jQuery Datepicker issue with one digit month no leading zero - jquery-ui

I have a date format that is a one digit month (no leading zero), followed by a two digit day and 2 digit year.
January 1st, 2013 = 10113
I'm trying to attach a datepicker with the correct format to it but it doesn't format correctly.
http://jsfiddle.net/S3KyX/1/
<input type="text" id="date" value="10113">
$("#date").datepicker({ dateFormat: 'mddy'});
When the date picker pops up it has the date Oct 11, 2003.

Unfortunately, your format won't work. if you look at the implementation of datepicker formatDate, you can see that for m it will try to match 1 or 2 digits, because it can't tell that you mean January and not October ahead of time,
This method is being called like this, getNumber('m'):
var getNumber = function(match) {
var isDoubled = lookAhead(match);
var size = (match == '#' ? 14 : (match == '!' ? 20 :
(match == 'y' && isDoubled ? 4 : (match == 'o' ? 3 : 2))));
var digits = new RegExp('^\\d{1,' + size + '}');
var num = value.substring(iValue).match(digits);
if (!num)
throw 'Missing number at position ' + iValue;
iValue += num[0].length;
return parseInt(num[0], 10);
};
as you can see size will be 2 for 'm', therefore it will match 2 digits! When in doubt put break points in the code and see what happens (that's what I just did, using chrome developer tools)

Related

Convert desimal number to 7 bit binary number in dart

I need to convert desimal number to 7 bit binary. Like the binary of 11(desimal) is 001011.
I did found a solution in starkoverflow but it doesn't work like i want
String dec2bin(int dec) {
var bin = '';
while (dec > 0) {
bin = (dec % 2 == 0 ? '0' : '1') + bin;
dec ~/= 2;
}
return bin;
}
it returns 1011, cuts all the zeros before 1011
How can i solve that?
Just use padLeft on the String before returning it to make sure it is a minimum length prefixed with zeros. Also, your dec2bin method can be simplified into just using toRadixString(2) on your input integer. So something like this:
String dec2bin(int dec) => dec.toRadixString(2).padLeft(6, '0');
void main() {
print(dec2bin(11)); // 001011
}

How to convert a string timestamp into an array in Lua (or other languages)

