[英]Laravel - Join with a table with composite primary key
My problem is to join 2 tables in Laravel framework. 我的问题是在Laravel框架中加入2个表。 One is dynamic name table (it's a variable) and second has composite primary key .
一个是动态名称表 (它是一个变量),第二个是复合主键 。 I have to use query builder instead of where().
我必须使用查询生成器而不是where()。 Please view my following for details:
请查看以下详细信息:
I have 2 tables: 我有2张桌子:
CREATE TABLE `details` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`source_id` int(10) unsigned NOT NULL,
`brand_id` int(10) DEFAULT NULL,
PRIMARY KEY (`id`)
);
CREATE TABLE `links` (
`source_id` int(10) unsigned NOT NULL,
`brand_id` tinyint(3) unsigned NOT NULL DEFAULT '1',
PRIMARY KEY (`source_id`,`brand_id`)
);
Now, I need to join 2 these tables, I use this code: 现在,我需要加入2这些表,我使用这段代码:
<?php $results = \DB::table('details')
->join('links', function($join)
{
$join->on('details.source_id', '=', 'links.source_id');
$join->on('details.brand_id','=', 'links.brand_id');
})
->get();?>
It's quite simple to join these table, OK. 加入这些表非常简单,好的。 But my problem is the table name is dynamic.
但我的问题是表名是动态的。
<?php
$type = Input::get('type', null);
$table = $type . '_details';
$results = \DB::table($table)
->join('links', function($join)
{
// the following code will show errors undefined $table
$join->on($table . '.source_id', '=', 'links.source_id');
$join->on($table . '.brand_id','=', 'links.brand_id');
})
->get();
?>
Please help me to solve this problem. 请帮我解决这个问题。 Many thanks!!!
非常感谢!!!
You need to import variables from the local scope to the anonymous function's scope, this is how: 您需要将变量从本地范围导入到匿名函数的范围,这是如何:
$results = \DB::table($table)
->join('links', function($join) use ($table)
{
$join->on($table . '.source_id', '=', 'links.source_id');
$join->on($table . '.brand_id','=', 'links.brand_id');
})
->get();
Notice the line: 注意这一行:
->join('links', function($join) use ($table)
Problem is the anonymous function doesn't know about the variable $table
, so you tell it about the variable using use
. 问题是匿名函数不知道变量
$table
,所以你告诉它使用变量use
。
Please try : 请试试 :
<?php
$type = Input::get('type', null);
$table = $type . '_details';
$joinFunction = function($join) use ($table)
{
$join->on($table . '.source_id', '=', 'links.source_id');
$join->on($table . '.brand_id','=', 'links.brand_id');
}
$results = \DB::table($table)
->join('links',$joinFunction )
->get();
?>
The problem was that the function doesn't see the $table variable inside it. 问题是该函数没有在其中看到$ table变量。 That's why you need to use the "use" statement .
这就是你需要使用“use”语句的原因。
Read more about anonymous functions in php here 在这里阅读有关php中匿名函数的更多信息
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