MySQL Query In PHP to get a field out of a table..

[Razorster]

Hi there,

I am in need of desperate help. I am using a PHP calendar script. I am trying to display an image for each date in the calendar.

See my attempt here:

http://proudcabaret.com/calendar/cal...h=11&year=2009

Its not quite right. The image field is called "promoimage" which is part of every record in the table of events. This is a URL. My code is as follows:

PHP Code:

<?

include('admin/dbconn.php');
include('functions/functions.php');

include('header.php');

$query= 'SELECT * FROM calendar_event ORDER BY day';
$result=mysql_query($query);
$num=mysql_numrows($result);

$type = CAL_GREGORIAN;

$month = $_GET['month'];
if(!$_GET['month']) { $month = date('n'); } // Month ID, 1 through to 12.

$year = $_GET['year'];
if(!$_GET['month']) { $year = date('Y'); } // Year in 4 digit 2009 format.

$today = date('Y/n/d');

$day_count = cal_days_in_month($type, $month, $year); // Get the amount of days in the chosen month to give to our function.

echo "<div id='calendar'>";
echo "<div id='calendar_wrap'>";
   
// Function for year change. //
   
$last_month = $month - 1;
$next_month = $month + 1;

$last_year = $year - 1;
$next_year = $year + 1;

   if($month == 12) {
   $change_year = $year;
   $change_month  = $last_month;
   } elseif($month == 1) {
   $change_year = $last_year;
   $change_month  = '12';
   } else {
   $change_year = $year;
   $change_month  = $last_month;
   }
   
   if($month == 1) {
   $change_year_next = $year;
   $change_month_next  = $next_month;
   } elseif($month == 12) {
   $change_year_next = $next_year;
   $change_month_next  = '1';
   } else {
   $change_year_next = $year;
   $change_month_next  = $next_month;
   }

// Do NOT edit the above. //

      echo "<div class='title_bar'>";
      
      echo "<a href='/calendar/calendar.php?month=". $change_month ."&year=". $change_year ."'><div class='previous'></div></a>";
      echo "<a href='/calendar/calendar.php?month=". $change_month_next ."&year=". $change_year_next ."'><div class='next'></div></a>";
      echo "<h2 class='month'>" . date('F',  mktime(0,0,0,$month,1)) . "&nbsp;" . $year . "</h2>";
      
      
      echo "</div>";

for($i=1; $i<= $day_count AND $num; $i++) { // Start of for $i

   $date = $year.'/'.$month.'/'.$i;
   
   $get_name = date('l', strtotime($date));
   $month_name = date('F', strtotime($date));
   $day_name = substr($get_name, 0, 3); // Trim day name to 3 chars
   $day_st = date('S', strtotime($date));
   
   $count = count_events($i, $month,$year);
      
   echo "<a href='day_view.php?day=$i&month=$month&year=$year' title='$i $month_name' rel='day_view'>";
   echo "<div class='calwrap'>"; // wrapper
   echo "<div class='cal_day'>"; // Calendar Day
      
      echo "<div class='day_heading'>" . $day_name . "</div>";
      
      if($today == $date) {
         echo "<div class='day_number today'>" . $i . $day_st ."</div>";
      } else {
         echo "<div class='day_number'>" . $i . $day_st ."</div>";
      }   
      
      $promoimage=mysql_result($result, $i, "promoimage");
         
      if($count >= 1) {
      
         if (empty($promoimage)) {
            echo "<span class='event'>Event</span>";

         } else {
             echo "<span class='event'><img src='". $promoimage ."'></span>"; }
      
      }
      
   echo "</div>";
   echo "</div>";
   echo "</a>";
   
} // EOF for $i

   echo "</div>";
?>

<script type="text/javascript">
$(document).ready(function(){ $("a[rel='day_view']").colorbox({width:"500px", height:"450px", iframe:true}); });
</script>

If anyone can help, id be eternally greatful. I think my coding is wrong.

[toxic_brain]

Its not quite right

I think my coding is wrong.

Please mention what is the expected result and the actual result.

[Razorster]

Hi there mate,

The expected output is for each day to show its image URL - logo. The first 14 show, but they are shifted to the right (wrong order). Then last 7 should show, but they do not.

Any ideas?

[Weedpacket]

PHP Code:

      $promoimage=mysql_result($result, $i, "promoimage");


One problem I see is that you've forgotten that result row numbers start from 0, while $i starts from 1.

[Razorster]

hi mate,

thanks for your reply! I appreciate it.

$i starts from 1 because if it were to start from 0 the first date of the month would be the 0st, which doesnt exist lol. I am so confused! PHP is not my expertise, any other ideas mate?

[Weedpacket]

mysql_result

Row numbers start at 0.

So .... ooh, this is difficult ....

PHP Code:

$promoimage=mysql_result($result, $i - 1, "promoimage");


[Razorster]

thank you! that didnt seem to solve it though? it didnt change anything.

[Razorster]

anyone?

[woodeye]

I'm trying to figure out how you have your database set up. From what it looks like, unless your database has an entry for EVERY day, then it's not going to work.

Have you tried looking the results from the query? What is the result if you do this:

PHP Code:

$query= 'SELECT * FROM calendar_event ORDER BY day';
$result=mysql_query($query);

print "<pre>";
print_r($result);
print "</pre>";


I'm thinking that the result of the query isn't what you're expecting.