I'm having this problem for a while and I can't find a good way to resolve it. So I have a timestamp like this for example : local t = "00:00:0.031" and I'm trying to convert it into an array like this :
local ft = {
hours = 0,
minutes = 0,
seconds = 0,
milliseconds = 31
}
This is pretty much the same no matter the language so if you have an idea on how to solve this but don't know Lua you can submit your answer anyway in any language. I tried solving it myself using regex and I'm quite sure it's possible this way...
Thank you for the interest to my question, have a good day (:
You could use string.match to extract the substrings in a first place. In a second time, you could use the function tonumber to convert it into numbers.
function ParseTimestampString (TimestampString)
local Hours, Minutes, Seconds, Milliseconds = string.match(TimestampString, "(%d+)%:(%d+)%:(%d+)%.(%d+)")
local Result
if Hours and Minutes and Seconds and Milliseconds then
Result = {
hours = tonumber(Hours),
minutes = tonumber(Minutes),
seconds = tonumber(Seconds),
milliseconds = tonumber(Milliseconds)
}
end
return Result
end
With the following code, you could get the results you want:
Result = ParseTimestampString("00:00:0.031")
print(Result.hours)
print(Result.minutes)
print(Result.seconds)
print(Result.milliseconds)
This should returns:
> Result = ParseTimestampString("00:00:0.031")
>
> print(Result.hours)
0
> print(Result.minutes)
0
> print(Result.seconds)
0
> print(Result.milliseconds)
31
Here is a not bad way using string.gmatch which is splitting by regex in Lua. Here the value is being split by either ":" or ".". Then there is a counter in place to match the index for the resulting table.
local t = "00:00:0.031"
local ft = {
hours = 0,
minutes = 0,
seconds = 0,
milliseconds = 0
}
local count = 1
for str in string.gmatch(t, "([^:|.]+)") do
ft[count] = tonumber(str)
count = count + 1
end
You can do a printing loop afterwards to check the results
for i = 1, 4 do
print(ft[i])
end
Output:
0
0
0
31
The main problem I have found with my solution is that it does not save the values under the keys listed but instead the numbers 1 2 3 4.
My Simplest Answer Will Be Just Split the given string by your regex, in This Case For HOUR:MIN:SEC.MS
first Split By (:) To Get HOUR MIN & SEC+MS, Then Split SEC+MS by (.) To separate Both seconds And milliseconds
Below is my answer in java
import java.util.*;
class timeX {
long hours = 0,
minutes = 0,
seconds = 0,
milliseconds = 31;
//Convert Given Time String To Vars
timeX(String input) {
//Split Input By (:) For Hour, Minutes & Seconds+Miliseconds
String[] splitted=input.split(":");
this.hours=Long.parseLong(splitted[0]);
this.minutes=Long.parseLong(splitted[1]);
//Split Again For Seconds And Miliseconds By (.)
String[] splittedMandS=splitted[2].split("\\.");
this.seconds=Long.parseLong(splittedMandS[0]);
this.milliseconds=Long.parseLong(splittedMandS[1]);
}
}
public class Main
{
public static void main(String[] args)
{
timeX tmp = new timeX("30:20:2.031");
System.out.println("H: "+tmp.hours+" M: "+tmp.minutes+" S: "+tmp.seconds+" MS: "+tmp.milliseconds);
}
}
With Lua you can do...
The os.date() can be a format tool for seconds.
...but depends on Operating System.
This works on Linux but not (as i know so far) on MS-Windows.
print(os.date('%H:%M:%S',0-3600)) -- puts out: 00:00:00
print(os.date('%H:%M:%S',300-3600)) -- puts out: 00:05:00
Also it can output the date/time as a table.
> tshow=function(tab) for k,v in pairs(tab) do print(k,'=',v) end end
> tshow(os.date('*t'))
day = 4
year = 2021
month = 11
hour = 11
yday = 308
isdst = false
min = 23
wday = 5
sec = 51
...and unfortunally it has no milliseconds.
If the table output of os.date() is saved as a table...
> ttable=os.date('*t')
> os.time(ttable)
1636021672
> os.date(_,os.time(ttable))
Thu Nov 4 11:27:52 2021
> os.date('%H:%M:%S',os.time(ttable))
11:27:52
...then its key/value pairs can be used for: os.time()
Further code do nearly what you expect when in ttable key 1 is your time with milliseconds as a string...
local tshow=function(tab) for k,v in pairs(tab) do print(k,'=',v) end end
local ttable=os.date('*t') -- Create a time table
ttable[1]='0:0:0.31' -- Numbered keys in sequence are ignored by os.tim()
ttable[2]=ttable[1]:gsub('(%d+):(%d+):(%d+)[.](%d+)','return {year=ttable.year,month=ttable.month,day=ttable.day,hour=%1,min=%2,sec=%3,milliseconds=%4}')
-- That creates ttable[2] with the value:
--- return {year=ttable.year,month=ttable.month,day=ttable.day,hour=0,min=0,sec=0,milliseconds=31}
-- Lets convert it now to a table with...
ttable[2]=load(ttable[2])()
-- Using tshow() to look inside
tshow(ttable[2])
That will output...
milliseconds = 31
day = 4
year = 2021
hour = 0
month = 11
min = 0
sec = 0
And this will put it out formated with os.date()
print(os.date('%H:%M:%S.'..ttable[2].milliseconds,os.time(ttable[2])))
-- Output: 00:00:00.31

How to get each individual digit of a given number in Basic?

