cakephp , the subquery
Cakephp 框架帮我们做了很多的工作,的确省了我们很多工作,提高了效率。
但是,碰到一些比较复杂的查询时,还是有些问题,官方的cookbook api 有说明一些详细的用法,但感觉还是不太够,有些地方讲的不够清楚。
StackOverflow上的这个问题不错
http://stackoverflow.com/questions/5770032/cakephp-and-subquery
The Question:
How can I write SQL subquery using cake syntax. I know how to code simple query, but I can't handle subquery.
This is the original query:
1 2 3 4 5 6 7 8 | SELECT Assumption.id, Referee.id, Referee.first_name, Referee.second_name FROM referees AS Referee INNER JOIN ( SELECT a.id, a.referee_id FROM assumptions a WHERE a.season_id =7 ) AS Assumption ON Referee.id = Assumption.referee_id |
The answer
Since you didn't understand the syntax, this is the actual query:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | $records = $ this ->Referee->find( 'all' , array( 'fields' => array( 'Assumption.id' , 'Referee.id' , 'Referee.first_name' , 'Referee.second_name' ), 'joins' => array( array( 'table' => 'assumptions' , 'alias' => 'Assumption' , 'type' => 'INNER' , 'foreignKey' => false , 'conditions' => array( 'Referee.id = Assumption.referee_id' , 'Assumption.season_id = 7' ), ), ), ) ); |
Which produces this query:
1 2 3 4 5 6 7 8 9 | SELECT `Assumption`.`id`, `Referee`.`id`, `Referee`.`first_name`, `Referee`.`second_name` FROM `referees` AS `Referee` INNER JOIN assumptions AS `Assumption` ON (`Referee`.`id` = `Assumption`.`referee_id` AND `Assumption`.`season_id` = 7) |
Which provide the results you are looking for.
Sample output:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | Array ( [0] => Array ( [Assumption] => Array ( [id] => 1 [0] => Array ( [id] => 1 [season_id] => 7 [referee_id] => 1 [name] => SomeAssumpton ) ) [Referee] => Array ( [id] => 1 [first_name] => Ref [second_name] => one ) ) ) |
a
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