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/calendar.php?month=11&year=2009

 

Its not quite right. The image is called "promoimage" in the table. My code is as follows:

 

<?

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)) . " " . $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.

[Bricktop]

Hi Razorster,

 

You're trying to output $promoimage using the following code:

 

echo "<span class='event'><img src='". $promoimage ."'></span>"; 

 

What does $promoimage contain?  It should contain something like:

 

"images/image.jpg"

 

For your code to work.

 

Hope this helps.

[Razorster]

Hi Bricktop, thanks for your reply. $promoimage does contain a URL to an image i.e. http://proudcabaret.com/img/folly.jpg.

 

The images arent correct for each day on the calendar view, they are jumbled up. I think its my query?

[Razorster]

Ive managed to order the images, but its still not coming up right on the calendar? Its all shifted to the right.. I dont get it! Anyone have an ideas? I think the query is wrong?

[Bricktop]

Hi Razorster,

 

Just to clarify.

 

Do you want  $promoimage to always output the same image.  Or does it need to output a separate image for every day of the week.  Or does it need to output a particular image on a particular date?

[Razorster]

Hi mate,

 

The image url is different for each Day of the month. So each record in the database has a different image url. Do yeah it does need to output a particular image on a particular date.

 

Thanks!

[Razorster]

P.S

 

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.

[Razorster]

Anyone got any ideas!!!! I really need help!