convert string to time format

Similar Questions

• Return the string Date/time format incorrect output...

  Here's a small sub - VI that I create out the day of the week, day of month, month and year in the form of a table of U16:

  

  The Date/time string VI get is supposed to produce a string of date in the format "DD/MM/YY" when the enum 'short' is selected. Instead, I get this:

  

  It seems that the VI is "broken" or documentation is incorrect. Someone at - it to another incountered?

  I'm under LabView 2015 on Windows 7.

  Much too complicated.

  

• Convert number to time Format

  Hello
  
  I'm trying to find how to display a number in the format hour, Minute, and second.
  
  I have a data in my table that store the number for example. 145.586956521739
  
  I want to convert in HOUR, Minute Secode report format.
  
  Any help is greatly appreciated.
  
  Thank you
  Poojak

  Say total = 189,89 minutes

  hours = round(total/60)

  minutes = round (mod(total,60))

  second round = (mod(total,1) * 60)

  Sorry if I misinterpreted your question.

  Thank you
  Jalila

• Convert String performed far and format her

  Hello

  I am new to Adobe and javascript. I have a string value that stores the Date of birth in yyyy-mm-dd format, as a string. I need to convert this Date and then place it on a PDF field in the format mm/dd/yyyy.

  Can anyone help?

  Thank you

  You can convert a string representing a date in a date object using the util.scand JavaScript method and and then format a date to a string object by using the util.printd method. For more information, see:

  http://livedocs.Adobe.com/acrobat_sdk/9.1/Acrobat9_1_HTMLHelp/JS_API_AcroJS.88.1254.html

  http://livedocs.Adobe.com/acrobat_sdk/9.1/Acrobat9_1_HTMLHelp/JS_API_AcroJS.88.1251.html

  In your case, the code would be something like:

  sDate var = '2013-01-10;

  Convert string to date

  Kai var = util.scand ("yyyy-mm-dd", sDate);

  Convert date to a new string

  var sDate2 = util.printd ("mm/dd/yyyy", anyway);

  Set a field value

  getField("date2").value = sDate2;

  The exact code you use depends on where you place the script and where you get the original date string, but this should help you get started.

• question of activeX cw graph3D time format

  With regard to the component works / ComponentWorks 3D chart control ActiveX OCX:

  I have a table of LabVIEW timestamp ("the default value is zero seconds since 12:00, Friday, January 1, 1904, universal time [01/01/1904 00:00:00] '") that I convert it to an array of Doubles.

  I would like to pass this array to the OCX (after conversion) as my y-axis. I have configured the OCX to display DD/MM/YYYY HH: mmusing the property editor. In addition, if you dig into the documentation of OCX, in the section that details the formatting strings, it is said:

  "The ActiveX control, the date is implemented as a floating-point value, measuring days from midnight, 30 December 1899. Thus, midnight, 31 December 1899, is represented by 1.0. Similarly, 06:00, January 1, 1900, is represented by 2.25, and midnight, 29 December 1899, is - 1.0. 06:00, 29 December 1899 is however - 1.25. To interpret the time portion, take the absolute value of the fractional part of the number. Thus, a second is equal to 1 / 24 hours / 60 minutes / 60 seconds, which is 1/86400 or about 1.157407e - 5. »

  Maybe my coffee is not strong enough today, but I don't see an easy way to convert the format of LabVIEW in the "ActiveX format" while properly accounting for leap years etc...

  My first hope is that someone already has, my backup plan is to have a good brainstorming on the forum on how to do this?

  Thanks for any input on this!

  I made a few changes that might be useful for others, or in the forum.

  In my case, I have an array of strings, which gives the date and time for when the data has been captured.

  The date and time strings were captured in another time zone of the computer where I run the analysis and visualization. Therefore, you must take care so that you don't end up adjust your time incorrectly based on your analysis computer-time zone schedules vs. where the data has been captured.

  I don't know if there may be a more smooth way and more optimized to do this, but I compared it and for data sets that I'm loading, the time required to perform this conversion is fairly trivial.

  I guess the point key to looking people should take away the format "chain" and "format of the string" VI is extremely powerful and his is worth the effort of learning of the March bump in the way to create the format specifier strings...

