Question

我允许我的用户选择英里半径数和位置，然后我试图获得在该半径范围内的结果。网上有很多教程，其中有一个似乎有完美的查询

'SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance FROM markers HAVING distance < 25 ORDER BY distance LIMIT 0 , 20;'

但我正在使用cakephp并且我想使用分页组件，因此我需要将该查询转换为条件数组，该数组选择所有标记和相关数据。谁能帮我吗？我是cakephp的新手，并且在最好的时候对sql绝对没用。

HAVING位特别令人困惑......

修改

我现在正在使用此代码

$this->Paginator->settings = array(
            'fields'=>array('id','( 3959 * acos( cos( radians(37) ) * cos(radians( '.$lat.' ) ) * cos( radians( '.$lng.' ) - radians(-122) ) + sin( radians(37) ) * sin(radians( '.$lat.' ) ) ) ) AS distance '),
                                'limit' => 20,
                                'conditions'=>array('distance < '.$miles)
        );
$data = $this->Paginator->paginate('Job');
$this->set(compact('data'));

至少在运行查询，但我收到错误

Unknown column 'distance' in 'where clause'

有什么想法吗？

更多编辑

日志正在显示

SELECT `Job`.`id`, ( 3959 * acos( cos( radians(37) ) * cos(radians( 54.59728500000001 ) ) * cos( radians( -5.930119999999988 ) - radians(-122) ) + sin( radians(37) ) * sin(radians( 54.59728500000001 ) ) ) ) AS distance FROM `trainnation`.`jobs` AS `Job` LEFT JOIN `trainnation`.`users` AS `User` ON (`Job`.`user_id` = `User`.`id`) LEFT JOIN `trainnation`.`courses` AS `Course` ON (`Job`.`course_id` = `Course`.`id`) WHERE 1 = 1 GROUP BY distance having distance < 49 LIMIT 20

看起来正确，但它现在没有从DB中提取任何东西。这些选项应该至少有9个

Answer 1

试试这个

$this->paginate = array('fields'=>array('id','( 3959 * acos( cos( radians(37) ) * cos(radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin(radians( lat ) ) ) ) AS distance '),
'limit' => 20,
'group'=>'distance having distance < 25 ')
);
$marker_data = $this->paginate('Marker');

将复杂的查询转换为cakephp分页条件

1 个答案: