Date$() problem/bug?

[tenochtitlanuk]

On Yahoo Groups there's been a thread in which date$ behaved strangely with 'and' conditional.

For me the varied forms of date$( were just right to calculate UNIX dates for my tide-predictor.

I had to use the first version, as the second errors out. The third version works...


'#1  works OK
d$                  = date$( "mm/dd/yyyy")
UnixTimeNowSecs     =(  date$( d$) -date$( "01/01/1970")) *24 *60 *60  +time$( "seconds") ' -3600 -adjust for British Summer Time
print "<"; UnixTimeNowSecs; ">"


#2 FAILS
UnixTimeNowSecs =( date$( "mm/dd/yyyy")) -date$( "01/01/1970")) *24 *60 *60 +time$( "seconds") ' -3600 -adjust for British Summer Time
'rint "<"; UnixTimeNowSecs; ">"
here


'#3 works OK
UnixTimeNowSecs     =(  date$( chr$( 34) +date$( "mm/dd/yyyy") +chr$( 34)) -date$( "01/01/1970")) *24 *60 *60  +time$( "seconds") ' -3600 -adjust for British Summer Time
print "<"; UnixTimeNowSecs; ">"re

[Brandon Parker]

In my opinion the Date$() function in Liberty BASIC is kind of "Super-Overloaded"; it's offers a large array of possibilities for working with dates. I believe I have run into this in the past at some point and I think it is either a parsing error or type casting/ conversion error on the back-end.

In order to get #2 to work as expected you need to pass the date$("mm/dd/yyyy") back to the date$() function to return the number of days that values is since that is what date$("01/01/1970") is actually doing (all based on LB's value for when "time started"). The hiccup is that when you pass something to the date$() function like this is should be a string value and that is not exactly what date$() seems to return. So.....just wrap the date$("mm/dd/yyyy") in str$() and return it just as is shown below.

UnixTimeNowSecs =( date$(str$(date$( "mm/dd/yyyy"))) -date$( "01/01/1970")) *24 *60 *60 +time$( "seconds") ' -3600 -adjust for British Summer Time
Print "<"; UnixTimeNowSecs; ">"

I hope this helps although I cannot completely explain the inner workings completely since I'm not privy to it.


{:0)

Brandon R. Parker

[tsh73]

So.
d$ = date$( "mm/dd/yyyy")
print date$( d$)
works, while
print date$( date$( "mm/dd/yyyy")) is not.
I agree this is a bug.
But.
LB has actually special command to get number of days for "NOW" and it is
print date$("days")

[Brandon Parker]

I am 99% positive the issue arises because the Date$() function is expecting a literal string to be passes to it. The Date$() function is returning, at least from what it looks like to me, a pointer to a string. I'm sure this could be classified as a minor bug if need be, but it really isn't stated in the documentation that you can feed the return value of the Date$() function as a parameter to the Date$() function itself. That is the part that people just make an assumption about given that it prints out as a string and can be assigned to a string variable.

When assigning the Date$() function to a PTR member of a Struct you get the pointer to the string being held in the Struct member; like below....
Struct myStruct, varPtr As ptr

Print date$("mm/dd/yyyy")
var$ = date$("mm/dd/yyyy")
Print date$(var$)

'Print date$(date$("mm/dd/yyyy"))
Print date$(str$(date$("mm/dd/yyyy")))

myStruct.varPtr.struct = date$("mm/dd/yyyy")

Print myStruct.varPtr.struct
Print Winstring(myStruct.varPtr.struct)

Print date$(Winstring(myStruct.varPtr.struct))


For me, this is the bigger of the issues..... Since there are some "things" that could be improved with the Date$() function this one should be the first thing on the agenda. Once again, I'm betting it is because the function is overload like crazy and there is something that just requires tweaking with it behind the scenes. Without further rambling, here is my main issue with it....

You can actually store a string value into a numerical variable using the Date$() function....
I'm pretty sure this has been brought up previously...
'Storing a string from the Date$() function into a numerical variable
var = date$("mm/dd/yyyy")
'Print said variable and tadaaaaa; it actually points to the string date
Print var
'Print the value using the str$() function and it is returned with single-quotes surrounding it
Print str$(var)
'store the numerical var variable into a string var$ variable and print it; stores with the single-quotes
var$ = str$(var)
Print var$


For me these things are not huge issues though....


