Answers

• tangerine Posts: 3,365Questions: 39Answers: 395

  March 2020

  I think you need $d where you have $date - that is, you're passing $d to the formatter which is formatting $date.
• bbrindza Posts: 316Questions: 73Answers: 1

  March 2020

  Made the change and still returns false.

  array( 'db' => 'TIME_LOG_DATE','dt' => 'time_log_date',
  'formatter' => function( $d, $row ) {
  return date_format($d, 'l F d, Y');
  }
  ),
• tangerine Posts: 3,365Questions: 39Answers: 395

  March 2020

  What is the value of $d at this point: function( $d, $row )?
• bbrindza Posts: 316Questions: 73Answers: 1

  March 2020

  $d = {"time_log_date":"2020-03-23"}
  $row = {"time_log_date":"2020-03-23"}
• kthorngren Posts: 21,327Questions: 26Answers: 4,949

  March 2020

  I'm not familiar with PHP but do you need to do something along the lines of
  return date_format($d["time_log_date"], 'l F d, Y');

  Kevin
• bbrindza Posts: 316Questions: 73Answers: 1

  March 2020

  Thanks for the suggestion Kevin, but that returns false as well.

  {"time_log_date":false,"}
• rf1234 Posts: 2,990Questions: 87Answers: 421

  March 2020 edited March 2020

  Hmmm ... I am using Data Tables not with the SSP class but with Editor and PHP and MySQL.

  Here is something on getFormatters for dates that might be helpful:
  https://datatables.net/forums/discussion/comment/146135/#Comment_146135

  Assuming your date comes from an SQL database in the usual format:
  "2020-03-24 00:00:00" or as "2020-03-24".
  And you just want one front-end date format regardless of the user language. Then it is really easy just using plain PHP, no built-in formatters:

  Field::inst( 'yourDateField' )
      ->getFormatter( function($val, $data, $opts) {
          $date = new DateTime($val);
          return $date->format("d/m/Y"); //or any other format you require
      }),  
  
• rf1234 Posts: 2,990Questions: 87Answers: 421

  March 2020 edited March 2020 Answer ✓

  if $d = {"time_log_date":"2020-03-23"}
  which means it is an object ?! (I am not so sure ...)
  this should work unless it is not an object OR there is another bug in there.

  array( 'db' => 'TIME_LOG_DATE','dt' => 'time_log_date',
     'formatter' => function( $d, $row ) {
        //convert obj to assoc array
        $dateArr = json_decode(json_encode($d), true);
        $date = new DateTime($dateArr["time_log_date"]);
        return $date->format("d/m/Y"); //or any other format you require
      }
  ),
  

  I just tried this in PHP sandbox and it worked. Please note: I had to create the object with PHP which is different syntax then what you present. So in the code below $d is definitely a PHP object. And this object is being converted to an associative array

  $d = (object)["time_log_date" => "2020-03-23"];
  //convert obj to assoc array
  $dateArr = json_decode(json_encode($d), true);
  $date = new DateTime($dateArr["time_log_date"]);
  echo $date->format("d/m/Y"); //or any other format you require
  

  Result: 23/03/2020

  This is what you need to check whether your variable really is an object:
  https://www.php.net/manual/en/function.is-object.php

  

  So if $d = ["time_log_date" => "2020-03-23"] which means it is an associative array you can simply skip the conversion:

  array( 'db' => 'TIME_LOG_DATE','dt' => 'time_log_date',
     'formatter' => function( $d, $row ) {
        $date = new DateTime($d["time_log_date"]);
        return $date->format("d/m/Y"); //or any other format you require
      }
  ),
  
• rf1234 Posts: 2,990Questions: 87Answers: 421

  March 2020 edited March 2020

  forgot to mention that this also works if it is an object ... sorry. When working with Editor I am so used to assoc arrays that I didn't even think about this simple way ...

  $d = (object)["time_log_date" => "2020-03-23"];
  $date = new DateTime($d->time_log_date);
  echo $date->format("d/m/Y"); //or any other format you require
  
• bbrindza Posts: 316Questions: 73Answers: 1

  March 2020

  rf123 - you where close and helpful. However I need to use $row not $d and removed the comments for the column name. It work!

  array( 'db' => 'TIME_LOG_DATE','dt' => 'time_log_date',
      'formatter' => function( $d, $row ) {
      $date = new DateTime($row[time_log_date]);
      return $date->format("d/m/Y"); //or any other format you require
      }
      ),
  

  Using $d in DateTime function threw this error on my PHP server.

  Uncaught Exception: DateTime::__construct(): Failed to parse time string (2) at position 0 (2): Unexpected character in /www/zendphp7/htdocs/TimeLog/ssp_TimeLogTable.php:33 Stack trace: #0 /www/zendphp7/htdocs/TimeLog/ssp_TimeLogTable.php(33): DateTime->__construct('2') #1 /www/zendphp7/htdocs/nwfFunctions/ssp.class.php(35): {closure}('2020-03-24', Array) #2 /www/zendphp7/htdocs/nwfFunctions/ssp.class.php(269): SSP::data_output(Array, Array) #3 /www/zendphp7/htdocs/TimeLog/ssp_TimeLogTable.php(70): SSP::simple(Array, Resource id #1, NULL, 'NWFF.TIME_LOG', 'EMPLOYEE_NUMBER', Array, '') #4 {main} thrown

  **THANK YOU ** so much for your insight,