  

  The code in the image shows how to convert a string to the time format format OLE/ActiveX without changing time I have in the original string array. I can then feed what (in my case) table of axis y for the CW 3-d chart where my axis has been configured to display 'time' (container, right-click ActiveX, "CWGRAPH 3D--> properties", go to "format" tab, select your axis and choose the display format of your choice.)

  as by chance, doing the same use the xControl turns into a nightmare when the only solution I found was to use pairs of value and a lot of code to manage the positioning and the decimation of the value pairs to work correctly... Overall, my treatment has now passed a lot for next to nothing... and as a bonus, the CW graphic cursors use the same formatting settings, so that my date/times on the slider too, on the xControl, the use of cursors the part number of a pair value, so your cursor does not match your axis...

• How to convert String to Date

  hell there everyone,

  I have a question which is:

  How to convert a string that we get by reading an XML in the instance of date value type?

  So there is a XML that modifiedDate as an attribute on a node of the XML and how I put this value in an instance variable date?

  the timeformat in the XML file is: mm/dd/yyyy hh: mm: = > 31/12/2010 23:57:46

  I chose this time format because I want to use the method . parse() which always causes an error, but that's another topic...,.

  hope I was clear on the issue,

  Thanks in advance for the help.

  First of all, you should watch the Date object constructor to find out how to specify different parameters.

  public void Date (yearOrTimevalue:Object, month:number, date:number = 1, hour:number = 0, minute:number = 0, second:number = 0, millisecond:number = 0)

  Then you need to determne how to break the string by using the String.Split method.  The first split() should be the space between the date and time, the second for the slashes in the date and the third for the colon in time.

  Then you come to take each of these paintings of substrings and feed them in the constructor as the Number() values.

  If you remain unsure how to do this, look at the different terms that I have identified in the help documents and you will find them all explained with examples.

• Date & time format to_timestamp

  Hello
  
  I struggled for a while now with the time format. It seems to be interpreted as a date as well. I just want it to be like in the original (with the difference
  that date should be dd/mm/yyyy.
  
  
  -original format 0307 10/14/2009
  
  Select to_timestamp (substr (ALLTEXT, 84.14), ' YYYY/MM/DDHH24MI') AS DATEOFCRE
  of csv_euro
  WHERE
  substr (ALLTEXT, 6.2) = "HH." -HEADER
  
  -result 14 October 09 03.07.00.000000000
  
  You can see that the time becomes something weird like 03.07.00.000000000. that is not really what I'm looking for. I'm looking for 0307.
  
  can someone help me with this problem?

  If you want to insert it in a date field, you must convert it to a date

  INSERT INTO EUROHEADER(
  DATEOFCRE
  )
  select to_date(substr(ALLTEXT, 84,14) , 'YYYY/MM/DDHH24MI')
  from csv_euro
  WHERE
  substr(ALLTEXT, 6,2) = 'HH'; 
  

  You get errors in your example, because your database's default date format does not match the string that extract you from your csv_euro.

  Concerning
  Marcus

  Edited by: Marwim the 03.12.2009 13:50

• When the value to display in the time format double digital indicator... Changes in value.

  Hi all

  I can't display the values that I want. I have a value double say 320 sent to a digital indicator where I edited the properties of it to display in a time format that contains only minutes and seconds as Yes, 3:20 '. Unfortunately, it takes the value and change it to another value and display that. I don't know what is the issue. If anyone knows the solution... Please post!

  There is an image attatched!

  Thank you.

  I believe only then due to the fact that when I convert a timestamp value to double as 03:20, he gave me 320. So I thought that the reverse can be trying. Should I just enter the total amount of seconds and that will show the Minuteseconds I want?

  Thanks for the quick response.

• convert string to float

  Hello

  I would like to convert string to float. I have a text which float to string contents of the file number.

  for example:

  10,000 2 of. 1.32323 0000 these channels I would like to convert float or double number.

  Help, please!

  I use the string of spreadsheet in table vi and I need to properly adjust the format string.

  %.; %f

  Thank you all!

