What's wrong with this mySQL IF Statement? - php

Question

I'm trying to do this IF statement in a mySQL query which I learnt from a YouTube video. I'm not too sure what's going wrong. I do get the following mysql error:
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near 'IF(Cuisine != 'Cuisine', WHERE Cuisine='Cuisine') AS ORDER BY
restaurantID' at line 2
Ok, sorry about the lack of detail. Let me explain this a little more.
On my page, I have a HTML form which has 3 drop-downs which act as 'filters'. The default option for one of these 'filters' is Cuisine, which acts as a title, and if it hasn't been changed it means that the user does not want to use the Cuisine as a filter for their search. However if it has changed to say 'Western', then obviously the user wants to use it.
Now, the above problem is quite simple to solve because there is only one filter at a time in place in the above scenario. However, when there are multiple filters being used at once, this is where it gets complicated for me and I don't know how to address this problem.
My solution was to go and search Google for some sort of IF statement in mySQL. I came across this video (which is probably quite good, however since I was very rushed at the time, probably misinterpreted it). Here is the video: http://www.youtube.com/watch?v=3xK5KKQx-J0
I figured that if I could use the condition and try it for the cuisine, I could research and modify it and work on it some more to get it to completely get the filter system to work.
In the code below, my objective is to check what a PHP variable is = to in SQL, and if it's = to 'Cuisine' then I don't want to execute the 'WHERE Cuisine = $cuisine' part of the query. $cuisine is a variable which is taken from a simple HTML/AJAX form dropdown menu using the 'POST' method.
<?php
$result = mysql_query("SELECT * FROM restaurants
IF($cuisine != 'Cuisine', WHERE Cuisine='$cuisine')
ORDER BY restaurantID
")
or die(mysql_error());
?>
P.S I'm not sure if this is the right approach to solving my problem, however I have now described my train of thought and my problem to you above.
I understand your frustration when I left no detail, and once again I apologise, for wasting your time with a poorly written question I will remember to ensure my future questions/answers are more detailed.

Answer 1

I would move the conditional from the SQL query to PHP where the correct query would be built.
if( $cuisine == 'Cuisine' ) ) {
$conditions = '1'; // "WHERE 1" matches every record
}
else {
$conditions = "Cuisine='$cuisine'";
}
$result = mysql_query( "SELECT * FROM restaurants
WHERE $conditions
ORDER BY restaurantID
") or die(mysql_error());
The above assumes that $cuisine is correctly sanitized and escaped.

Answer 2

What are you trying to do? If you want to select all rows where the Cuisine column is not 'Cuisine', use the WHERE clause:
SELECT * FROM restaurants
WHERE Cuisine != 'Cuisine'
ORDER BY restaurantID

Answer 3

Did not fully understand your question, but if you want to select all restaurants by given cuisine and order them by restaurant ID then you can use:
$result = mysql_query("SELECT * FROM restaurants WHERE Cuisine = '$cuisine' ORDER BY restaurantID")

Answer 4

I see multiple problems, which can only be answered if you provide more information. As of now the error in SQL syntax is ,
The syntax of IF condition is
IF(<condition>, <value if true>, <value if false>)
which is troubling you (you have only two parameters for you IF).