I have one program downloaded from internet and need to get each digit printed out from a three digit number. For example:
Input: 123
Expected Output:
1
2
3
I have 598
Need to Get:
5
9
8
I try using this formula but the problem is when number is with decimal function failed:
FIRST_DIGIT = (number mod 1000) / 100
SECOND_DIGIT = (number mod 100) / 10
THIRD_DIGIT = (number mod 10)
Where number is the above example so here is calulation:
FIRST_DIGIT = (598 mod 1000) / 100 = 5,98 <== FAILED...i need to get 5 but my program shows 0 because i have decimal point
SECOND_DIGIT = (598 mod 100) / 10 = 9,8 <== FAILED...i need to get 9 but my program shows 0 because i have decimal point
THIRD_DIGIT = (598 mod 10) = 8 <== CORRECT...i get from program output number 8 and this digit is correct.
So my question is is there sample or more efficient code that get each digit from number without decimal point? I don't want to use round to round nearest number because sometime it fill failed if number is larger that .5.
Thanks
The simplest solution is to use integer division (\) instead of floating point division (/).
If you replace each one of your examples with the backslash (\) instead of forward slash (/) they will return integer values.
FIRST_DIGIT = (598 mod 1000) \ 100 = 5
SECOND_DIGIT = (598 mod 100) \ 10 = 9
THIRD_DIGIT = (598 mod 10) = 8
You don't have to do any fancy integer calculations as long as you pull it apart from a string:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
PRINT MID$(X$, Z, 1)
NEXT
Then, for example, you could act upon each string element:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
NEXT
Additionally, you could tear apart the string character by character:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " ", ".", "+", "-", "E", "D"
' special char
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
I'm no expert in Basic but looks like you have to convert floating point number to Integer. A quick google search told me that you have to use Int(floating_point_number) to convert float to integer.
So
Int((number mod 100)/ 10)
should probably the one you are looking for.
And, finally, all string elements could be parsed:
INPUT X
X$ = STR$(X)
PRINT X$
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " "
' nul
CASE "E", "D"
Exponent = -1
CASE "."
Decimal = -1
CASE "+"
UnaryPlus = -1
CASE "-"
UnaryNegative = -1
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
IF Exponent THEN PRINT "There was an exponent."
IF Decimal THEN PRINT "There was a decimal."
IF UnaryPlus THEN PRINT "There was a plus sign."
IF UnaryNegative THEN PRINT "There was a negative sign."

How can I better parse variable time stamp information in Fortran?

