排除在搜索结果中首先获取的结果

Question

I'm working on a property project where user searches for properties in some locality.
After fetching the relevant results I would like to display some suggestive results which should not contain the results previous displayed.

Table schema is:

property table:

+-----+-------------+-----------+-------------+------+----------+
| id  | property    | amenities | description | city | locality |
+-----+-------------+-----------+-------------+------+----------+
| 1   | House/Villa |           | some desc   | cit1 | loc1     |
+-----+-------------+-----------+-------------+------+----------+
| 2   | Office      |   ....    | desc....    | cit1 | loc2     |
+-----+-------------+-----------+-------------+------+----------+
| 3   | House/Villa | .....     | desc....    | cit1 | loc3     |
+-----+-------------+-----------+-------------+------+----------+

I want, if the user searches for House/Villa in city city1 and locality loc1 then along with the available results I would like to suggest him for other properties in same city with other locality .
What would be the SQL query. I've to use it in PHP

Answer 1

let your first query is

select * from property where city="city1" and locality="loc1" and property ="House/Villa";

then your query for desired result should be

select * from property where  property ="House/Villa" and city="city1" and locality NOT IN(loc1);

In will be better if you are showing him result for multiple location in first time, and want to omit that then you can pass an array by makin an comma separated string,

$notIn = implode(", ", array("loc1","loc2"));

so NOT IN will fetch result that would not include passed location.

Step to optimize this query.

1. select only required column, do not use *

2. use limit eg:- select * from property where property ="House/Villa" and city="city1" and locality NOT IN(loc1) limit 0,30; // this will fetch only top 30 result. you can set according to your way.

3. select filtered data by using proper queries and subqueries

4. Not In is slow for large result b/c mysql first scan full table then apply Not IN, so you can go for NOT EXISTS() instead of NOT IN().

Answer 2

Try this :

For proper answer:

SELECT DISTINCT property, amenities, description, city, locality FROM property
WHERE property = 'House/Villa' AND city = 'city1' AND locality = 'loc1'

For suggestions use:

 SELECT DISTINCT property, amenities, description, city, locality FROM property
    WHERE property = 'House/Villa' AND city = 'city1' AND locality != 'loc1'

Answer 3

$property = 'House/Villa';
$city = 'city1';
$locality = 'loc1';

$q = "SELECT * FROM property WHERE property = '" . $property . "'AND city = '" . $city . "'";

$stmt = pdoObject->prepare($q);
$stmt->execute() or die(print_r($stmt->errorInfo()));
$result = $stmt->fetchAll(PDO::FETCH_BOTH);
foreach($resukt as $res)
{
   if($res['locality'] == $locality)
    echo "your search result";
} 

foreach($resukt as $res)
{
   if($res['locality'] != $locality)
    echo "this is suggestion";
}

I assume you are using PDO.

Answer 4

You should be able to do this with just one SQL query

$query = "SELECT *, IF(locality='" . mysqli_real_escape_string($locality) . "', 1, 0) as locality_match FROM property WHERE property = '" . mysqli_real_escape_string($property) . "'AND city = '" . mysqli_real_escape_string($city) . "' ORDER BY locality_match DESC";

This puts the locality if statement in an if statement as part of the select. So it doesn't limit the results but instead gives us a value to sort by, so the items that match the locality will be first and suggestions will be at the end.

I also added an escape function around each parameter, because you definitely want to sanitize any input from the user.