How to Convert Yyyymmddhhmmss to a Date Readable by 'Date'

How to convert YYYYMMDDHHMMSS to a date readable by `date`

date doesn't allow "YYYYMMDDHHMMSS", but it does "YYYYMMDD HH:MM:SS", so:

D="20100101123456"
date -d "${D:0:8} ${D:8:2}:${D:10:2}:${D:12:2}"

Converting YYYYMMDDHHMMSS to a date and time class in r

In Base R:

data.frame(datetime2=as.POSIXct(as.character(df$datetime),
                                format="%Y%m%d%H%M%S"), 
           cycle)

Output:

            datetime2 cycle
1 2021-08-30 10:20:55    25
2 2021-08-30 10:23:12    30
3 2021-08-30 10:23:35    35

The key here is as.POSIXct with format:

> as.POSIXct(as.character(df$datetime), format="%Y%m%d%H%M%S")
[1] "2021-08-30 10:20:55 CST" "2021-08-30 10:23:12 CST" "2021-08-30 10:23:35 CST"
>

unix yyyymmddhhmmss format conversion to specific date format

The internal string conversion functions are too limited, so we use sed and tr when needed.

## The "readable" format yyyy.mm.dd.hh.mm.ss isn’t understood by date. 
## yyyy-mm-dd hh:mm:ss is. So we first produce the latter.

# Note how to extract the last 14 characters of ${folder} and that, since
# we know (or should have checked somewhere else) that they are all digits,
# we match them with a simple dot instead of the more precise but less
# readable [0-9] or [[:digit:]]
# -E selects regexp dialect where grouping is done with simple () with no
# backslashes.
d="$(sed -Ee's/(....)(..)(..)(..)(..)(..)/\1-\2-\3 \4:\5:\6/'<<<"${folder:(-14)}")"

# Print the UTF date (for Linux and other systems with GNU date)
date -u -d "$d"

# Convert to your preferred "readable" format  
# echo "${d//[: -]/.}" would have the same effect, avoiding tr
tr ': -' '.'<<<"$d"

For systems with BSD date (notably MacOS), use

date -juf'%Y-%m-%d %H:%M:%S' "$d"

instead of the date command given above. Of course, in this case the simplest way would be:

# Convert to readable
d="$(sed -Ee's/(....)(..)(..)(..)(..)(..)/\1.\2.\3.\4.\5.\6/'<<<"${folder:(-14)}")"
# Convert to UTF
date -juf'%Y.%m.%d.%H.%M.%S' "$d"

Convert date from yyyymmddHHMMSS format

You can import as a structured array as follow:

mydate, myvar = np.loadtxt('infile.txt', comments="#", skiprows=1, usecols=(0,4), unpack=True, dtype=[('date', '|S10'), ('floatmio', float)])

it will import the date as str in mydate array. And then you can use datetime library on the single array entries to handle them as follow:

import datetime
dates = [datetime.datetime.strptime(x, '%Y%m%d%H') for x in mydate]

Convert 'yyyymmddhhmmss' to MM/DD/YYYY HH:MM (AM/PM) in Snowflake

Please check below select try_to_timestamp('20220120035900', 'YYYYMMDDHHMISS');

Sample Image

UTC to CST:

SELECT try_to_timestamp('20220120035900', 'YYYYMMDDHHMISS') as t_str,
    t_str::timestamp_ntz as UTC_tz,
    convert_timezone('UTC','America/Chicago', UTC_tz) chicago_time;

How to convert date-time string in the form (yyyymmddhhss) to dd/mm/yyyy in Javascript?

You could try using the Luxon module (an evolution of Moment.js), it will do this kind of thing very nicely:

const { DateTime } = luxon;

const timestamp = "20201127075434";
const dt = DateTime.fromFormat(timestamp, "yyyyMMddHHmmss");

console.log("dd/mm/yyyy:", dt.toFormat("dd/MM/yyyy"));
console.log("hh:mm:ss:", dt.toFormat("HH:mm:ss"));

<script src="https://cdnjs.cloudflare.com/ajax/libs/luxon/1.25.0/luxon.min.js" integrity="sha512-OyrI249ZRX2hY/1CAD+edQR90flhuXqYqjNYFJAiflsKsMxpUYg5kbDDAVA8Vp0HMlPG/aAl1tFASi1h4eRoQw==" crossorigin="anonymous"></script>

Convert a string with 'YYYYMMDDHHMMSS' format to datetime

You can use the STUFF() method to insert characters into your string to format it in to a value SQL Server will be able to understand:

DECLARE @datestring NVARCHAR(20) = '20120225143620'

-- desired format: '20120225 14:36:20'
SET @datestring = STUFF(STUFF(STUFF(@datestring,13,0,':'),11,0,':'),9,0,' ')

SELECT CONVERT(DATETIME, @datestring) AS FormattedDate

Output:

FormattedDate
=======================
2012-02-25 14:36:20.000

This approach will work if your string is always the same length and format, and it works from the end of the string to the start to produce a value in this format: YYYYMMDD HH:MM:SS

For this, you don't need to separate the date portion in anyway, as SQL Server will be able to understand it as it's formatted.

Related Reading:

STUFF (Transact-SQL)

The STUFF function inserts a string into another string. It deletes a specified length of characters in the first string at the start position and then inserts the second string into the first string at the start position.
STUFF ( character_expression , start , length , replaceWith_expression )

unable to convert string date in Format yyyyMMddHHmmss to DateTime dart

intl DateFormat can't cope with your input string as it doesn't have any separators. The whole string gets consumed as the year. However DateTime.parse does cope with this (nearly). It happens to expect precisely the format you have (again, nearly).

One of the acceptable styles to parse is 20120227T132700, which just differs by the T date/time separator.

Try this:

String date = '20180626170555';
String dateWithT = date.substring(0, 8) + 'T' + date.substring(8);
DateTime dateTime = DateTime.parse(dateWithT);

Convert String value format of YYYYMMDDHHMMSS to C# DateTime

Define your own parse format string to use.

string formatString = "yyyyMMddHHmmss";
string sample = "20100611221912";
DateTime dt = DateTime.ParseExact(sample,formatString,null);

In case you got a datetime having milliseconds, use the following formatString

string format = "yyyyMMddHHmmssfff"
string dateTime = "20140123205803252";
DateTime.ParseExact(dateTime ,format,CultureInfo.InvariantCulture);

Thanks

How to Convert Yyyymmddhhmmss to a Date Readable by 'Date'