iekrasot dienas

[destroy_all]

foreach nočakarē visu tabulas struktūru.

ok, pamēģināsu šito variantu.

[bubu]

foreach nočakarē visu tabulas struktūru.

wtf? foreach neko nečakarē, ja vien pats tu nemaini datus. Tikpat labi arī while/for/if "čakarē visu tabulas struktūru" (lai ko arī tas nozīmētu).

[destroy_all]

jā, nenocakarē gan. http://paste.php.lv/4334

 

- atkal tā pati problēma.

Edited August 24, 2006 by destroy_all

[bubu]

Dumāku for/if kombināciju sen nebiju redzējis. Dzēs visu ārā, un sāc zīmēt blokshēmas. Citādi palīdzēt vairs nevaru.

[Kristabs]

es dariitu taa:

 

select visas vajadziigaas dienas {
  $from_till[$sql['no']] = $sql['lidz'];
}

for(){ //ziimeejam to datu tabulu
  foreach($from_till as $no => $liidz){
  if($cur_day >= $no || $cur_day <= $liidz){
	  //iekraasojam
  }
 }
}

[Delfins]

Nē... es darītu tā..

 

foreach/while ( $room = $query->get() )
{
print '<tr>'; 
$activities = array();
while ($activity)
{
   $activities[$activity->diena] = $activity->diena;
}

for ($diena=DienaNo;$diena<=DienaLidz;$diena++)
{
	print "<td class="'.(empty($activities[$diena])?'free':'used').'" ... > ... ";
}
print '</tr>'; 
}

Edited August 25, 2006 by Delfins

[Gundars]

$days=array_fill(0,31,'green'); //uztaisam array ar vajadziigo dienu skaitu

 

 

for($i=$day_from;$i<=$day_to;$i++) $days[$i]='pink'; //ieziimejam arraya aiznjemtas dienas

 

echo '<tr>';

foreach ($days as $day) echo "<td class=\"$day"\ ></td>"; /izvadam attieciigo array, kaa vien veelamies

echo '</tr>';

[bubu]

Delfin: kāda jēga inicializēt masīvu $activities ciklā, ja tas ir invariants pret to?

[Delfins]

bubu, a iedomaajies N periodus

$activities tiek inizializēts no datubazes...

 

ACT:_E_E_F_F_F_E_E_E_E_F_
DAY:_D_D_D_D_D_D_D_D_D_D_
HTM:_0_0_1_1_1_0_0_0_0_1_

 

pseidokods

while ($activities_row)
{
 // add to array
}

for (diena = [diena_from .. diena_to])
{
  stils = (empty(activity[diena])) ? "FREE" : "USED";
  print "<td STILS>DIENA</td>"
}

Edited August 25, 2006 by Delfins

[john.brown]

IMHO, tam tā vajadzētu izskatīties. Testējis gan neesmu, tak pēc loģikas, vajadzētu iet :)

<?php

$days_this_month = 31; // te dabunam konkreta menesa dienu skaitu

$query = "SELECT room_id FROM rooms";
$res = mysql_query($query);
?>
<table>
<?
while($row = mysql_fetch_array($res)) {
$occupied = get_occupied_dates($row['room_id']);  

echo '<tr>';

for($i = 1; $i <= $days_this_month; $i++) {
	$class = in_array($i,$occupied)? 'occupied' : 'free';

	echo '<td class="'.$class.'">'.$i.'</td>';
}
echo '</tr>';	  
}
?>
</table>
<?
function get_occupied_dates($room_id) {
$occupied = array();
$query = "SELECT DAY(oc_from) as oc_from, DAY(oc_to) as oc_to FROM activity WHERE room_id=$room_id";
$res = mysql_query($query);
while($row = mysql_fetch_array($res)) {
	$occ = range($row['oc_from'],$row['oc_to']);
	$occupied = array_merge($occupied,$occ);
}
return $occupied;
}
?>

[destroy_all]

john.brown, paldies, viss strādā!