If not

[Anon]

ok I am a PHP4 guy and I have been doing some work on a php3 server... in php 4 I do this
if(!mysql_query("Select * from table where number = '5'"))
{
echo("if there are results it displays this");
} else {
echo("If there are not results it displays this");
}

but that WILL NOT work on a php3 server Any clues how to do it?
Thanks

[znouza]

probably you're missing (I know it's optional) second argument in mysql_query function.

$link=mysql_connect(..)
if (!@("query",$link))
{
echo "all ok";
} else {
echo "something wrong";
}

mysql_query didn't tell ya how many records it returns. Use mysql_num_rows instead, eg:

$rows=mysql_query("select query",$link)||die("error");
if (mysql_num_rows($rows))
{
echo "something to display";
} else {
echo "empty set";
}