Insert value if recieved value is null - php

Question

so, i'm trying to show a certain value in a table row, if the value recieved by post is null.
So far i've got this as my small database (it's in spanish):
DROP DATABASE IF EXISTS fantasmas;
CREATE DATABASE fantasmas;
USE fantasmas;
CREATE TABLE tipos(
ID tinyint(2) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
TIPO VARCHAR(45) UNIQUE
);
CREATE TABLE datos(
ID tinyint(2) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
NOMBRE VARCHAR(50) NOT NULL,
ARCHIVO VARCHAR(30),
AVISTAMIENTO VARCHAR(50) NOT NULL,
LOCALIDAD VARCHAR(50) NOT NULL,
INFO VARCHAR(200),
FKTIPOS TINYINT(2) UNSIGNED,
FOREIGN KEY (FKTIPOS) REFERENCES tipos(ID)
);
INSERT INTO tipos (TIPO) VALUES('Vapor'), ('Forma Animal'), ('Forma Humanoide'), ('Dios/Semidios'),('No Catalogado');
Those inserted values into "tipos" are then showed in options like this:
<select name="tipos">
<option value="sintipo">---------</option>
<?php
while( $row = mysqli_fetch_assoc($rta)):
$tipo = $row['TIPO'];
echo '<option value="'.$row['ID'].'">'.$tipo.'</option>';
endwhile;
?>
</select>
After, when the form is sent, the inset to the table, looks like this, it assigns null value if someone select the "------" option:
$query = "INSERT INTO datos SET NOMBRE='$nombre', AVISTAMIENTO='$lugar', LOCALIDAD='$localidad', INFO='$info', ARCHIVO='$ruta', FKTIPOS = NULLIF('$tipo','sintipo')";
All the data filled in the form, is then showed in a different row of a table.
What I need now, is that, if someone selects the "------" option, the value shown in screen is "No catalogado"
So far I have not been able to do it.
Can anyone help me?

Answer 1

Before the insert happens you can do a null check on the PHP side. Or... you can just use SQL's built in ISNULL()
You can also use CASE statements in SQL to accomplish this behavior across multiple values.
I know this isn't the most lengthy answer, but it should work.

Answer 2

Is this what your asking for?
$query = "INSERT INTO datos SET NOMBRE='$nombre', AVISTAMIENTO='$lugar', LOCALIDAD='$localidad', INFO='$info', ARCHIVO='$ruta', FKTIPOS = " . ($tipo == 'sintipo' ? 'NULL' : "'$tipo'");
Be aware that you have very dirty code, you don't even filter and escape income values, as I guess. Try to google php mysql escape values.