i have a simple question, i need to convert back the dateformat in coldfusion, for example i transformed the dateformat in this way: #dateformat(attributes.start_date,'dd/mm/yyyy')# and the value was looking like: 15/01/2012 and the original is: {ts '2012-01-15 00:00:00'} all i need is to convert it back, in the original one from this 15/01/2012 to {ts '2012-01-15 00:00:00'} is it possible?
Thank you all for the help!
Use ParseDateTime() (or LSParseDateTime) function: http://livedocs.adobe.com/coldfusion/8/htmldocs/help.html?content=functions_m-r_11.html
Related
I'm writing an application with Flutter. I read the times and dates from a source. The date and time format string sent by the resource is:
(Day, Month, Year, Hour, Minute, Second)
07.04.2021 13:30:00
03.04.2021 11:30:00
04.04.2021 17:30:00
03.04.2021 17:30:00
I want to convert this date and time format to DateTime data type with DateTime.parse() function. Here are some examples of what this function accepts as strings and what I need:
"2012-02-27 13:27:00"
"20120227 13:27:00"
"20120227T132700"
I have to convert the string type data coming to me from the source into one of these formats. But in Dart language I couldn't create the Regular Expression needed to do this and couldn't find it anywhere.
I would be very grateful if anyeone could help me understand what I should do.
If you have to play a lot with the dates, you could use the Jiffy package to ease your development.
DateTime yourDatetime = Jiffy("07.04.2021 13:30:00", "dd.MM.yyyy hh:mm:ss").dateTime;
This is a piece a cake by using regular expressions:
var regExp = RegExp(r'(\d{4}-?\d\d-?\d\d(\s|T)\d\d:?\d\d:?\d\d)');
use DateFormat.parse and DateFormat.format from intl package:
https://api.flutter.dev/flutter/intl/DateFormat/parse.html
https://api.flutter.dev/flutter/intl/DateFormat/format.html
final date = DateFormat("yyyy.MM.dd HH:mm:ss").parse("07.04.2021 13:30:00");
DateFormat("yyyy-MM-dd HH:mm:ss").format(date);
DateTime.parse accepts only a subset of ISO 8601 formats: https://api.flutter.dev/flutter/dart-core/DateTime/parse.html
i have been trying to convert a string to date using
${test:toString():toDate('dd-MMM-yy HH.mm.ss.SSSSSSSSS'):format('dd-MMM-yy HH.mm.ss.SSSSSSSSS')}
my value for test attribute is like 13-MAR-20 15.50.41.396000000
when i'm using the above mentioned expression to convert the string to Date, it actually is changing the date as below:
test (input value):
13-MAR-20 15.50.41.396000000
time (output value)
18-Mar-20 05.50.41.000000000
please advise!
I ran into a similar issue with date time encoded in ISO 8601.
The problem is, that the digits after the second are defined as fragment of a second, not milliseconds. If it has 3 digits, it equivalent to milliseconds. If more than 3, the toDate() function parse the fragment of a second as milliseconds. In your case 396000000 milliseconds = 4,58333333 days.
I solved my issue with replaceAll() by cutting digits to first 3.
${test:replaceAll('(\.[0-9]{3})([0-9]+)','$1')}
But my value was formatted as 18-06-20T05:50:41.396000000, so maybe you have to adjust the regex.
I have a google sheet script that fetches youtube's video durations. The problem is the time data is in the ISO 8601 format.
For example:
PT3M23S
The formula I'm using right now does a good job converting this into a more readable format.
=iferror(REGEXREPLACE(getYoutubeTime(B20),"(PT)(\d+)M(\d+)S","$2:$3"))
It converts the above into a more readable format 3:23
Now the issue at hand is if the duration of the video is exactly 3 minutes or if the video is shorter than 1 minute regexreplace doesn't reformat it.
Instead it reads
PT4M OR PT53S
Is there a way to edit the formula to address each variant that potential could occur?
Where it would format PT4M into 4:00 or PT53S into 0:53
Lastly, if the seconds in the duration are between 1-9 the API returns a single digit value for the seconds. Which means the formula above will look wrong. For example, PT1M1S is formatted into 1:1 when it should read 1:01
It would be great if the formula could account for the first 9 seconds and add a 0 to make it more readable.
Thanks for reading this far, if anyone could help me out I'd very much appreciate it.
