Advanced Search Form - PART II

Question

In the insert page: People introduce the birth date of the
dog.
In the search page:
I have a pulldown menu with the values "Baby"; "Young";
"Adult" and finally, "Oldies".
For each of this values I will need to have a range:
Baby - until 6 months.
Young - From 6 Months until 3 years.
Adult - From more then 3 until 8 years.
Oldies - more then 8 years.
I have a SQL script that gives me the Age in years:
SELECT CURDATE(), (YEAR(CURDATE())-YEAR(data_nasc_an)) -
(RIGHT(CURDATE(),5)<RIGHT(data_nasc_an,5)) AS Age
The problem is that if the little dog as less then 12 months
he will return 0. This will be not a big issue, but dogs grow a lot
in the first year, so we cannot put a dog with 2 months and a dog
with 11 months like being the same.
1) How can we show the age in a way that the first year can
be show in months?
2) How can we, for this case of Age, get a range of values
associated with the combo box?
Any feedback on this please?
Thanks in advance

Answer 1

I'm getting the following error:
Notice: Undefined index: idade in
/home/cantinho/www/www/admin/cantinho/encontrar_animal.php on line
86
And this line have:
switch($_POST['idade']) {
I have tried to use if (isset($_POST['idade'])) to try to
avoid this error, like this:
<?php
if (isset($_POST['idade'])) {
switch($_POST['idade']) {
case 'bebe':
$sql = 'SELECT * FROM dogs WHERE data_nasc_an <
SUBDATE(NOW(),
INTERVAL 6 MONTH)';
break;
case 'jovem':
$sql = 'SELECT * FROM dogs WHERE data_nasc_an >
SUBDATE(NOW(),
INTERVAL 6 MONTH) AND data_nasc_an < SUBDATE(NOW(),
INTERVAL 3 YEAR)';
break;
case 'adulto':
$sql = 'SELECT * FROM dogs WHERE data_nasc_an >
SUBDATE(NOW(),
INTERVAL 3 YEAR) AND data_nasc_an < SUBDATE(NOW(),
INTERVAL 8 YEAR)';
break;
case 'velhotes':
$sql = 'SELECT * FROM dogs WHERE data_nasc_an >
SUBDATE(NOW(),
INTERVAL 8 YEAR)';
else {
echo '$_POST[idade] not valid or something the point is, I
have no idea what to put here :(';
but no luck either.
My drop-down menu are defined like this:
<select name="idade" id="idade">
<option value="%" <?php if(isset($_POST['idade'])
&& $_POST['idade'] == "%") echo 'selected="selected"';
?>>Qualquer</option>
<option value="bebe" <?php if(isset($_POST['idade'])
&& $_POST['idade'] == "bebe") echo 'selected="selected"';
?>>Bebé</option>
<option value="jovem" <?php if(isset($_POST['idade'])
&& $_POST['idade'] == "jovem") echo 'selected="selected"';
?>>Jovem</option>
<option value="adulto"<?php if(isset($_POST['idade'])
&& $_POST['idade'] == "adulto") echo 'selected="selected"';
?>>Adulto</option>
<option value="velhote" <?php
if(isset($_POST['idade']) && $_POST['idade'] == "velhote")
echo 'selected="selected"'; ?>>Velhote</option>
So it should not give this error.
This will not be a big issue since its only a notice, but the
query does not work either - When I press Submit he gives me all
always all ages - so maybe its related somehow?
Thanks again.
ADD-ONS to this post :
I have test the querys in a new f"aketest" recordset but they
are not accurate.
I have change the querys for this:
Select CURDATE(), (YEAR(CURDATE())-YEAR(data_nasc_an)) -
(RIGHT(CURDATE(),5)<RIGHT(data_nasc_an,5)) as Idade FROM animal
HAVING Idade < 3
It does not solve the months problem, but It was only for
testing proposes, and this query worked ok in the fakerecorset BUT
the form stills ignore the value of the drop-down because no filter
was made.
So, before que mysql querys, the drop-down in the form must
be connected with the values of the database, and that's not the
case right now.
Anyone please?...