PHP MySQL joins and outputting as array - php

Question

I was wondering if someone could help me please.
I have the following two tables, positions and polygons.
positions:
----------------------------------------------
| positionID | name | information
----------------------------------------------
| 1 | Southampton | text here
| 2 | London | text here
.
polygons:
-------------------------------------------------------------------
| polygonID | positionID | lat | lon
-------------------------------------------------------------------
| 1 | 1 | 434343.034343 | -43.4343434
| 2 | 1 | 343343.034343 | -43.4343434
| 3 | 1 | 434343.034343 | -54.4343434
I want to join the two tables and output them as an array in the following format.
$positions = array(
array( positionID => 1,
name => 'Southampton',
information => 'text here',
polys => array( array( Lat => 54.54299483853406,
Lng => -6.5224456787109375,
),
array( Lat => 54.648809788121866,
Lng => -6.405029296875,
),
array( Lat => 54.54020652089137,
Lng => -6.39129638671875,
),
),
),
);
I have wrote the following join statement however I'm really stuck on the next step.
SELECT * FROM (`positions`) LEFT JOIN `polygons` ON `positions`.`positionID` = `positions`.`positions`
The lat/lon data from the joins must be stored on the 'polys' key in the array.
Any help or advice would be much appreciated.
Thanks a million

Answer 1

If you have more than one polygonID per positionID, you going to fetch a lot of copies of positions.
I recommend to do 2 queries instead:
$positions = array();
$query_1 = "SELECT * FROM positions";
$query_2 = "SELECT * FROM polygons";
$resource_1 = mysql_query($query_1);
while($position = mysql_fetch_assoc($resource_1))
{
$positions[$position['positionID']] = $position;
$positions[$position['positionID']]['polys'] = array();
}
$resource_2 = mysql_query($query_2);
while($polygon = mysql_fetch_assoc($resource_2))
{
$positions[$polygon['positionID']]['polys'][$polygon['polygonID']]['Lat'] = $polygon['lat'];
$positions[$polygon['positionID']]['polys'][$polygon['polygonID']]['Lon'] = $polygon['lon'];
}
(Sorry for using deprecated functions like mysql_query in my exemple)

Answer 2

You only need one query, but you still need to do some processing with the result in PHP:
$sql = "SELECT * FROM positions
LEFT JOIN polygons ON positions.positionID = polygons.positionID";
$result = mysql_query($sql);
if (mysql_num_rows($result) > 0) {
$positions = array();
while ($row = mysql_fetch_assoc($result)) {
if (!isset($positions[$row['name']])) {
$positions[$row['name']] = array(
'positionID' => $row['positionID'],
'information' => $row['information'],
'polys' => array()
);
}
$positions[$row['name']]['polys'][] = array(
'lat' => $row['lat'],
'lon' => $row['lon']
);
}
print_r($positions);
}
This will also return positions which don't actually have any polygon information. To prevent that, simply change the LEFT JOIN to JOIN.
As for which is most efficient, I'm afraid I don't know the answer to that. Best bet is to benchmark both this method and the one advised by Puggan Se if you can.