Question

我正试图用以下的法典行文来过滤“谁”的“谁”具有某些(无关的)条件的所有员额。

c 美元=6美元;

然而,我不是仅仅在标的岗位上才有结果,而是在没有“谁”的帽子上,每个职位都有空档。

I know my question has come up before, but the solutions posted have not worked with my code. I have tried adapting the code here to my own: http://web-development-blog.co.uk/2010/09/14/cakephp-habtm-find-with-conditions-and-containable-behavior/, but I get an SQL error stating "Tag.post_id" not found in on clause.

请提供帮助。

试图从某些链接中执行守则的错误信息:

SELECT `Post`.`id`, `Post`.`title`, `Post`.`body`, `Post`.`created`,       `Post`.`modified`, `Tag`.`id`, `Tag`.`tag`, `Tag`.`created`, `Tag`.`modified` FROM `posts` AS `Post` LEFT JOIN `tags` AS `Tag` ON (`Tag`.`post_id` = `Post`.`id`) WHERE `Tag`.`tag` =  who     1054: Unknown column  Tag.post_id  in  on clause

原因如下:

$this->Post->bindModel(array( hasOne  => array( Tag )));
$this->Post->contain(array(
   Tag 
));
$com=$this->Post->find( all , array(
   conditions =>array( Tag.tag => who ) 
));

Answer 1

你的错误是,你正在使用束缚式的阴道,说出外国钥匙或东西,你在模式上建立了协会,并在严格需要时使用这种手法。

you need to tell cake which foreignkey of not cake will use the default, in your case it will try Tag.post_id.

Also you should use a has many association or at least i think so. I said this beacause i suppose a post may have multiple tags.

一俟您在模型中统一,删除这一行文

(a) bin; 1 => 阵列(Tag));

it should work perfectly

http://book.cakephp。

EDIT:

职业介绍会读写错误...... 有许多协会没有加入,因此,你必须使用其中的文字来说明该联合会的状况或附加条件,这里指的是,你的法典应如何考虑添加包含条件的条件。

$contain = array(
   Tag  => array(
        conditions => array( Tag.tag => who ) 
   )
);
$com=$this->Post->find( all , array(
   contain =>$contain 
));

会员资格包含超越协会条件的条件! 如果你想有额外条件,你就必须在这样的传单上加添:

$this->Post->hasMany[ Tag ][ conditions ] += array( Tag.tag => who );

将传单上的协会改名为“光顾”组织。

Also you may do a hasmany alias for different types (remember to define always the classname and foreignkey).

希望它能发挥这一作用,甚至不只评论看它的位置。

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