Just in case its easier to do this within the script itself here's the custom script that retrieves the video duration.
function getYoutubeTime(videoId){
var url = "https://www.googleapis.com/youtube/v3/videos?part=contentDetails&id=" + videoId;
url = url + "&key=";
var videoListResponse = UrlFetchApp.fetch(url);
var json = JSON.parse(videoListResponse.getContentText());
return json["items"][0]["contentDetails"]["duration"];
}
Is very ugly, but seems to work for the examples provided:
=iferror(left(mid(A1,3,len(A1)-2),find("M",mid(A1,3,len(A1)-2))-1)*60,0)+substitute(REGEXreplace(mid(A1,3,len(A1)-2),"(.+M)",""),"S","")
Outputs seconds, eg 203 from PT3M23S. To change to 00:03:23 wap the above formula in ( ... )/86400 and format result as Time.
Script Solution:
function iso8601HMparse(str) {
return str.replace(/PT(\d+(?=M))?M?(\d+(?=S))?S?/g,function(mm,p1,p2){//regex to get M and S value
return [0,p1,p2].map(function(e){
e = e ? e:0;
return ("00"+e).substr(-2); //fix them to 2 chars
}).join(':');
})
}
Splice it in your script like:
return iso8601HMparse(json["items"][0]["contentDetails"]["duration"].toString());
Spreadsheet Function:
=TEXT(1*REGEXREPLACE(REGEXREPLACE(A1,"PT(\d+M)?(\d+?S)?","00:00$1:00$2"),"[MS]",),"MM:SS")
Late to the party, but here's what I'm using:
=TIMEVALUE(
IFERROR(TEXT(MID(A1,3,FIND("H",A1)-3),"00"),"00")&":"&
IFERROR(TEXT(MID(A1,IFERROR(FIND("H",A1)+1,3),FIND("M",A1)-IFERROR(FIND("H",A1)+1,3)),"00"),"00")&":"&
IFERROR(TEXT(MID(A1,IFERROR(FIND("M",A1)+1,3),FIND("S",A1)-IFERROR(FIND("M",A1)+1,3)),"00"),"00"))
You can also stick ARRAYFORMULA in front of this and change A1 to a a column to get values for a whole list of them.
Here is my scenario. I am getting this date from a database:
11-AUG-15 10.38.00.000000000 AM
Is there any way to format this string to look something similar to mm/dd/yy?
So far I have tried the following with no luck:
DateFormat()
CreateODBCDate()
LSParseDateTime()
Every time I use one of the above, I get the following error:
11-AUG-15 10.38.00.000000000 AM is an invalid date or time string.
Any advise will be greatly appreciated.
Thanks!
parseDateTime("11-AUG-15 10.38.00.000000000 AM", "dd-MMM-yy hh.mm.ss.S aa");
Run me: http://trycf.com/gist/aac6d63777ae1b0e9aa3/acf?theme=monokai
Then you are free to use DateFormat() or DateTimeFormat()to format the date object.
I'm a newbie in Coldfusion. I'm trying to change a string that looks like "Jun 11, 2014, 8:50 PM" to "Jun 11, 2014". I tried to use
<cfset album[currentrow]['date'] = ListGetAt(album[currentrow]['date'], 2, ",")>
It gives me 2014. If I change 2 to 1, I get Jun 11. Can someone please give me some advice on if there's a method to extra everything before the second ","? Your help is greatly appreciated.
Messing with strings isn't tough, but since you are working with a date, lets use the Date Functions.
If you have a string, the best thing to do is convert it into a Date Object, using
<cfset myDate = ParseDateTime(string)>
Then once it is a Date Object, you can do whatever you want with it. Use Dateformat to manipulate it as needed.
<cfoutput>#dateformat(myDate, "mmm dd, yyyy")#</cfoutput>
ParseDateTime documentation here
DateFormat documentation here
Edit - Using Strings instead.
You can use LEFT to get the left piece of the string. To know how many characters your want, you need to find the location of the second "," comma. Assuming the format is consistent, the first comma should be no more than 8 characters in, so we use FIND to look for "," in the string, starting at position 8.
<cfset theLoc = find(",", album[currentrow]['date'], 8) >
Then we use the left function to get the characters, but we do not want the comma, so we take 1 off of it.
<cfset theDate = left( album[currentrow]['date'], theLoc- 1 )>
<cfoutput>#theDate#</cfoutput>
You can do it all inline, but its a little messier
<cfset theDate = left( album[currentrow]['date'], find(",", album[currentrow]['date'], 8)- 1 )>
If you can guarantee that the string is always formed of three comma-separated parts, and the date is always the first two, then you can do this:
dateTimeString = "Jun 2, 2014, 6:20 PM";
dateString = listDeleteAt(dateTimeString, 3);
date = parseDateTime(dateString);
dateFormattedForOutput = dateFormat(date, "mmm d, yyyy");
Note that you should only ever use dateFormattedForOutput for output; if you're storing the date or manipulating it, use date.