我的SQL语句有什么问题？

Question

I'm making a site where the user can select a rental property that has two fields they can select from, Furnished and Pets. The options in both select boxes is Yes and No.

<select name="furnished">
  <option value="">
  <option value="yes">Yes</option>
  <option value="no">No</option>
</select>

<select name="pets">
  <option value="">
  <option value="yes">Yes</option>
  <option value="no">No</option>
</select>

I am writing a SQL statement based on what the user picks in those fields.

$sql = 'SELECT *
       FROM properties
       WHERE num_bedrooms >= ' . $_GET['num_bedrooms'] . 
       ' AND num_bathrooms >= ' . $_GET['num_bathrooms'];

if($_GET['furnished'] == 'yes') { //if the furnished is set to yes
  $sql .= ' AND furnished = "yes" OR furnished = "partially"';
} else if($_GET['furnished'] == 'no') { //if the furnished is set to no
  $sql .= ' AND furnished = "no" OR furnished = "description"';
} 

if($_GET['pets'] == 'yes') { //if the pets is set to yes
  $sql .= ' AND pets = "yes" OR pets = "cats" OR pets = "dogs"';
} else if($_GET['pets'] == 'no') { //if the pets is set to no
  $sql .= ' AND pets = "no" OR pets = "description"';
}

If the user chooses yes for furnished, I want it to display all properties that are furnished (yes) OR that are partially furnished(partially). If the user chooses No, I want it to display all properties that are not furnished(no) OR that have a special description(description).

If the user chooses yes for pets, I want it to display all properties that allow all pets (yes) OR that allow only cats(cats) OR that allow only dogs(dogs). If the user chooses No, I want it to display all properties that don't allow pets(no) OR that have a special description(description).

As an example, this is the outputted SQL statement if the user chooses Yes on Furnished and Yes on Pets:

SELECT * FROM properties 
WHERE num_bedrooms >= 1 
AND num_bathrooms >= 1 
AND furnished = "yes" OR furnished = "partially" 
AND pets = "yes" OR pets = "cats" OR pets = "dogs"

The problem is that it will currently return results that meet EITHER of the furnished or pets requirements in the where clause, whereas I need it to return results that meet BOTH of the furnished and pets requirements in the where clause.

So it will currently return a result that says has the Furnished set to yes, and the pets set to no. How can I have it return all results with furnished set to yes AND pets set to yes?

What is wrong with my SQL statement?

Answer 1

AND comes before OR in operator precedence, so your query reads like

SELECT * FROM properties 
WHERE 
(num_bedrooms >= 1 AND num_bathrooms >= 1 AND furnished = "yes")
OR (furnished = "partially" AND pets = "yes")
OR pets = "cats" 
OR pets = "dogs"

which is not quite what you expected. Try

SELECT * FROM properties 
WHERE num_bedrooms >= 1 
AND num_bathrooms >= 1 
AND (furnished = "yes" OR furnished = "partially")
AND (pets = "yes" OR pets = "cats" OR pets = "dogs")

or even better

SELECT * FROM properties 
WHERE num_bedrooms >= 1 
AND num_bathrooms >= 1 
AND furnished IN ("yes", "partially")
AND pets IN ("yes", "cats", "dogs")

You shouldn't do SELECT * try naming the columns you need!

Answer 2

Put parens around the various OR 'd things.

$sql .= ' AND (furnished = "yes" OR furnished = "partially")';

$sql .= ' AND (pets = "yes" OR pets = "cats" OR pets = "dogs")';

et cetera.