laravel 5.3 查询生成器中的内部查询

Question

我尝试使用 Laravel 内部查询来获取数据，但我遇到了一个问题，我将值变量作为值传递给我的查询如下所示。

    $seller_name = $request->seller_name;
    $supplier_address = DB::table('supplier_addresses')
        ->where('company_main_name', function($query)
        {
            $query->select('company_main_name')
                ->where('company_name',$seller_name)
                ->from('supplier_addresses')
                ->get();
        })
        ->get();

如果我传递的是静态值而不是 $seller_name 那么它就可以正常工作，我检查了 $seller_name 不为空，它显示错误变量未定义。

Answer 1

你必须像

一样使用use

$seller_name = $request->seller_name;
$supplier_address = DB::table('supplier_addresses')
    ->where('company_main_name', function($query) use ($seller_name)
    {
        $query->select('company_main_name')
            ->where('company_name',$seller_name)
            ->from('supplier_addresses')->get();
    })
    ->get();

Answer 2

您必须使用 use ($seller_name)

将变量传递给匿名函数

$seller_name = $request->seller_name;
$supplier_address = DB::table('supplier_addresses')
    ->where('company_main_name', function($query) use ($seller_name)
    {
        $query->select('company_main_name')
            ->where('company_name',$seller_name)
            ->from('supplier_addresses')
            ->get();
    })
    ->get();

Answer 3

试试这个，你必须像这样用 use 传递你的 $variable after 函数

$seller_name = $request->seller_name;
$supplier_address = DB::table('supplier_addresses')
        ->where('company_main_name', function($query ) use($seller_name)
        {
            $query->select('company_main_name')
                ->where('company_name',$seller_name)
                ->from('supplier_addresses')
                ->get();
        })
        ->get();

Answer 4

$seller_name = $request->seller_name;
$supplier_address = DB::table('supplier_addresses')
        ->where('company_main_name', function($query ) use($seller_name)
        {
            $query->select('company_main_name')
                ->where('company_name',$seller_name)
                ->from('supplier_addresses')
                ->get();
        })
        ->get();

您忘记使用 use

Answer 5

您需要使用 use 将 $seller_name 传递给闭包。不要在闭包查询中使用 get() 来通过单个数据库查询获得结果。

$supplier_address = DB::table('supplier_addresses')
    ->where('company_main_name', function($query) use ($seller_name) {
        $query->select('company_main_name')
            ->where('company_name', $seller_name)
            ->from('supplier_addresses');
    })
    ->get();

您的代码也与此相同。

$supplier_address = DB::table('supplier_addresses')
    ->where('company_name', $seller_name)
    ->get();

Answer 6

您必须在函数中传递变量 使用 ($seller_name)

$seller_name = $request->seller_name; 
$supplier_address = 
    DB::table('supplier_addresses')
       ->where('company_main_name', function($query) use ($seller_name) 
       {
              $query->select('company_main_name')
                    ->where('company_name',$seller_name) 
                    ->from('supplier_addresses') ->get(); 
       }) 
    ->get();

您可以在这里参考文档： https://laravel.com/docs/5.4/queries#conditional-clauses

laravel 5.3 查询生成器中的内部查询

inner query in laravel 5.3 query builder

inner-query

laravel

laravel-query-builder