Forward and backward navigation

Anon

Does anyone have any idea how to write a script that will allow a user to navigate through a calendar? What I am trying to do is to allow users to navigate to what ever day they wish to view events in. Right navigation button adds a day, and the left subtracts a day, all within the calendar month. I can't seem to figure out how to add or subtract while calling the same page. Am I making sense? Please help!

Anon

I have this codeing of my calender for my site .
Just copy paste and run it.
Hope this helps you.

Chanderjit.

  <table width="10%">
    <tr> 
      <td width = "10%"> <?php

$currentday = date("j", time());
$currentmonth = date("m", time());
$currentyear = date("Y", time());

$lastday = 1; 


if (!$month)
{
	$month = date("m", time());
	$year = date("Y", time());
}






$firstday = date( 'w', mktime(0,0,0,$month,1,$year));
while (checkdate($month,$lastday,$year))
{
        $lastday++;
}      

$nextmonth = $month+1;
$nextyear = $year;
if ($nextmonth == 13)
{
	$nextmonth = 1;
	$nextyear = $year + 1;
}
$lastmonth = $month-1;
$lastyear = $year;
if ($lastmonth == 0)
{
	$lastmonth = 12;
	$lastyear = $year-1;
}
echo"<center>";
echo "<table><tr>";     
    echo "<td><form method=post action=checkappointment.php3><input type=submit value='<<'>
            <input type=hidden name=month value=$lastmonth>
            <input type=hidden name=year value=$lastyear></form></td>";
    echo "<td><center><font face=Arial size=5><b>";
echo date('F', mktime(0,0,0,$month,1,$year));
echo " $year<p><font size=3></b><p></td>";
    echo "<td><form method=post action=checkappointment.php3><input type=submit value='>>'>
            <input type=hidden name=month value=$nextmonth> 
            <input type=hidden name=year value=$nextyear></form></td></tr></table></center>";


echo "<table width=100% cellpadding=5 cellspacing=5 border=1>";
echo "<tr><td width=14%><b>Sunday</td><td width=14%><b>Monday</td>
	  <td width=14%><b>Tuesday</td><td width=14%><b>Wednesday</td>
	  <td width=14%><b>Thursday</td><td width=16%><b>Friday</td>
	  <td width=14%><b>Saturday</td></tr>";

for ($i=0; $i<7; $i++)
{
	if ($i < $firstday)
	{
		echo "<td></td>";
	}
	else
	{
		$thisday = ($i+1)-$firstday;

		if ($currentyear > $year)
		{
			echo "<td valign=top bgcolor=dddddd>";
		}
		else if ($currentmonth > $month && $currentyear == $year)
		{
			echo "<td valign=top bgcolor=dddddd>";
		}
		else if ($currentmonth == $month && $currentday > $thisday && $currentyear == $year)
		{
			echo "<td valign=top bgcolor=dddddd>";
		}
		else
		{
			echo "<td valign=top bgcolor=white>";
		}
		echo "<a href=checkappointment2.php3?selectday=$thisday&selectmonth=$month&selectyear=$year>$thisday</a><br><hr>
			<font size=2>";
		echo "</td>";
	}
}

echo "</tr>\n";
$nextday = ($i+1)-$firstday;

for ($j = 0; $j<5; $j++)
{
	echo "<tr>";
	for ($k = 0; $k<7; $k++)
	{
		if ($nextday < $lastday)
		{
			if ($currentyear > $year)
                            {       
                                    echo "<td valign=top bgcolor=dddddd>";
                            }
			else if ($currentmonth > $month && $currentyear == $year)
                    	{       
                           		echo "<td valign=top bgcolor=dddddd>";
                    	}               
                    	else if ($currentmonth == $month && $currentday > $nextday && $currentyear == $year)
                    	{
                            	echo "<td valign=top bgcolor=dddddd>";
                    	}
                    	else    
                    	{
                           		echo "<td valign=top bgcolor=white>";
                    	}
			echo "<a href=checkappointment2.php3?selectday=$nextday&selectmonth=$month&selectyear=$year>$nextday</a><br><hr>
				<font size=2>";
			echo "</td>";
			$nextday++;
		}
	}
	echo "</tr>\n";
}

echo "</table><font size=3>";

chowdary

Hi,
I am not clear with what exactly you want.As i understood you have the list of events in the database and you want to list the events i.e by default today and back and forth am i right?
If that is the scenario then nothing to worry.....Try to follow this

first of all findout the number of days in that month(I think you can do it) and say it is $no_of_days
Also findout the todays day(4 only not 04)
say it is $cur_day
Now you need to set the values for $days_prev && $days_next

if(!$days_prev){$days_prev=1;}
elseif($cur_day-$days_prev>1){$days_prev--;}

if(!$days_next){$days_next=1;}
elseif($cur_day+$days_next<$no_of_days){$days_next++;}

Now pass two variables $act && $days through query string for both prev and next
...abc.php?$act=sub&$days=$days_prev
for previous
....abc.php?$act=add&$days=$days_next
for next

While getting data from MySQL database you can use this qry

if($act=="add")
{

$qry="select............ where $date_field_name=DATE_SUB(CURDATE(), INTERVAL $days DAY)"

}
elseif($act=="sub")
{

$qry="select............ where $date_field_name=DATE_SUB(CURDATE(), INTERVAL $days DAY)"

}
else
{
$qry="select............ where $date_field_name=CURDATE()"
}

I am sorry.........I have no time.....If you couldn't follow let me know i will try to explain in detail.

eliotkh

Hi, my name is Eliot and I am taking JAVA class and I am having hard time doing one of the projects,

The project is :

A program that accepts a date D (say 3/27/1995) and a number N of days (say 50), then prints the dates that are N days backward from D and N days forward from D:

Exp: 50 days forward from Monday the 27th of March 1995 is Tuesday the 16th of May 1995

      50 days back from Monday the 27th of March 1995 is Sunday the 5th of February 1995

also we have to write a method int GetValue (String S, int LoBound, int HiBound) to ask the user (using S as the prompt message) for a value. Trap the value if it is not in range, and then return it. If the program asks for year, then month, then day, the program calls can be

        year = GetValue (“year”, 1700 , 2200);

        month = GetValue (“month”, 1, 12);

        day = GetValue (“day of month”, 1 , DaysInMonth (year, month));

the first assignment statement would first issue the prompt:

        Please enter year between 1700 and 2200:

And would then accept an integer in the range 1700 and 2200.

I would really appreciate it if you could help me on this.

Thanks

Eliot