I have a column of times in ALteryx like the following:
Time
=====
8:05 AM
8:07 AM
8:11 AM
8:12 AM
8:16 AM
...
They are currently stored as a String, but I want to convert them to the Time format. So far, I have tried using the LEFT and TRIM function to isolate the time itself as such:
Trim(Left([Time], 5)," ")
However, when I try to convert this to a Time datatype, I receive type conversion errors. How can I convert something like 8:46 AM from a String to a Time datatype in Alteryx?
I did some more digging and trying different things out, and I was able to figure it out. For the reference of others, here is what I did:
DateTimeParse([VisitTime] + " " + [PartofDay],"%I:%M %P")
Where [VisitTime] is the time expressed like 8:30 and [PartofDay] is AM or PM.
After this, I used a Select tile to change the datatype from String to Time without type conversion errors.
Related
In my power automate flow, I have an action that give time output in this format: 2022-12-01T18:52:50.0000000Z
How can I take this output and format as yyyy/mm/dd .
I want to use the time output as string for a folder structure.
https://learn.microsoft.com/en-us/azure/logic-apps/workflow-definition-language-functions-reference#formatDateTime
Assuming you have a variable called DateTime, you would create another variable and use this expression ...
formatDateTime(Variables('DateTime'), 'yyyy/MM/dd')
Result
This is pretty straight forward.
Use the Convert datetime to text function
In the Format to use dropdown select Custom
and specify the format how you would like it to appear.
yyyy/MM/hh will give you 2022/12/01
HH-mm-ss will give you 18-52-50
i am currently working with VB6, and i have this value of date that is like this:
2022-02-26T12:06:10+02:00
I followed this url too
VB6: How to remove the Time part from Date type
but doesnt work especially the last one still shows the date as
2022-02-26T12:06:10+02:00
This is my code
Dim tdate As String
tdate = format$("2022-02-26T12:06:10+02:00" , "m/d/yyyy")
and the output is still 2022-02-26T12:06:10+02:00
Your input is a string in ISO8601 format. As its a string in a fixed format the easiest way is to just chop off the first 10 characters.
isoDateTimeString = "2022-02-26T12:06:10+02:00"
To get the date part as another string:
Dim dateAsString As String
dateAsString = Left$(isoDateTimeString, 10)
'// for 2022-02-26
Or to get it as a Date type:
Dim dateAsDateType As Date
dateAsDateType = CDate(Left$(isoDateTimeString, 10))
'// for 26/02/2022 (or whatever your locale format is)
In VB6 I've always found that the easiest way to deal with time/date values is to cast them as a Double (simply declare a variable as type Double, then assign the value from whatever source). Then, simply deal with either the integer part (days) or the fractional part (fractional days). For example, seconds are just TimeStamp/86400.0, etc. When a variable is declared as Date, it's actually stored as a Double, so I just use that as my basic TimeStamp type. VB is pretty good about re-formatting into a time/date string when printing, and it makes time-based calculations really straight-forward.
In business central there is this data type named DateFormula, in it you can define things like '1D + 2H + 3S' (1 day, 2 hours and 3 seconds). I found out that i can convert this time range to a Date using CalcDate() however, Date objects dont contain the time information, which I do need.
There is no CalcDateTime() function in BC, nor does it mention converting DateFormula's to DateTime anywhere online it seems.
How can I convert a DateFormula to a DateTime object in BC?
OK so apparently it turns out that DateFormula datatype does not even support things like hours and seconds, only dates and no times.
Solved!
Problem:
In a field called $Detailed Decription$ sometimes dateformat 08/09/2021 is enterd and this need to be converted to swedish format 2022-02-11
I'am going to use BMC Developer studio and make a filter but i cant find a fitting solution for it. Replacing it wont work (i think) becaus it need to have a value to replace it with.
Maby there can be a function that reads regex (\d{2})/(\d{1,2})/(\d{4}) but how can i convert it?
If it's sometimes - look at AR System User Preferencje form. Check certain user's locale and date time config.
Also is important where the data comes from. Could be a browser setting or java script mod.
1- Using Set fields action, copy the date value from Detailed Description to a Date/Time field (i.e. z1D_DateTime01).
2- Using Set fields action and Functions (MONTH, YEAR, DAY, HOUR, MINUTE, SECOND) you can parse the date/time and convert it to format you like.
Something like this:
SwedishDate = YEAR($z1D_DateTime01$) + "-" + MONTH($z1D_DateTime01$) + "-" + DAY($z1D_DateTime01$)
This will capture the parts of date and combine them with "-" in the middle and in the order you want.
Can someone please guide me on how to convert more than 6 characters into int? Because I need to do sum after convert to int. I tried so many ways like CInt, CLng, etc still throw exponential value.
Stroutput = 2018050302216556
Sum = Stroutput + 1
I tried to divide into sveral chuck using right function but it doesnt look good. Can be manage but I need another option. Thanks
You seem to be working with a Date Structure, which as VBS says - hard to represent as numbers only. Use the CDate to get a date object from the string (If needed change the representation of that string to (YYYY-mm-dd ...). With the DateAdd method you can add days, years etc; and finally the FormatDateTime will create an output of your wish.