如何使用 db::raw 对 Laravel fluent 进行更新查询
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How to do update query on laravel fluent using db::raw
提问by Bryan P
This is related to one of my question earlier where:
这与我之前的一个问题有关：
Update table1 field with table2 field value in join laravel fluent
在join laravel fluent中使用table2字段值更新table1字段
But since this is a different approach now, I will just ask another question:
但由于现在这是一种不同的方法，我将问另一个问题：
How do you properly do an update using DB:raw?
你如何正确地使用更新DB:raw？
I want to update the favorite_contents.typewith the value of contents.type, but it doesn't do anything, the static setting of 1 to favorite_contents.expiredis working.
我想favorite_contents.type用 contents.type 的值更新，但它没有做任何事情，1 to 的静态设置favorite_contents.expired正在工作。
This is my code which still doesn't update the type even when the DB::rawwas used:
这是我的代码，即使使用了它仍然不会更新类型DB::raw：
$table = 'favorite_contents';
$contents = DB::table($table)
->join('contents', function($join) use($table){
$join->on("$table.content_id", '=', 'contents.id');
})
->whereIn("$table.content_id",$ids)
->update(array(
"$table.expired" => 1
));
DB::raw("UPDATE favorite_contents, contents SET favorite_contents.type = contents.type where favorite_contents.content_id = contents.id");
This is the first code that doesn't update before I resorted to the above code that doesn't work as well:
这是在我使用上述不起作用的代码之前没有更新的第一个代码：
$table = 'favorite_contents';
$contents = DB::table($table)
->join('contents', function($join) use($table){
$join->on("$table.content_id", '=', 'contents.id');
})
->whereIn("$table.content_id",$ids)
->update(array(
"$table.expired" => 1,
"$table.type" => "contents.type"
));
P.S: This is working when done on an sql editor:
PS：这是在 sql 编辑器上完成的工作：
UPDATE favorite_contents, contents SET favorite_contents.type = contents.type where favorite_contents.content_id = contents.id
UPDATE favorite_contents, contents SET favorite_contents.type = contents.type where favorite_contents.content_id = contents.id
回答by AMIB
code raw updates like this:
代码原始更新是这样的：
...->update( array(
'column' => DB::raw( 'column * 2' )
) );
回答by Ironwind
DB::statement("UPDATE favorite_contents, contents SET favorite_contents.type = contents.type where favorite_contents.content_id = contents.id");
Try DB::statementfor raw queries that does not involve outputting something (select).
尝试DB::statement不涉及输出内容的原始查询（选择）。
回答by Ivan Pirus
And will be work such similar, simple realization in Laravel 5.2 , Query Builder:
并将在 Laravel 5.2 Query Builder 中实现类似的简单实现：
DB::table('stores')
->where('id', $request)
->update(['visibility' =>DB::raw($value)]);
This response is tested real site and working properly
此响应经过真实站点测试并正常工作
