[英]How to change the value in laravel 5 collection
I have some problems here. 我这里有些问题。 I get a collection which like:
我得到一个像这样的集合:
$VenderData=Vender::where('active', '=', '1')->get();
Inside the collection i have a column called ' type ' and the data looks like 'A1,A2,A3,'
or 'A1,A3,'
I want to transfer those codes to real names from another table ' vender_type '. 在集合内部,我有一列称为“ type ”,数据看起来像
'A1,A2,A3,'
或'A1,A3,'
我想将这些代码从另一个表“ vender_type ”传输为实名。
code name
A1 XX
A2 OO
A3 ZZ
then add a new column into the original collection $VenderData. 然后在原始集合$ VenderData中添加一个新列。
How can i do this? 我怎样才能做到这一点?
You must split the type attribute (using array explode(string $delimiter , string $string)
) , and get the name of each one from vender_type
table, try this : 您必须拆分type属性(使用
array explode(string $delimiter , string $string)
),并从vender_type
表中获取每个名称,请尝试以下操作:
$VenderData = Vender::where('active', '=', '1')->get();
foreach($VenderData as $vend)
{
$vend->new_type = array();
$types = explode(",", $vend->type);
foreach($types as $type)
{
$t = vender_type::where('code', $type)->first();
if($t)
$vend->name_type[] = $t->name;
// or for example : $vend->name_type[$type] = $t->name;
}
}
I hope that will help you. 希望对您有所帮助。
I think that you need better query than modifying each value after. 我认为与修改每个值之后相比,您需要更好的查询。 This
vender_type
is probably some foreign key in you database so you can get that value in query builder. 此
vender_type
可能是数据库中的某些外键,因此您可以在查询生成器中获取该值。 To find out about foreign keys and joins in laravel eloquent check this . 要了解有关外键和laravel雄辩的加入,请检查此 。
Basically you need something like this 基本上你需要这样的东西
$users = DB::table('table1')
->join('table2', 'table1.vender_type', '=', 'table2.vender_type')
->select('table1.code', 'table1.name', 'table2.vender_type')
->where('table1.active', '=', '1')
->get();
Just replace table1
and table2
with values of you table names in database and you're done. 只需将
table1
和table2
替换为数据库中表名的值即可。
Hope it helps 希望能帮助到你
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