filter results from a check box

[patrioticcow]

hello.

i have a checkbox like this

<input name="request" type="checkbox" id="request" value="1" checked="checked" />

and i send the value 1 or 0 into my sql database.

how can i filter the results depending if the box is 0 or 1

i got this code but it doesn't return results when $request='0'

if($type == 'like' && $request='1' && $request='0')
  	{
		$sql = 'SELECT * FROM topic WHERE (test LIKE '."'%$search%') AND request='1'";
  	}
	elseif($type == 'like' && $request='0')
  	{
		$sql = 'SELECT * FROM topic WHERE (test LIKE '."'%$search%') AND request='0'";
  	}

thanks

[zorgi]

if($type == 'like' && $request='1' && $request='0')

You are assigning value to your $request variable ... Check out Comparison Operators

[patrioticcow]

this

if($type == 'like' && $request =='1' && $request =='0')

gives a syntax error

[zorgi]

this

if($type == 'like' && $request =='1' && $request =='0')

gives a syntax error

I don't see syntax issues but I do see huge logic issue as your condition will NEVER be true.

if($request =='1' && $request =='0')

Read that as IF $request IS 1 AND IF $request IS 0 ... at the same time!? Impossible ... it will never happen

[patrioticcow]

oh. i see.
but even if i leave it like this it still wont work

    if($type == 'like' && $request='1' && $request='0')
          {
            $sql = 'SELECT * FROM topic WHERE (test LIKE '."'%$search%') AND request='1'";
          }
        elseif($type == 'like' && $request='0')
          {
            $sql = 'SELECT * FROM topic WHERE (test LIKE '."'%$search%') AND request='0'";
          }

i mean it only shows the results that have 1 in the database

[zorgi]

Now you are assigning values again.... Maybe this helps

[patrioticcow]

$request='1' is a checkbox that has values 1 or 0.

so what i am saying id that if the checkbox is checked display the results from SEARCH and REQUEST.

  if($type == 'like' && $request='1')
           {
             $sql = 'SELECT * FROM topic WHERE (test LIKE '."'%$search%') AND request='1'";
           }
         elseif($type == 'like' && $request='0')
#           {
             $sql = 'SELECT * FROM topic WHERE (test LIKE '."'%$search%') AND request='0'";
           }

or u think i need to use status="'.$acti ve.'"'