• Convert String to DATE

  Hello

  There is a '20091002' chain, and I need to convert it to this day.

  20091002 is in format YYYYMMDD - the VARCHAR2 data type

  I need the above in DD/MM/YYYY - DATE datattype

  '20091002' (VARCHAR2) - 10/02 / 2009 "(DATE)

  Thank you

  Aman

  Hi, Aman,

  user13115886 wrote:

  Hello

  There is a '20091002' chain, and I need to convert it to this day.

  20091002 is in format YYYYMMDD - the VARCHAR2 data type

  I need the above in DD/MM/YYYY - DATE datattype

  ...

  You are contradicting yourself.  All dATEs have the same format, and it's nothing like YYYYMMDD or DD/MM/YYYY.

  If you need a DATE, use TO_DATE:

  To_date (str, 'YYYYMMDD')

  but it will not be in the format MM/DD/YYYY

  If you want to convert a string into a format to another string in another format, and then use the string manipulation functions to reorganize secondary channels, for example

  REGEXP_REPLACE (str

  '(\d{4})(\d{2})(\d{2})'

  , "\2/\3/\1".

  )

  or

  SUBSTR (str, 5, 2) | '/' ||

  SUBSTR (str, 7, 2) | '/' ||

  SUBSTR (str, 1, 4)

  but it will not be a DATE.

  If you were certain that str contains valid data, then you can call TO_DATE to convert a DATE and then call TO_CHAR to convert the DATE to a string in the format MM/DD/YYYY.  I do not recommend, because it is likely that the strings would not be valid.  It's one of the reasons why you should use DATEs to store the date information.

• The analysis of string to xml format

  Dear all,
  
  I have an OSB service that returns an xml string output form.
  I call this OSB service in BPEL but it fails because of the format of the string received by BPEL.
  
  How can I avoid it. I also try to convert string to xml but failing to do so.
  
  The pointers to that would be useful
  
  Thank you and best regards,
  MITS

  try to set the response from transport HTTP header
  http:content-type=text/XML
  application/xml or (even better)

  (and of course, you must make sure you have fn - bea: inlinedXml() function to convert the string in an XmlObject)

• How can I convert string to xml in CEP

  I made a simple CEP process that reads from a queue. The CEP has the following components:
  1 input adapter named MktEvtInputadapter.
  2 channel MktEvntInchannel
  3. a MktEvntProcessorprocessor processor.
  4. a MktEvntOutChannel output.
  5. a MarketEventBean bean.
  
  code for MarketEvent.java:
  package com.bea.wlevs.example.algotrading.event;
  
  public class MarketEvent {}
  
  Private Sub Long;
  symbol of the private channel;
  price Double private;
  volume of Long private;
  private Long latencyTimestamp;
  private string text;
  
  /**
  * @return the latencyTimestamp
  */
  {} public Long getLatencyTimestamp()
  Return latencyTimestamp;
  }
  /**
  @param latencyTimestamp the latencyTimestamp to
  */
  {} public void setLatencyTimestamp (Long latencyTimestamp)
  this.latencyTimestamp = latencyTimestamp;
  }
  /**
  * the timestamp @return
  */
  {} public Long geteventTimeStamp()
  return Sub;
  }
  /**
  @param timestamp the timestamp to set
  */
  {} public void seteventTimeStamp (Long Sub)
  this.eventTimeStamp = Sub;
  }
  /**
  * @return the symbol
  */
  public String getSymbol() {}
  return the symbol;
  }
  /**
  @param symbol the symbol to define
  */
  {} public void setSymbol (symbol of the chain)
  This.Symbol = symbol;
  }
  /**
  * @return the price
  */
  {} public Double getPrice()
  return of the price;
  }
  /**
  @param price TBD
  */
  {} public void setPrice (Double rate)
  This.Price = price;
  }
  /**
  * @return the volume
  */
  {} public Long getVolume()
  return of volume;
  }
  /**
  @param volume the volume to set
  */
  {} public void setVolume (Long volume)
  This.volume = volume;
  }
  public void setString_1 (String text) {}
  This.Text = text;
  TODO self-generating method stub
  
