Page Break, Table and MySQL

[sharapov]

I am retreaving my data from the MySQL database and display it inside HTML table. What I need to do is display every sixth row of the table on the new page (when user prints it). I know that it is possible with CSS if I do something like this: <tr style=\"page-break-after: always;\"> My problem is how can I make my code to insert this line (<tr style=\"page-break-after: always;\">) every sixth row? I tried to do the following but it doesn't really work correctly. Can anybody help?


. . .

$num_results = mysql_num_rows($result);

for ($i=0; $i <$num_results; $i++) {

if($i%6==5){

echo"<tr style=\"page-break-after: always;\">"; }

else {

echo" <tr>"; }

. . .

[pyro]

What isn't working?

<?PHP
$num_results = 24;

for ($i=0; $i <$num_results; $i++) {
if($i%6==5){
echo"break here<br>";
}
else {
echo"Row<br>";
}
}
?>Will give me:

Row
Row
Row
Row
Row
break here
Row
Row
Row
Row
Row
break here
Row
Row
Row
Row
Row
break here
Row
Row
Row
Row
Row
break here

[202Eg]

plz i want it 2

[crtech]

Hi.. I was needing some help also.. The above is exactly what i need, and i can get the above code working but, my problem is it when it gets data from the data base it just keeps pulling the same row of data, I dont know how to make it move to the next row of data until all rows are displayed.. any help please.. this is what i have...

<?php
include("conf.php");
$connection = mysql_connect($server, $user, $pass);
mysql_select_db($db) or die ("Unable to select database!");
$query = "SELECT * FROM users WHERE (status='Active')";
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
$row = mysql_fetch_row($result);

$num_results = 19;
for ($i=0; $i <$num_results; $i++) {
if($i%5==4){
echo "next<br>";
}
else {
echo "User:$row[3]>";
}
}

mysql_free_result($result);
mysql_close($connection);

im sure its easy for someone that knows what their doing, im just know trying to learn php...