Displaying records in a cyclic order

[minds_gifts]

Hello,

I use the following code in-order to display records based on the week day.My intention behind writing this one was to display 90 records in a cyclic fashion.(i.e, first 90 on monday, next 90 on tue and so on...)

$today = getdate();  
$day = $today['weekday']; 

switch($day)  
{  
case "Monday":  
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break;  
case "Tuesday":  
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 90,90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break;  
case "Wednesday": 
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 180,90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break;  
case "Thursday": 
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 270,90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break; 
case "Friday": 
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 360,90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break; 
case "Saturday": 
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 450,90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break; 
case "Sunday": 
    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit 540,90";  

    $numresults=mysql_query($query,$connection);  
    $numrows=mysql_num_rows($numresults); 
break;         

} 

My question is - what happens if there are 1000 records???
How can i get the second weekday(Monday) displaying results from 631 onwards and so on??The condition is I want to display every day 90 records

or in other words, if i have 100 records in the table, I like to display first 90 today and from 91 to 100 and from 1 to 80 tomorrow

Is there any way to modify my code or does some body have any other alternative??

Thanks

[MelloPie]

Well at first make it smaller

$day = date("W");
$offset = $day*90;

$query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit ". $offset . ",90";  
$numresults=mysql_query($query,$connection);

by the way $numrows=mysql_num_rows($numresults); will be 90 in any case...

[MelloPie]

for more than 1000 records you should think of a new column

[minds_gifts]

Mellopie,
Is it really that simple as you said.I was wondering that i would be bangging my head all day today in order to make it working.Anyways, let me make it clear once again with less no of records in-order to avoid confusion.

Say, I want to display 5 records everyday.I have only 8 records in a table.Ok, today, I'll show first 5, i.e(1,2,3,4,5) records and tomorrow the records should be (6,7,8,1,2).So, this will be a cyclic process.

Well, thanks again.I'll give a try.

[minds_gifts]

Mellopie,
Sorry that has'nt helped me.It always returns there are no records.
I have 12 records in a table and then i'm trying to show the first 7 records today and then the next 5 plus the first 2 tomorrow(all together 7).

At first, I get the day and then set the off-set.

$day = date("W");
$offset = $day*7; 

and heres the query:

// Lets select the record fields from our database table and check for results 

    $query="SELECT id,introtext,zipcode,image,telephone,description FROM tbl_personals ORDER BY id Limit ". $offset . ",7";  
    $numresults=mysql_query($query,$connection); 
    $numrows=mysql_num_rows($numresults); 
    
// get results

    $result = mysql_query($query) or die("Couldn't execute query");

// We have no results 

    if ($numrows == 0) { 
    echo "Sorry,there are no records"; 
    } 

// We carry out the table population using a 'while' loop 

Any ideas??
Thanks once again.

[MelloPie]

PLS check content of $offset date('W') works with php >= 4.1.0
I havent checked that.
Well this cycling is a tricky thing maybe you shoul make a
SELECT COUNT(*) FROM tbl_personals
first to get the amount of records
Then something like
if ( $offset + $limit > $result_of_count)
get the first x records

where $limit is you amount of records you get per datafetch
that my idea

[mrhappiness]

you might be able to do it like this

$test = array('News 1','News 2','News 3','News 4','News 5','News 6', 'News 7');
$day=date('w'); //Numeric representation of the day of the week  0 (for Sunday) through 6 (for Saturday)
$key=each($test);
while ($key['key']!=$day)
    $key=each($test);
prev($test);
prev($test);
for ($i=0;$i<7;$i++)
{
    $news = (next($test)) ? current($test) : reset($test);
    echo $news.'<br />';
} 

just a short example but i think it should do what you want it to do.

of course it would be better to ask the database for the newsid only since you have to fetch all records