  }
  public String getString_1() {}
  TODO self-generating method stub
  return (text);
  }
  
  }
  
  Code for MarketEventBean:
  package com.bea.wlevs.example.algotrading;
  
  import com.bea.wlevs.ede.api.StreamSink;
  import com.bea.wlevs.example.algotrading.event.MarketEvent;
  
  
  / public class MarketEventBean implements StreamSink {}
  {} public void onInsertEvent (event object)
  If (event instanceof MarketEvent) {}
  MarketEvent marketEvent = (MarketEvent) event;
  
  System.out.println ("msg:" + marketEvent.getString_1 ());
  }
  }
  
  }
  
  I put an xml doc in my queue for entry:
  <? XML version = "1.0" encoding = "UTF-8"? >
  < MarketEvent >
  < sub > 201100501 < / sub >
  < symbol > ABCD < / symbol >
  < price > 258474.56 < / price >
  < volume > 4657892 < / volume >
  < latencyTimestamp > 201100501 < / latencyTimestamp >
  < / MarketEvent >
  
  My class MarketEventBean will print the msg as string. Now, I have to convert it to xml format to get the value of individual items.
  How can I do?

  I think you need to do this:

  Events.Add (obj);

  in this way, you add the event unmarshaled flow rather than just a string of text

• How convert string to number

  How convert string to number
  
  
  my character
  
  
  Select form double to_number ('1,6,9,4'); my putting here 1,6,9,4
  
  get the character of number conversion error.
  
  concerning
  
  Published by: vr on March 31, 2011 23:59
  
  Published by: vr on March 31, 2011 23:59

  VR says:
  Getting error while the conversion of character to the number in instring is possible to get the position of the channels at one time

  I have the following table, called pack

  Acct_No pack2 pack3 pack1
  1000 10101011 111111101 01010101111

  Select * Pack
  where instr (pack2, ' 1') and * (to_number (replace('1,6,9,4',','))); * - getting values of function as string '1,6,9,4'

  If you want to retrieve lines where the position of the first "1" in the pack2 chain is one of the positions in your list number e.g. position 1, 6, 9, or 4?

  Something with regexp_like can help...

  SQL> ed
  Wrote file afiedt.buf
  
    1  with pack as (select '1000' as acct_no, '10101011' as pack1, '111111101' as pack2, '01010101111' as pack3 from dual union all
    2                select '1001', '10101011', '000011111', '10101010110' from dual union all
    3                select '1001', '10101011', '000000001', '10101010110' from dual)
    4  --
    5  -- end of test data
    6  --
    7  select pack.*, case when regexp_like('1,6,9,4','(^|,)'||to_char(instr(pack2,'1'),'fm9999')||'(,|$)') then 'Yes' else 'No' end as x
    8* from pack
  SQL> /
  
  ACCT PACK1    PACK2     PACK3       X
  ---- -------- --------- ----------- ---
  1000 10101011 111111101 01010101111 Yes
  1001 10101011 000011111 10101010110 No
  1001 10101011 000000001 10101010110 Yes
  

• String to XML format

  Hello

  How can I convert a string in XML format if I use tags?

  sergei_developer wrote:

  You must use:

  var x: XML =blaaaaaaat;

  var s:String = x.info.@title;

  This code is incorrect and will not work. 'title' is not an attribute but a node if you need to access it this way and also, the node root (called 'info') is considered implicit, which means you acceding the node 'title' in this way: var s:String = x.titel;

  Good luck

  Barni

• type conversion of numeric value in the time format

  Hi guys,.
  I have the column called need .i (time in seconds) to convert the numeric value of a column in time format HH. for example 3012 = 00: 50.12.
  How can I proceed even in the front.
  
  
  Thank you and best regards,
  Nandu

  Hi Nelly,

  Pls go through the below formula:
  This deffenitely will help you resolve the issue.

  Concat (concat (concat (concat (cast (floor ((COLUMN_NAME) / 3600) as char), ':'), cast (floor (mod ((COLUMN_NAME), 3600) / 60) as char)),': '), cast (floor (mod (mod ((COLUMN_NAME), 3600), 60)) as char))

  Thank you
  Cognet