in the code snippet above (the one used in hint 3), especially considering years 1, 3, 5, 7, 8, 10, and 12 are not between 1000 and 1999. You need to be really carefull not make these kind of mistakes when coding, because they are by far the hardest to track down and fix later since they don't really throw up any flags or anything at compile or run time. Take your time and think thuroughly about each line of code while coding it, it will save you tons of time in the long run.
5. What exactly do you expect statements like "month = month;" to do? That translates as make month equal to month. Do you go to the bank and say " I have exactly $3.56 in my pocket, so I would like to deposite all $3.56 and then withdraw $3.56 and put it back in my pocket"? How do you think the teller would look at you? Teller would probably do it, but the teller would feel like he/she wasted time with you and that you are not really right in the head. Java feels the same way when you make it do the same thing, and you love to do it.
6. Code like the following is all wrong, and for more reasons than pointed out in hint 5.
if (month >= 1)
month = month;
else
if (month <= 12)
month = month;
else;Let's say someone put 13 in as the month. It passes the first check because 13 is greater than or equal to 1. so month which is 13, now gets set to 13 (gee that was effective). Now we hit the else and things get confusing because you didn't use brackets or proper indentation (hint 1) so we don't know what your real intent was. Did you mean else do nothing, and the next if statement is then executed, or did you mean to just run the next if statement if the else condition was met? Fortunatly it doesn't matter here because the next if statement is failed anyways since 13 is not less than or equal to 12.
So, we leave this code with month ebing 13, wait when did we add a 13th month to the calendar? Are you using the Jewish calendar? Could be, except even if I put 1234567 as the month your code would except it as valid, and I know no calendar with that many months. Try writing this in english first and translating it to jave, like i would probably say "if the month is greater than or equal to 1 and less than or equal to 12 then the month is valid." Course now what do you do if it is invalid? Hmm, maybe I would actually say "while the month is less than 1 or greater than 12 ask the user for the month" until they get it right.
There are a few other problems, but most of them are probably things you haven't learned yet, and they are not show stoppers so we will let them fly. You already have a lot of work to do to make this better. But I do have one more really really big usefull hint for you:
Never, ever, under any circumstances, should you ever ask in any way or even hint at asking for someone else to provide code solutions to your problems. So "so if you do hint me with some code then I'd appreciate it if it was stuff that's not too advanced " was a very bad thing to do, but fortunatly for you you followed it with proof you were trying to write the code yourself. Had the code you provided not been so full of problems it was obvious a beginner wrote it, you would probably have gotten much less cordial responses. I would seriously consider avoiding any implication of wanting code, at least until you become a regular poster here and people know you are not just looking to get your homework done for you.
Hope some of this helps.
JSG -
7970G, CM 3.3.3 - leap year issue
Hi,
We`re having a issue with 7970G phones, which are not registering correctly. As its a leap year we expect the issue is related to that. All other phones are OK.
Has anyone else come across this ?
Appload - 70.63.0-1-4
JVM 70.603ES1R4.bin
OS 70.63.0-1.4.sbn
Boot 7970_64060118.binthanks, we got round this by upgrading to the latest firmware. Changing the date on the server would habe probably caused a lot of other issues...at least for us anyway.
Tip for everyone...don`t give the execs latest and greatest :) -
Hi friends,
I have a problem in one of the existing codes.
The cancellation amount is calculated for a range of dates.
Say the active line item has a start date of 01.03.2006 and end date of 01.03.2012 say (B).
Now if a rejection/Cancellation is put on date 01.05.2007 (A), the current program is calculating a value
based on the diffrence of days (B - A).
then this cancelled amnt = the no of cancelled days * (a constn value * 12 / 365).
Now I want to modify my logic in such a way that the leap is also accounted.
If I use the same dates , i realise that 2008 is the leap yr.
So how can i modify my calculation for the range if it contains a leap yr.
Please suggest something on this, as it will be really helpful.
Thanks,
Sapna AgarwalHi Sapna,
try to use the FM /SDF/CMO_DATETIME_DIFFERENCE, it will take care of the leap year, just pass the two dates for which you want the difference.
