[SOLVED] mysql based menu problem !

[nita]

Hi

I'm building photo gallery .. and i got stuck at one point.

 

There is a code..

 

$session=$_GET['session'];
$result = mysql_query("SELECT * FROM session WHERE session='$session' LIMIT 1");
// printing out country of the current session			
while ($row = mysql_fetch_array($result)) {
    echo "	<span class='info1'>";
echo $row['country'];
echo "<br><br></span>";
}

// this is a part im not sure of ..

$row = mysql_fetch_array($result);
$row['country'] = $country;

// printing out all the sessions in current country

$result = mysql_query("SELECT * FROM session WHERE country='$country' ORDER BY year DESC") or die(mysql_error());
while($row = mysql_fetch_array( $result )) 
{
echo "<span class='links1'><A href='ontheroad.php?session=$row[session]'>$row[session]</a></span><br>";
}

 

Wht i want to do is .. when i choose photo session, browser will print the menu -

1st query containg name of the country, based on selected session

2nd query will print all the sessions that are in selected country.

There is no problem to print out country, but i think is something wrong with passing varaible $country

 

What do i do Wrong?

 

Help Please.

 

Nita

[csplrj]

What is this code for?

 

$row['country'] = $country;

 

The First thing is that you have transversed through all the records and so $row I think has to have false value rather than array

 

i.e. $row has value of false

 

The second thing I think you required

 

$country=$row['country'];

 

 

Please check this out if it helps

 

Bye for now

 

CSJakharia

[nita]

i did manage to get what i wanted.

 

new code:

 

$session=$_GET['session'];
			$result = mysql_query("SELECT country FROM session WHERE session='$session' LIMIT 1");    
			$row = mysql_fetch_array($result);
			echo "    <span class='info1'>";
			echo $row['country'];
			echo "<br><br></span>";
			$result2 = mysql_query("SELECT * FROM session WHERE country='$row[country]' ORDER BY year DESC") or die(mysql_error());    
			while($rows = mysql_fetch_array( $result2 ))     
			{    
			echo "<span class='links1'>
			<A href='ontheroad.php?session=$rows[session]'>$rows[session]</a></span><br>";    
			}

 

Thanks csplrj for advice !

 

Nita