假设我想显示 type="color" 的完整奖项列表:
Let's say I want to show a full list of awards with type="color":
Awards Type 2013 Winner
====== ==== ===========
Blue Award color Tom
Red Award color
Green Award color Dan
为了达到这个结果,我可以在 Laravel 中进行这样的查询:
To achieve this result I could have a query in Laravel like this:
$year = '2013';
$awards = DB::table('awards')
->leftJoin('winners', function($join) use ($year)
{
$join->on('awards.id','=','winners.award_id');
$join->on('winners.year','=',DB::raw("'".$year."'"));
}
->where('awards.type','color')
->get();
如果您输出 Laravel 生成的 SQL,您将看到只有 WHERE 子句 被参数化,并且 ON 子句 中的 $year 容易受到 sql 注入,如果我从不受信任的来源获取它.此外,查询的缓存潜力也会降低,因为 $year 会经常更改.注意:如果您认为我只是将第二个左连接条件添加到查询的 WHERE,这些不一样.
If you output the SQL that Laravel generates you will see that only the WHERE clause is parameterized and $year in the ON clause is left vulnerable to sql injection if I get it from an untrusted source. Also the query's caching potential is reduced because $year will change often. Note: In case you were thinking that I just add the second left join condition to the WHERE of the query, these are not the same.
关于如何将查询的 $year 部分参数化有什么想法吗?
Any ideas on how to get the $year part of the query parameterized?
这里有一个奇怪的解决方法(不想扩展 Builder 和 JoinClause 类):
注意:这会破坏 ->
的查询链,所以请注意 where
在下面被分隔.
Here's an odd work-around (didn't want to extend the Builder and JoinClause classes):
Notice: This will break query chaining with ->
so notice the where
was seperated below.
$query = DB::table('awards')
->leftJoin('winners', function($join)
{
$join->on('awards.id','=','winners.award_id');
$join->on('winners.year','=',DB::raw('?'));
}
->setBindings(array_merge($query->getBindings(),array($year)));
$query->where('awards.type','color');
$awards = $query->get();
更新:泰勒添加 joinWhere
, leftJoinWhere
... 他说如果你有一个函数连接,只需使用 ->where
和 ->orWhere
从封闭内."不过我还没有尝试过.
UPDATE: Taylor added joinWhere
, leftJoinWhere
... he says that "if you have a function join just use ->where
and ->orWhere
from within the Closure." I've yet to try this though.
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