CALL FUNCTION '/SDF/CMO_DATETIME_DIFFERENCE'
EXPORTING
DATE1 = s_date
* TIME1 =
DATE2 = e_date
* TIME2 =
IMPORTING
DATEDIFF = DATEDIFF
* TIMEDIFF =
* EARLIEST =
EXCEPTIONS
INVALID_DATETIME = 1
OTHERS = 2
IF sy-subrc <> 0.
MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
ENDIF.
if the gap is for more one year than you have to do the calculation year wise only, in this case use the fm '/SAPNEA/JSC_LEAP_YEAR' that which year is a leap year. And than go for a IF condition while calculating either it should be 365 or 366 days.
With luck,
Pritam. -
Leap year without date objects
I need to check to see how many leap years are between two user-inputted years and I can't use date objects. I figure I need to start with a for loop, but I get lost when to by the mod 4, mod 100, or mod 400. Any help?
for(int l=cYear; l <= eYear; l++){
if(l%4 == 0){
if(l%100 == 0){
}also, if anyone is bored and really wants to give of themselves, I could use some more help with a pretty challenging-for-my-level program. Let me know.
Message was edited by:
kalugayeah, he said like 20 times "one big main method, no
other methods"He's a moron and should not be teaching CS or Java.
Problems: 1. I am having trouble wrapping my mind
around the logic,Meaning what? You described it--or was that just copy/paste of what's in the asisgnent? I gave you some pseudocode that's practically Java already.
Just try it. Take your best shot and write some test cases and see how close you are.
2. I don't know if this (below)
syntax works.Try it. If the syntax is wrong, it won't compile.
Once you get it to compile, then test it. Here's one reason why you'd want a separate method--you could test just that method by itself. As it stands, the easiest way to test it is to always give it just one year and see if you end up with the proper count--zero or one. If you give it many, you could end up with the right number by having the same number of false positives and false negatives.
Once you know it works for one year, feed it a few lists of 5 or 10.
btw jverd, I appreciate your helpYou're quite welcome. It looks like you're on the right track. Just don't be afraid to try stuff and see what happens, and then come back with specific questions about the specific problems you observed. :-)
---- Edit ----Or do like DrClap suggested and print each one. This way you can still test many at once.
Note to self: Duh, Jeff.
---- ---- ----Message was edited by:
jverd -
Leap year issue in depriciation
Hi All,
I have an asset which is acquired on 16th January 2008. 2008 being a leap year should calculate depriciation according to 366 days in a year. However it is calculation on the basis of 365 days in a year. Where should this setting of leap year be maintained so that it starts considering 366 days instead of 365.
Regards,
AjayHi Ajay,
In transaction OAVH:
Example
K4 01 2008 2 29 2
K4 01 2009 2 28 2
K4 01 2010 2 28 2
K4 01 2011 2 28 2
K4 01 2012 2 29 2
K4 01 2016 2 29 2
regards Bernhard -
how do you corr5ect leap year issues with calendar?
I have a reoccurring appoint that happens every wednesday. But I noticed yesterday that it put that appointment on Tuesday - Feb 28th instead of Feb 29th. When you open up the event it says:
"Tuesday, 28 Feb, 2012
from 6:30 PM to 8:30 PM
repeats every week on Wednesday"
I'm quite sure I didn't add a single instance of this event on the Tuesday. I'm a software developer and took a guess that the weekly event falling on Feb 29th of a leap year was the most likely culprit of this bug.
Note that I used the iCloud control panel to push out my events from Outlook to iCloud a couple of weeks ago. So the logic error could be in that software. Or in Outlook, or in Google (as I used to synch my google calendar to outlook before going to iCloud). But I doubt it was caused by Outlook or Google calendar. No proof of that, just my gut feeling.
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Problems setting up Epson Workforce 310
Hi, I am a new MAC user who got the Epson Workforce 310 free after rebate when purchasing the MAC is it true that I can't setup the printer unless I go and buy an ethernet cable or USB cable to set it up? This is what epson is telling me. Can some on