need to know if input already exist in field

[attaboy]

I have a field in a table named flights

I hope to find a 1 line statement to bypass inserting a new row if $flight_no is already in the field flight_no

btw flight_no is the primary key and won't  allow a duplicate anyway but I want to avoid the error message

 

this is what I tried but doesn't work

if(SELECT * FROM flights WHERE flight_no !LIKE  %$flight_no%) {

[EmperorJazzy]

Might I suggest;

 

$query = "SELECT FLIGHT_NO FROM FLIGHTS WHERE FLIGHT_NO = " . $flightno;

Call the function for the query results

If recordcount > 0 Then
insert
else
dont insert

 

It's not ONE LINE, but you could combine them together, however, it wouldn't be prudent to combine into one line as it'll make it harder to debug.

[SaCH]

Code will be this

 

$query = "SELECT `flights_no` FROM `flights` WHERE `flights_no` = '$flights_no'";
$result = mysql_num_rows(mysql_query($query));
if($result == "0") {

//Value is not existing do something..

}

else {

//value already existing, do something

}

 

[Pikachu2000]

I think I'd just run the INSERT query, check for an error that is not a duplicate key error, and handle it in the logic of the code if there is one.

 

$query = "INSERT INTO table ( flight_no, etcetera ) VALUES ( $flight_no, $etcetera )";
if( !$result = mysql_query( $query ) ) {
if( mysql_error_no() !== 1061 ) {
	// There was a db error other than a duplicate key, so handle it accordingly
} else {
// query executed
}

[attaboy]

I think I like this last one best.  So many people really know there s#$t here, I love this place!

[attaboy]

for anyone reading the last example is the best but he had 2 typos it should be

mysql_errno() !== 1062

[Pikachu2000]

Yeah, I guess I did. Glad you figured it out and got it working, though.