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i18nFormat with CakePHP 3 - Why does YYYY produce a different year than yyyy?


i18nFormat with CakePHP 3 - Why does YYYY produce a different year than yyyy?

By : Mark James
Date : November 17 2020, 11:52 AM
hop of those help? Uppercase Y is the year that the week of the timestamp in, in "Week of Year" based calendars. This week now is in 2015 so it returns 2015. Lowercase y is the current year for the timestamp, which for now is 2014.
Date Field Symbol Table
code :


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Extracting each year from an item in a list in the format [yyyy,yyyy]

Extracting each year from an item in a list in the format [yyyy,yyyy]


By : Connor K
Date : March 29 2020, 07:55 AM
Hope this helps I am trying to append each individual year from an item in the list in the format[yyyy,yyyy] to an empty list. I am trying to append the first year and second year as well as all the years between them. This is what I have so far, but it is not adding the correct years to the empty list. , There's a simpler way.
code :
startYear = 1986
endYear = 2014

empty_list = range(startYear, endYear+1)
Regular expression validator for Date(yyyy/MM/dd) or year(yyyy)

Regular expression validator for Date(yyyy/MM/dd) or year(yyyy)


By : Pepette
Date : March 29 2020, 07:55 AM
This might help you You don't need to use regex for that.
I would use DateTime.TryParseExact overload that takes string[] as a parameter.
code :
string s = "2012";
DateTime dt;
var formats = new[] {"yyyy", "yyyy-MM-dd"};
if(DateTime.TryParseExact(s, formats, CultureInfo.InvariantCulture,
                          DateTimeStyles.None, out dt))
{
    // Your string is a valid DateTime for yyyy or yyyy-MM-dd format.
}
How to extract year and change format to mm-dd-yyyy from yyyy/mm/dd in BigQuery

How to extract year and change format to mm-dd-yyyy from yyyy/mm/dd in BigQuery


By : Jian
Date : March 29 2020, 07:55 AM
Hope that helps Not able to extract year from this format(yyyy/mm/dd) and not able to change the format to mm-dd-yyyy . I have tried the following but failed. , Below is for BigQuery Standard SQL
code :
#standardSQL
WITH `project.dataset.table1` AS(
  SELECT "2017/1/1" AS doj UNION ALL
  SELECT "2017/2/3" AS doj UNION ALL
  SELECT "2017/3/5" AS doj UNION ALL
  SELECT "2017/4/7" AS doj UNION ALL
  SELECT "2017/6/9" AS doj 
)
SELECT 
  PARSE_DATE("%Y/%m/%d", doj) doj_as_date,
  EXTRACT(YEAR FROM PARSE_DATE("%Y/%m/%d", doj)) year
FROM `project.dataset.table1`
PHP preg_replace for All Types of Year Formats (YYYY, YYYY-YYYY, YYYY - YYYY)

PHP preg_replace for All Types of Year Formats (YYYY, YYYY-YYYY, YYYY - YYYY)


By : Cyle Dawson
Date : March 29 2020, 07:55 AM
hop of those help? preg_replace_callback() will let you validate/extract your targeted substrings and make conditional replacements in one pass.
A
must be written before all year/year-ranges so long as it is not the very start of the string. The first capture group is \s*, so it will capture zero or more white-space characters before your targeted year/year-range. This element will always exist as [1] in the $m array. The second capture group is the first or only year value. This is required to exist for the callback function to be implemented. This is [2] in the $m array. As for the optional secondary year value, it must follow zero or more white-spaces then a hyphen then zero or more white-spaces. Because there are no capture groups to follow this capture group, php will only generate an element for this group if it is found -- isset() is used to check if [3] exists. Originally, I used \b to ensure that the year values were not substrings of larger digital substrings, but the format of your strings allows the use of white-space matching to confirm an accurate match.
code :
$string = "2007 Chevy Silverado Pickup new body style models 2008-2013 Chevy Silverado All Models 2014 Chevy Silverado 2500HD 3500HD 2007 GMC Sierra Pickup new body style models 2008 - 2013 GMC Sierra All Models 2014 GMC Sierra 2500HD 3500HD 2007-2013 Chevy Tahoe 2007-2013 Chevy Suburban 2007   -   2013 Chevy Avalanche 2007-2013 GMC Yukon Yukon XL Yukon Denali";

echo preg_replace_callback('~(\s*)(19[5-9]\d|20[0-4]\d)(?:\s*-\s*(19[5-9]\d|20[0-4]\d))?(?=\s)~', function($m) {
    //var_export($m);  // un-comment if you want to see each $m array
    //echo "\n---\n";
        return (strlen($m[1]) ? "\n" : "")
                . "<strong>*{$m[2]}"
                . (isset($m[3]) ? " - {$m[3]}" : "")
                . "</strong>";
    },
    $string);
<strong>*2007</strong> Chevy Silverado Pickup new body style models
<strong>*2008 - 2013</strong> Chevy Silverado All Models
<strong>*2014</strong> Chevy Silverado 2500HD 3500HD
<strong>*2007</strong> GMC Sierra Pickup new body style models
<strong>*2008 - 2013</strong> GMC Sierra All Models
<strong>*2014</strong> GMC Sierra 2500HD 3500HD
<strong>*2007 - 2013</strong> Chevy Tahoe
<strong>*2007 - 2013</strong> Chevy Suburban
<strong>*2007 - 2013</strong> Chevy Avalanche
<strong>*2007 - 2013</strong> GMC Yukon Yukon XL Yukon Denali
jQuery masked input - format date as m/d/yyyy or m/dd/yyyy or mm/dd/yyyy or mm/d/yyyy

jQuery masked input - format date as m/d/yyyy or m/dd/yyyy or mm/dd/yyyy or mm/d/yyyy


By : kell
Date : March 29 2020, 07:55 AM
hope this fix your issue I had a similar requirement recently, and the only thing I could find that didn't require me to do a lot of work on tweaking the plugin was to use this modified date masking plugin that is based on the one you mentioned. It seems to be doing the job so far.
If you want to test it out, head to this page.
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