{:0)

Brandon R. Parker

[konijn]

Hello,

I have a question about date$
I have a string betaaldatum$ what gives for example "2018-12-21"
Now I want this date + 14 days. What should give "2019-01-04"
How can I made this?

Thanks!!

[tsh73]

'I have a string betaaldatum$ what gives for example "2018-12-21"
'Now I want this date + 14 days. What should give "2019-01-04"

betaaldatum$="2018-12-21"   'sorry, doesn't work.
'needs format "mm/dd/yyyy"
dt$=word$(betaaldatum$,2,"-")+"/"+word$(betaaldatum$,3,"-")+"/"+word$(betaaldatum$,1,"-")
dtOld= date$(dt$)
dtNew$ = date$(dtOld+14)
dtOut$=word$(dtNew$,3,"/")+"-"+word$(dtNew$,1,"/")+"-"+word$(dtNew$,2,"/")

print betaaldatum$
print dtOut$


[konijn]

Thanks tsh73
Now its working.
I dont understand this line well:
dt$=word$(betaaldatum$,2,"-")+"/"+word$(betaaldatum$,3,"-")+"/"+word$(betaaldatum$,1,"-")
First 2 and than 3 and as last 1. How does it work?

I use it to get the payment info out of a ubl invoice(xml) and safe it in a file. With a sepa file in excel I can make a sepa file wich I can uploud to the internet banking

dim a$(30000)
i =0
open "c:\ubl.csv" for input as #readFile
   while not( eof( #readFile))
       inputcsv #readFile, a$(i)
       'print               a$(i)
       i =i +1
   wend
close #readFile

naam$ = "cac:PartyName"

a=0
while naam$ <> a$(a) and a <30000
   a = a + 1
wend
naamafz$ = a$(a+4)
print naamafz$

iban$ = "cac:PayeeFinancialAccount"

b=0
while iban$ <> a$(b) and b <30000
   b = b + 1
wend
banknr$ = a$(b+4)
print banknr$

'bic$ = "cbc:ID schemeName=BIC"

'c=0
'while bic$ <> a$(c) and c <30000
   c = c + 1
'wend
'bicnr$ = a$(c+1)
'print bicnr$

factuur$ = "cac:AdditionalDocumentReference"

d=0
while factuur$ <> a$(d) and d <30000
   d = d + 1
wend
factuurnr$ = a$(d+4)
print factuurnr$

bedrag$ = "cbc:PayableAmount currencyID=EUR"

e=0
while bedrag$ <> a$(e) and e <30000
   e = e + 1
wend
bedragincl$ = a$(e+1)
print bedragincl$

betaaldat$ = "cbc:PaymentDueDate"

f=0
while betaaldat$ <> a$(f) and f <30000
   f = f + 1
wend
betaaldatum$ = a$(f+1)
print betaaldatum$
dt$=word$(betaaldatum$,2,"-")+"/"+word$(betaaldatum$,3,"-")+"/"+word$(betaaldatum$,1,"-")
dtOld= date$(dt$)
dtNew$ = date$(dtOld+21)
dtOut$=word$(dtNew$,3,"/")+"-"+word$(dtNew$,1,"/")+"-"+word$(dtNew$,2,"/")

print betaaldatum$
print dtOut$


open "c:\invoer SEPA-CT.txt" for append as #1
print #1, "your IBAN,your BIC,your name,"+banknr$+","+bicnr$+","+naamafz$+","+bedragincl$+",Factuurnr:"+factuurnr$+","+dtOut$

close #1


[konijn]

link

Ubl file. The <> repleaced with , and the "removed

[tsh73]

I dont understand this line well:
dt$=word$(betaaldatum$,2,"-")+"/"+word$(betaaldatum$,3,"-")+"/"+word$(betaaldatum$,1,"-")
First 2 and than 3 and as last 1. How does it work?

Your data goes if the form
betaaldatum$="2018-12-21"
that is yyyy-mm-dd
But for LB date$ to work it should be in the form
mm/dd/yyyy
So this line does the conversion.
word$(a$, i) gives i-th word from line a$, delimited with spaces
word$(a$, i, d$) gives i-th word from line a$, delimited with d$.
So I used "-" as a delimiter
and get 2nd (mm), 3rd (dd) and 1st (yyyy) words of yyyy-mm-dd (that is of 2018-12-21).
And glued them together with (+), putting (/) in between to get mm/dd/yyyy.