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[PHP] Something is wrong with "date" => 1970




Hello *,

the code sniplet is:

----[ 'energy_harvesting.php' ]-----------------------------------------

$LIST=shell_exec('ls /srv/CONFIG_www.miila-mahe-aed.eu/htdocs/renewable_energies/data/20*.data |sort');

$ARRAY=preg_split('/\s+/', $LIST, -1, PREG_SPLIT_NO_EMPTY);

$TITLE = "";

foreach ( $ARRAY AS $DAY) {
  include $DAY;
  $NEW_TITLE = date('Y', $date);

  if ( $NEW_TITLE != $TITLE ) {
    echo '<TABLE class="table_energy">';
    echo '<TR class="table_energy_row">';
    echo '    <TD class="energytable_title" colspan="6"><SPAN class="comments">' . $NEW_TITLE . '</SPAN></TD>';
    echo '</TR>';
    echo '<TR class="table_energy_row">';
    echo '    <TD class="energytable_num"     ><SPAN class="comments">No</SPAN></TD>';
    echo '    <TD class="energytable_date"    ><SPAN class="comments">Date</SPAN></TD>';
    echo '    <TD class="energytable_time"    ><SPAN class="comments">Time</SPAN></TD>';
    echo '    <TD class="energytable_wh_solar"><SPAN class="comments">Wh Solar</SPAN></TD>';
    echo '    <TD class="energytable_wh_wind" ><SPAN class="comments">Wh Wind</SPAN></TD>';
    echo '    <TD class="energytable_comment" ><SPAN class="comments">Comments</SPAN></TD>';
    echo '</TR>';
    $TITLE = $NEW_TITLE;
  }

echo "<PRE>";
echo $DAY . "</P>";
echo $date . "</P>";
echo $NEW_TITLE . "</P>";
echo "</PRE>";
}
------------------------------------------------------------------------

it does not work as expected.

$DAY is corect the filename.
$date is a value inside the (php) file and is correct e.g. 2017-08-31

However, $NEW_TITLE is not working and show always "1970 January"

Where is my thinking error?

Does date() not more accept an ISO Date of 2017-08-31?

I had already used it and it was working properly.

Thanks in avance

-- 
Michelle Konzack        Miila ITSystems @ TDnet
GNU/Linux Developer     00372-54541400

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