I am writing code in gfortran to separate a variable time stamp into its separate parts of year, month, and day. I have written this code so the user can input what the time stamp format will be (ie. YEAR/MON/DAY, DAY/MON/YEAR, etc). This creates a total of 6 possible combinations. I have written code that attempts to deal with this, but I believe it to be ugly and poorly done.
My current code uses a slew of 'if' and 'goto' statements. The user provides 'tsfo', the time stamp format. 'ts' is a character array containing the time stamp data (as many as 100,000 time stamps). 'tsdelim' is the delimiter between the year, month, and day. I must loop from 'frd' (the first time stamp) to 'nlines' (the last time stamp).
Here is the relevant code.
* Choose which case to go to.
first = INDEX(tsfo,tsdelim)
second = INDEX(tsfo(first+1:),tsdelim) + first
if (INDEX(tsfo(1:first-1),'YYYY') .ne. 0) THEN
if (INDEX(tsfo(first+1:second-1),'MM') .ne. 0) THEN
goto 1001
else
goto 1002
end if
else if (INDEX(tsfo(1:first-1),'MM') .ne. 0) THEN
if (INDEX(tsfo(first+1:second-1),'DD') .ne. 0) THEN
goto 1003
else
goto 1004
end if
else if (INDEX(tsfo(1:first-1),'DD') .ne. 0) THEN
if (INDEX(tsfo(first+1:second-1),'MM') .ne. 0) THEN
goto 1005
else
goto 1006
end if
end if
first = 0
second = 0
* Obtain the Julian Day number of each data entry.
* Acquire the year, month, and day of the time stamp.
* Find 'first' and 'second' and act accordingly.
* Case 1: YYYY/MM/DD
1001 do i = frd,nlines
first = INDEX(ts(i),tsdelim)
second = INDEX(ts(i)(first+1:),tsdelim) + first
read (ts(i)(1:first-1), '(i4)') Y
read (ts(i)(first+1:second-1), '(i2)') M
read (ts(i)(second+1:second+2), '(i2)') D
* Calculate the Julian Day number using a function.
temp1(i) = JLDYNUM(Y,M,D)
end do
goto 1200
* Case 2: YYYY/DD/MM
1002 do i = frd,nlines
first = INDEX(ts(i),tsdelim)
second = INDEX(ts(i)(first+1:),tsdelim) + first
read (ts(i)(1:first-1), '(i4)') Y
read (ts(i)(second+1:second+2), '(i2)') M
read (ts(i)(first+1:second-1), '(i2)') D
* Calculate the Julian Day number using a function.
temp1(i) = JLDYNUM(Y,M,D)
end do
goto 1200
* Onto the next part of the code
1200 blah blah blah
I believe this code will work, but I do not think it is a very good method. Is there a better way to go about this?
It is important to note that the indices 'first' and 'second' must be calculated for each time stamp as the month and day can both be represented by 1 or 2 integers. The year is always represented by 4.
With only six permutations to handle I would just build a look-up table with the whole tsfo string as the key and the positions of year, month and day (1st, 2nd or 3rd) as the values. Any unsupported formats should produce an error, which I haven't coded below. When subsequently you loop though your ts list and split an item you know which positions to cast to the year, month and day integer variables:
PROGRAM timestamp
IMPLICIT NONE
CHARACTER(len=10) :: ts1(3) = ["2000/3/4 ","2000/25/12","2000/31/07"]
CHARACTER(len=10) :: ts2(3) = ["3/4/2000 ","25/12/2000","31/07/2000"]
CALL parse("YYYY/DD/MM",ts1)
print*
CALL parse("DD/MM/YYYY",ts2)
CONTAINS
SUBROUTINE parse(tsfo,ts)
IMPLICIT NONE
CHARACTER(len=*),INTENT(in) :: tsfo, ts(:)
TYPE sti
CHARACTER(len=10) :: stamp = "1234567890"
INTEGER :: iy = -1, im = -1, id = -1
END TYPE sti
TYPE(sti),PARAMETER :: stamps(6) = [sti("YYYY/MM/DD",1,2,3), sti("YYYY/DD/MM",1,3,2),&
sti("MM/DD/YYYY",2,3,1), sti("DD/MM/YYYY",3,2,1),&
sti("MM/YYYY/DD",2,1,3), sti("DD/YYYY/MM",3,1,2)]
TYPE(sti) :: thisTsfo
INTEGER :: k, k1, k2
INTEGER :: y, m, d
CHARACTER(len=10) :: cc(3)
DO k=1,SIZE(stamps)
IF(TRIM(tsfo) == stamps(k)%stamp) THEN
thisTsfo = stamps(k)
EXIT
ENDIF
ENDDO
print*,thisTsfo
DO k=1,SIZE(ts)
k1 = INDEX(ts(k),"/")
k2 = INDEX(ts(k),"/",BACK=.TRUE.)
cc(1) = ts(k)(:k1-1)
cc(2) = ts(k)(k1+1:k2-1)
cc(3) = ts(k)(k2+1:)
READ(cc(thisTsfo%iy),'(i4)') y
READ(cc(thisTsfo%im),'(i2)') m
READ(cc(thisTsfo%id),'(i2)') d
PRINT*,ts(k),y,m,d
ENDDO
END SUBROUTINE parse
END PROGRAM timestamp
I would encode the different cases in another way, like this:
module foo
implicit none
private
public encode_datecode
contains
integer function encode_datecode(datestr, sep)
character(len=*), intent(in) :: datestr, sep
integer :: first, second
character(len=1) :: c1, c2, c3
first = index(datestr, sep)
second = index(datestr(first+1:), sep) + first
c1 = datestr(1:1)
c2 = datestr(first+1:first+1)
c3 = datestr(second+1:second+1)
foo = num(c1) + 3*num(c2) + 9*num(c3)
end function encode_datecode
integer function num(c)
character(len=1) :: c
if (c == 'Y') then
num = 0
else if (c == 'M') then
num = 1
else if (c == 'D') then
num = 2
else
stop "Illegal character"
end if
end function num
end module foo
and then handle the legal cases (21, 15, 19, 7, 11, 5) in a SELECT statement.
This takes advantage of the fact that there won't be a 'YDDY/MY/YM' format.
If you prefer better binary or decimal readability, you can also multiply by four or by 10 instead of 3.

ruby on rails: need a function to calculate the time and display it in multi language

I need a custom function to calculate the times given from a certain point in the future X
It should countdown from 1 hour to zero
1 hour
59 min
59 min
..
10 min
9 min
..
1 min
59 sec
58 sec
57 sec
..
1 sec
Im currently using the
= clean_text distance_of_time_in_words(Time.now, real_time, true)
With the clean_text function:
def clean_text(string)
bad_words = ["less than", "about"]
change_words = ["hours", "minutes"]
bad_words.each do |bad|
string.gsub!(bad + " ", '')
end
string = string.gsub("hours", "hrs")
string = string.gsub("minutes", "min")
string = string.gsub("minute", "min")
string = string.gsub("half a min", "30 sec")
string = string.gsub("seconds", "sec")
return string
end
But this is insufficient since Im rewriting "minutes" to "min" etc. What would be a good start to write such a custom function for time? I guess it will be quite complex but wondered if there are any rails internal functions I could use?
I need
countdown like listed above
it has to be i18n compatible so I can do 1 hour / 1 houra, etc
Anyone an idea or will it be a custom function thats really neccesairly here?

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