A better way to use } else if {...

ATS16805

i'm trying to create a page where i can add records into my "days_of_week" table as shown here

what would have been a better, more efficient way to check my db for a "day of week" id # to tell a form which "day" we're adding:

$check_day_id = "select day_id from days_table"
	$check_day_id_res = mysql_query($check_day_id) or die(mysql_error());
	// LOGIC= IF RESULTS <= 0 THEN DAY IS MONDAY ETC
	if $check_day_id_res == 0 {
		$day_of_week = "Monday";
	} else if $check_day_id_res == 1 {
		$day_of_week = "Tuesday";
	} else if $check_day_id_res == 2 {
		$day_of_week = "Wednesday";
	} else if $check_day_id_res == 3 { 
		$day_of_week = "Thursday";
	} else if $check_day_id_res == 4 {
		$day_of_week = "Friday";
	} else if $check_day_id_res == 5 {
		$day_of_week = "Saturday";
	} else if $check_day_is_res == 6 {
		$day_of_week = "Sunday";

thanks!

thorpe

<?php

  $weekarray = array('Monday' => 0,'Tuesday' => 2,'Wednesday' => 3,'Thursday' => 4,'Friday' => 5,'Saturday' => 6,'Sunday' => 7);
  foreach ($weekarray as $key => $val) {
    if ($check_day_id_res == $val) {
      $day_of_week = $key;
      break;
    }
  }
?>

ATS16805

Thorpe,

cool, thanks!

hey, will my primary key start out as 0 or 1? it will be 1, correct? not sure why i was using the zero in there to start with...

do you think this is solid way to develop this element of my event scheduler app? i suppose i needn't provide the option (not that it's an option, as it seems it must start on monday and go through till sunday) to set days details individually, and in days-of-the-week order. i just feel there was a better way to do it to ensure i've got records for each day. maybe i'm wrong-- maybe i don't need this-- after all, i've got logic on another page checking for the records' existence.

the advantage (i think) to my approach is that a concert date can be scheduled on a monday once the monday "specials" have been entered into that table-- thereby allowing the user to move forward w/ scheduling before his "specials" table is completely populated.