[snip]
I keep getting the following error when I try to run an "if" statement
Warning: 2 is not a valid MySQL-Link resource in then give the filename
Here is what I am trying to do.
if ($bumpnumber<4) {
print ("display this");
}else {
print ("display that");
}
mysql_close ($Link);
Anyone have an
I keep getting the following error when I try to run an "if" statement
Warning: 2 is not a valid MySQL-Link resource in then give the filename
Here is what I am trying to do.
if ($bumpnumber<4) {
print ("display this");
}else {
print ("display that");
}
mysql_close ($Link);
Anyone have any idea
