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Parse XSL Date Format


Parse XSL Date Format

By : Akash Panda
Date : November 17 2020, 04:28 AM
help you fix your problem One way to do that would be, just as example for matching the complete node
code :
<xsl:template match="created-at">
  <xsl:copy>
   <xsl:attribute name="type" select="@type"/>
     <xsl:value-of select="substring(translate(.,'T', ' '),1,string-length()-6)"/>
  </xsl:copy>
</xsl:template>
Order Date: <xsl:value-of select="created-at" />
Order Date: <xsl:value-of select="substring(translate(created-at,'T', ' '),1,19)" />


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Trouble bringing in a string with d-MMM-yy format parse.exact to date but keeping date format and date type for sorting

Trouble bringing in a string with d-MMM-yy format parse.exact to date but keeping date format and date type for sorting


By : Sam
Date : March 29 2020, 07:55 AM
will help you SCENARIO: I am passing a datatable back as a DataGrids datasource. I am passing in a string field that is a date. I want to convert it to a date so that I can sort by it but I also want to maintain the same format that it was on the string that came in.
ISSUE: From some of my research it seems that the DateTime type is exactly that a datatype in whatever datatable that you put it in and is not formatable. So even though I bring in an unormal string date and convert it to a datetime via the DateTime.ParseExact, when I put it into my datetime field and try to format it (newRow["colX"] = colXDate.ToString("d-MMM-yy"); //as Scott had above) it still goes in as a set datetime format with hours...etc.
Java : How to parse date format to show specific output format?

Java : How to parse date format to show specific output format?


By : Bharath Reddy Seelam
Date : March 29 2020, 07:55 AM
This might help you In my app, i retrieve date from my database in a specific format. (Generated by PHP) , Try out this Code:
code :
  DateFormat inDF = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); // inputFormat
    DateFormat TodayDF = new SimpleDateFormat("HH'h'mm"); //OutputFormat For today and yesterday
    DateFormat FullDF = new SimpleDateFormat("dd MMM - HH'h'mm"); //Outputformat long

    Date inDate = inDF.parse("2014-06-05 17:50:50");
    //calendar for inputday
    Calendar inCal = new GregorianCalendar();
    inCal.setTime(inDate);
    //startOfToday
    Calendar cStartOfDate = new GregorianCalendar();
    cStartOfDate.set(Calendar.HOUR_OF_DAY, 0);
    cStartOfDate.set(Calendar.MINUTE, 0);
    cStartOfDate.set(Calendar.SECOND, 0);
    cStartOfDate.set(Calendar.MILLISECOND, 0);
    //endOfToday    
    Calendar cEndOfDate = new GregorianCalendar();
    cEndOfDate.set(Calendar.HOUR_OF_DAY, 23);
    cEndOfDate.set(Calendar.MINUTE, 59);
    cEndOfDate.set(Calendar.SECOND, 59);

     //startOfYesterday
    Calendar cStartOfYesterday = new GregorianCalendar();
    cStartOfYesterday.set(Calendar.HOUR_OF_DAY, 0);
    cStartOfYesterday.set(Calendar.MINUTE, 0);
    cStartOfYesterday.set(Calendar.SECOND, 0);
    cStartOfYesterday.set(Calendar.MILLISECOND, 0);

     //endOfYesterday
    Calendar cEndOfYesterday = new GregorianCalendar();
    cEndOfYesterday.set(Calendar.HOUR_OF_DAY, 23);
    cEndOfYesterday.set(Calendar.MINUTE, 59);
    cEndOfYesterday.set(Calendar.SECOND, 59);

    if (cStartOfDate.before(inCal) && cEndOfDate.after(inCal)){
      System.out.println("Today "+TodayDF.format(inDate));
    } else if (cStartOfYesterday.before(inCal) && cEndOfYesterday.after(inCal)){
      System.out.println("Yesterday"+TodayDF.format(inDate));
    }  else {

      System.out.println(FullDF.format(inDate));
    }
How to define DateTime parse format for general date format with optional time part?

How to define DateTime parse format for general date format with optional time part?


By : user2504633
Date : March 29 2020, 07:55 AM
help you fix your problem If your CurrentCulture supports MM/dd/yyyy h:mm:ss tt (I assume your LongTimePattern has h) and M/dd/yyyy (I assume your ShortDatePattern has M) as standard date and time format, using DateTime.TryParse(String, out DateTime) method can solve all your problems.
code :
string s = "";
DateTime dt;
if (DateTime.TryParse(s, out dt))
{
    // Your string parsed successfully.
}
string s = "";
string[] formats = { "MM/dd/yyyy h:mm:ss tt", "M/dd/yyyy" };
DateTime dt;
if (DateTime.TryParseExact(s, formats, CultureInfo.CurrentCulture,
                           DateTimeStyles.None, out dt))
{
    // Your string parsed with one of speficied format.
}
How to parse JSON format date string into date format

How to parse JSON format date string into date format


By : J. Davis
Date : March 29 2020, 07:55 AM
wish of those help If there is no issue in adding a dependency, then you can add moment.js and it will help you to format data in any format I am supposing that date from server is in this format '/Date(725828400000)/'
code :
var d = item.EMP_DOB;
result.push(moment(Number(d.match(/\d+/)[0])).format('MM/DD/YYYY'));
var date = new Date(Number(d.match(/\d+/)[0]));
var day = date.getDate();
day = day = (day < 10) ? ("0" + day) : day;
var month = date.getMonth() + 1);
month = (month < 10) ? ("0" + month) : month;
var dateStr = day + "-" + month + "-" + date.getFullYear();
result.push(dateStr);
python read_csv pandas parse column with format like 1990M01 to a montly date format

python read_csv pandas parse column with format like 1990M01 to a montly date format


By : Peter G.
Date : March 29 2020, 07:55 AM
Does that help I have a CSV file like that , Use:
code :
import pandas as pd
from io import StringIO

temp=u"""obs,yr30,tbill3m,ret3m
1990M01,7.98,7.8,0.028205
1990M02,8.44,8.02,0.007481
1990M03,8.61,8.08,-0.003713"""
#after testing replace 'pd.compat.StringIO(temp)' to 'filename.csv'
df = pd.read_csv(StringIO(temp),  
                 index_col=0)

print (df)
         yr30  tbill3m     ret3m
obs                             
1990M01  7.98     7.80  0.028205
1990M02  8.44     8.02  0.007481
1990M03  8.61     8.08 -0.003713
df.index = pd.to_datetime(df.index, format='%YM%m')
print (df)
            yr30  tbill3m     ret3m
obs                                
1990-01-01  7.98     7.80  0.028205
1990-02-01  8.44     8.02  0.007481
1990-03-01  8.61     8.08 -0.003713
df.index = pd.to_datetime(df.index, format='%YM%m').to_period('m')
print (df)
         yr30  tbill3m     ret3m
obs                             
1990-01  7.98     7.80  0.028205
1990-02  8.44     8.02  0.007481
1990-03  8.61     8.08 -0.003713
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