我该如何转换:
SELECT latitude, longitude, SQRT(
POW(69.1 * (latitude - [startlat]), 2) +
POW(69.1 * ([startlng] - longitude) * COS(latitude / 57.3), 2)) AS distance
FROM TableName HAVING distance < 25 ORDER BY distance;
到Zend_Table。
我在数据库中有两个键,lat和long,我想获取与用户关系密切的数据,查询工作由第一个用户提供,它将Zent_Table转换为但是如果我有人的数据< strong>在同一个城市它不会返回任何数据,我想查询需要改进,以便显示更大距离的结果
$this->select()->from(array('c' => 'content'))......
我不知道怎么做,你们能帮助我吗?
// LE
$data2->setIntegrityCheck(FALSE)
->from(array('c' => 'content'), array('location', 'id', 'user_id', 'date', 'attachment', 'content','lat','long'))
->columns(array(
'distance' => "SQRT(POW(69.1 * (lat - {$location['lat']}), 2) + POW(69.1 * ({$location['long']} - long) * COS(lat / 57.3), 2))"
))
->having('distance < ?', 25)
对于找到离你很近的用户来说,查询是错误的,这个应该可以解决这个问题
ASIN(
SQRT( POWER(SIN((@orig_lat -
abs(
dest.lat)) * pi()/180 / 2),2) + COS(@orig_lat * pi()/180 ) * COS(
abs
(dest.lat) * pi()/180) * POWER(SIN((@orig_lon – dest.lon) * pi()/180 / 2), 2) ))
as distance
答案 0 :(得分:3)
使用ZF 1.12.4-dev,PHP 5.3.26,MySQL 5.6.13进行测试
Zend_Db_Table::setDefaultAdapter($adapter);
$table = new Zend_Db_Table("TableName");
$startLat = 39.0;
$startLng = 122.0;
$select = $table->select()
->from($table, array("latitude", "longitude",
"distance" => "SQRT(
POW(69.1 * (latitude - $startLat), 2) +
POW(69.1 * ($startLng - longitude) * COS(latitude / 57.3), 2))"))
->having("`distance` < ?", 25)
->order("distance");
$rowset = $table->fetchAll($select);
print_r($rowset->toArray());
使用ZF 2.2.5-dev,PHP 5.3.26,MySQL 5.6.13进行测试
use Zend\Db\TableGateway\TableGateway;
use Zend\Db\Sql\Select;
use Zend\Db\Sql\Expression;
$table = new TableGateway("TableName", $adapter);
$startLat = 39.0;
$startLng = 122.0;
$rowset = $table->select(
function (Select $select) {
global $startLat, $startLng;
$select->columns(
array(
"latitude",
"longitude",
"distance" => new Expression("SQRT(
POW(69.1 * (latitude - ?), 2) +
POW(69.1 * (? - longitude) * COS(latitude / 57.3), 2))",
array($startLat, $startLng))
)
);
$select->having->lessThan("distance", 25);
$select->order("distance");
}
);
print_r($rowset->toArray());
测试数据并测试SQL查询(不含PHP):
use test;
drop table if exists TableName;
create table TableName (
id int auto_increment primary key,
latitude numeric(9,4) NOT NULL,
longitude numeric(9,4) NOT NULL,
key (latitude,longitude)
);
insert into TableName (latitude, longitude) values
(10.0, 10.0),
(20.0, 20.0),
(30.0, 30.0),
(40.0, 40.0),
(50.0, 50.0),
(60.0, 60.0),
(70.0, 70.0),
(39.2, 122.2);
set @startlat = 39.0;
set @startlng = 122.0;
SELECT latitude, longitude, SQRT(
POW(69.1 * (latitude - @startlat), 2) +
POW(69.1 * (@startlng - longitude) * COS(latitude / 57.3), 2)) AS distance
FROM TableName
HAVING distance < 25
ORDER BY distance;
SQL测试输出:
$ mysql -E < 19057187.sql
*************************** 1. row ***************************
latitude: 39.2000
longitude: 122.2000
distance: 17.484284526375415
PHP测试的输出:
$ php 19057187.php
Array
(
[0] => Array
(
[latitude] => 39.2000
[longitude] => 122.2000
[distance] => 17.484284526375415
)
)
Array
(
[0] => Array
(
[latitude] => 39.2000
[longitude] => 122.2000
[distance] => 17.48428452637566
)
)
答案 1 :(得分:1)
$db = Zend_Db_Table::getDefaultAdapter();
$select = $db->select();
$select->from('TableName', array('latitude', 'longitude'));
$select->columns(array(
'distance' => "SQRT(POW(69.1 * (latitude - {$startLat}), 2) + POW(69.1 * ({$startLong} - longitude) * COS(latitude / 57.3), 2))"
));
$select->having('distance < ?', 25);
$select->order('distance ASC');