CakePHP的使用像％％

Question

找到查詢我有我的CakePHP應用中的以下查找查詢：

$this->paginate = array(
'limit'=>5, 
'order'=>'Note.datetime DESC', 
'conditions' => array(
    'Note.status'=>1, 
    'OR' => array(
     'Note.title LIKE' => '%'. $q . '%', 
     'Note.content LIKE' => '%'. $q . '%' 
     ) 
    ) 
); 

這需要一個名爲$q做這樣兩個標題和內容查詢參數。

因此，舉例來說，如果我有以下幾點：

Title: Lorem ipsum Content: Lorem ipsum dolare

和搜索'lorem'

這將精細找到它。但是，如果我搜索'lorem dolare'它不會找到它。

我該怎麼做？

注：我不希望使用任何插件等

編輯：如果我使用FULLTEXT將這項工作？

$this->paginate = array(
'limit'=>5, 
'order'=>'Note.datetime DESC', 
'conditions' => array(
    'Note.status'=>1, 
    'OR' => array(
     'MATCH(Note.title) AGAINST('.$q.' IN BOOLEAN MODE)', 
     'MATCH(Note.content) AGAINST('.$q.' IN BOOLEAN MODE)' 
     ) 
    ) 
); 

編輯2：

收到此錯誤在嘗試上述：

Error: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'dolor IN BOOLEAN MODE)) OR (MATCH(`Note`.`content`) AGAINST(lorem dolor IN BOOLE' at line 1 

SQL Query: SELECT `Note`.`id`, `Note`.`title`, `Note`.`excerpt`, `Note`.`content`, `Note`.`datetime`, `Note`.`user_id`, `Note`.`slug`, `Note`.`status`, `Note`.`topic_id`, `User`.`id`, `User`.`email`, `User`.`firstname`, `User`.`lastname`, `User`.`password`, `User`.`status`, `Topic`.`id`, `Topic`.`title`, `Topic`.`slug` FROM `db52704_driz2013`.`notes` AS `Note` LEFT JOIN `db52704_driz2013`.`users` AS `User` ON (`Note`.`user_id` = `User`.`id`) LEFT JOIN `db52704_driz2013`.`topics` AS `Topic` ON (`Note`.`topic_id` = `Topic`.`id`) WHERE `Note`.`status` = 1 AND ((MATCH(`Note`.`title`) AGAINST(lorem dolor IN BOOLEAN MODE)) OR (MATCH(`Note`.`content`) AGAINST(lorem dolor IN BOOLEAN MODE))) ORDER BY `Note`.`datetime` DESC LIMIT 5 

Answer 1

試試這個

$this->paginate = array(
    'limit'=>5, 
    'order'=>'Note.datetime DESC', 
    'conditions' => array(
    'Note.status'=>1, 
    'OR' => array(
     "MATCH(Note.title) AGAINST('".$q."' IN BOOLEAN MODE)", 
     "MATCH(Note.content) AGAINST('".$q."' IN BOOLEAN MODE)" 
    ) 
) 

）;

換句話說，則會以引號

搜索條件

編輯NDM的建議是有道理的

$this->paginate = array(
    'limit'=>5, 
    'order'=>'Note.datetime DESC', 
    'conditions' => array(
    'Note.status'=>1, 
    'OR' => array(
     'MATCH(Note.title) AGAINST(? IN BOOLEAN MODE)' => $q, 
     'MATCH(Note.content) AGAINST(? IN BOOLEAN MODE)' => $q 
    ) 
) 

）;

Answer 2

LIKE不會幫助你在這裏。您的標題是Lorem ipsum，您的內容是Lorem ipsum dolare。您正在搜索lorem dolare，這不會通過LIKE運算符找到，因爲它不是這些列中的任何一個的子字符串。你需要考慮使用FULLTEXT，否則你需要使用類似Sphinx的東西。如果你想弄清楚實施什麼解決方案，請看this answer here。

Answer 3

你也可以試試這個：

$this->paginate = array(
    'limit'=>5, 
    'order'=>'Note.datetime DESC', 
    'conditions' => array(
     'Note.status'=>1, 
     'OR' => array(
      'Note.title LIKE' => "%$q%", 
      'Note.content LIKE' => "%$q%" 
     ) 
    ) 
); 

Answer 4

Please Try this: 

$this->paginate = array(
    'limit'=>5, 
    'order'=>'Note.datetime DESC', 
    'conditions' => array(
     'Note.status'=>1, 
     'OR' => array(
      'Note.title LIKE' => "%".$q."%", 
      'Note.content LIKE' => "%".$q."%" 
     ) 
    ) 
);