如何从Cake PHP中的两个不同表中获取数据 [英] How to get data from two different table in cake php
问题描述
但是它只会返回来自tab [1]的相同数据。.
i希望分别从tab [1]和tab [0]中获取数据
But it will give return same data from tab[1] only.. i want to get data from tab[1] and tab[0] distinctly
$ db = ConnectionManager :: getDataSource('default');
$db = ConnectionManager::getDataSource('default');
$ tab = $ db-> listSources();
$tab = $db->listSources();
echo '<br>';
$this->Form->useTable=$tab[1];
print_r($this->Form->find(`all'));
echo '<br>';
$this->Form->use Table=$tab[0];
print_r($this->Form->find('all'));
推荐答案
更改 Model-> useTable
在运行时无法正常工作,因为一旦模型初始化,CakePHP就会缓存数据库表的架构。
Changing Model->useTable
at runtime does not work properly because once a model has been initialised, CakePHP caches the schema of the database-table.
要切换到另一个表并清除缓存的架构,使用 Model-> setSource('tablename')
To switch to another table and clear the cached schema, use Model->setSource('tablename')
Documentation; http://api.cakephp.org/2.3/source-class -Model.html#1100-1125
Documentation; http://api.cakephp.org/2.3/source-class-Model.html#1100-1125
您的示例将如下所示;
Your example will then look like this;
echo '<br>';
$this->Form->setSource($tab[1]);
print_r($this->Form->find(`all'));
echo '<br>';
$this->Form->setSource($tab[0]);
print_r($this->Form->find('all'));
另外,请使用 debug()
代替 print_r()
输出调试结果。这将输出正确格式化的结果。 (您需要在app / Config / core.php配置中将debug设置为1或更高,才能使debug()起作用)
Also, please use debug()
to output results for debugging in stead of print_r()
. This will output the results properly formatted. (You'll need to set debug to 1 or higher inside your app/Config/core.php configuration for debug() to work)
但是
切换模型的源表通常是不好的做法,并且仅适用于非常特殊的情况。我会强烈建议为每个数据库表创建一个单独的模型。
Switching the sourcetable of a Model is generally bad practice and will only apply to very specific cases. I would strongly suggest to create a separate model for each database table.
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