IF statement problem with null

[suess0r]

Hi,

my table is set so if the field isn't filled out it's set as NULL, i want to only print out the records that aren't = to "NULL" but it prints out all of them and not just the ones that that have info in them and aren't = to NULL...

[quote]if (mysql_result($result_set, $i, "NS_Mon") != "NULL") {
$NSmon=mysql_result($result_set, $i, "NS_Mon");
echo'<br><b>Mon: </b>'.$NSmon.'';
} else {
echo'<br><b>No Specials Provided.</b>';
$NSmon="None";
}
if (mysql_result($result_set, $i, "NS_Tues") != "NULL") {
$NStues=mysql_result($result_set, $i, "NS_Tues");
echo'<br><b>Tue: </b>'.$NStues.'';
} else {
$NStues="None";
}
if (mysql_result($result_set, $i, "NS_Wed") != "NULL") {
$NSwed=mysql_result($result_set, $i, "NS_Wed");
echo'<br><b>Wed: </b>'.$NSwed.'';
} else {
$NSwed="None";
}
if (mysql_result($result_set, $i, "NS_Thur") != "NULL") {
$NSthur=mysql_result($result_set, $i, "NS_Thur");
echo'<br><b>Thr: </b>'.$NSthur.'';
} else {
$NSthur="None";
}
if (mysql_result($result_set, $i, "NS_Fri") != "NULL") {
$NSfri=mysql_result($result_set, $i, "NS_Fri");
echo'<br><b>Fri: </b>'.$NSfri.'';
} else {
$NSfri="None";
}
if (mysql_result($result_set, $i, "NS_Sat") != "NULL") {
$NSsat=mysql_result($result_set, $i, "NS_Sat");
echo'<br><b>Sat: </b>'.$NSsat.'';
} else {
$NSsat="None";
}
if (mysql_result($result_set, $i, "NS_Sun") != "NULL") {
$NSsun=mysql_result($result_set, $i, "NS_Sun");
echo'<br><b>Sun: </b>'.$NSsun.'';
} else {
$NSsun="None";
}[/quote]

[craygo]

Should just do it with the sql statement

[code]$sql = "SELECT * FROM table WHERE field IS NOT NULL";[/code]

Also in php if you put single quotes around 'NULL' it is actually looking for the word NULL. Just use NULL without any quotes.

[code]if (mysql_result($result_set, $i, "NS_Mon") != NULL)[/code]

Ray