filter results by using search box and drop down

[patrioticcow]

hello.

i have a search box that will display some results. I also have a drop down list with values "1" and "2".

the values for each results are stored into database in a "req" field.

 

<select name="req" id="req">
           <option value="1">yes</option>
           <option value="2">no</option>
</select>
$sql = 'SELECT * FROM topic WHERE (Title LIKE '."'%$search1%') AND req='1'

 

what am i missing.

thanks

[Pikachu2000]

It isn't really clear what your question is.

[patrioticcow]

 

i want the results to be filtered by choosing YES or NO. more specific 1 or 2.

each result has already the values 1 or 2 added to them in the database

 

 

[Pikachu2000]

You should be able to simply use the value of $_POST['req'] in the query string.

 

$_POST=['req'] = trim($_POST['req']);
if( $_POST['req'] == 1 || $_POST['req'] == 2 ) {
     $req = (int) $_POST['req'];
} else {
     // If field is empty, or not equal to 1 or 2, set a default (or whatever you'd like)
     $req = 1;
}

$sql = "SELECT * FROM topic WHERE (Title LIKE '%$search1%') AND req= $req";

[patrioticcow]

i've got a little problem integrating the code into my code

 

if($search1 == '')
{
	$smarty->assign('Error','Please enter any destination!');
	$z=1;
}
elseif($type == 'like' && $status='1')
  	{
$sql = "SELECT * FROM topic WHERE (Title LIKE '%$search1%') AND req= $req";
}

 

my html is

<select name="req" id="req">
           <option value="1">yes</option>
           <option value="2">no</option>
</select>

i am already putting the values into the database using this:

$sql = 'INSERT INTO topic (req) VALUES '." ('$_SESSION[EmpId]', $req)";

 

let me see if i got the right idea on how I should implement the code

if($search1 == '')
{
	$smarty->assign('Error','Please enter any destination!');
	$z=1;
}
if( $_POST['req'] == 1 || $_POST['req'] == 2 ) {
     $req = (int) $_POST['req'];
}
elseif($type == 'like' && $status='1')
  	{
$sql = "SELECT * FROM topic WHERE (Title LIKE '%$search1%') AND req= $req";
$req = 1;
}

what u think?

[Pikachu2000]

One type in yours that I noticed, should be status ==

 

elseif($type == 'like' && $status='1')

 

[patrioticcow]

i found the answer thanks to -->axelay

$req = intval ($_GET['req']);
if($type == 'like' && $status='1' && $req > 0 && $req < 3)
{
$sql = "SELECT * FROM topic WHERE (Title LIKE '%$search1%') AND req= $req"";
}

[Pikachu2000]

You still need to change $status = '1' to $status == '1' . . .