variable and query question

laserw

I would like to know if someone can advise me on two issues?
First, I have been trying to make use of parameters as much as
possible while learning mysql / php and I would like to know what is
wrong with the following.

$link='link="#0000FF"';
$vlink='vlink="#800080"';
$leftmargin='leftmargin=35';
$bgcolor='bgcolor="#F5EBDA"';

print $ttable;

And second, I perform a query on a database using an array and would
like to have a condition which basically is as follows but it produces
an error.

// Displays each row of the table in array & formatting in html table

while ( $row = mysql_fetch_array($result1) 
if $row["fn04"]<>new_year
	new_line ();
)
{ 
$bgcolor = ($i++ & 1) ? '#99CCCC' : '#CCCC99';
print("<tr bgcolor='$bgcolor'><td>"); 
printf("%s</td>\n", $row["fn01"]); 
printf("<td>%s</td>\n", $row["fn02"]); 
printf("<td>%s</td>\n", $row["fn03"]); 
printf("<td>%s</td>\n", $row["fn04"]); 
printf("<td>%s</td></tr>\n", $row["<a href='fn05'></a>"]);
} 
new_year=$row["fn04"];

print ("<tr>\n");print ("<br>\n");print ("</tr>\n");

Thanks in advance,
Peter

flipis

$link='link="#0000FF"';
$vlink='vlink="#800080"';
$leftmargin='leftmargin=35';
$bgcolor='bgcolor="#F5EBDA"';

print $ttable;

Escape the " character with a bar \

while ( $row = mysql_fetch_array($result1)
if $row["fn04"]<>new_year
new_line ();
)
{

I suggest this syntax:

while($row=mysql_fetch_array($result)){
if(row["fn04"] != $new_year){ //Is new_year a constant? Because you write it as new_year, without the $
new_line();

/rest of code/

}

Hope this helps

fLIPIS