Laravel 空间查询生成器

Question

在我的用户控制器的索引中，我有以下内容：

return QueryBuilder::for(User::class)
    ->with('phoneNumbers')
    ->allowedIncludes('trashed')
    ->get();

我希望像这样传递一个包含参数：

http://127.0.0.1:8000/v1/users?include=trashed

将 withTrashed() 全局范围附加到查询。

这可能吗？我很可能遗漏了一些明显的东西，我在测试中尝试了一些变体，通常以如下错误告终：

"message": "Call to a member function addEagerConstraints() on boolean",
    "exception": "Symfony\Component\Debug\Exception\FatalThrowableError",
    "file": "/Users/timwickstrom/Sites/Wavefire/api/vendor/laravel/framework/src/Illuminate/Database/Eloquent/Builder.php",
    "line": 522,

供参考： https://github.com/spatie/laravel-query-builder enter link description here

Answer 1

查看该图书馆后，这就是我所拥有的。

return QueryBuilder::for(User::class)
    ->with('phoneNumbers') // <-- I think you can use `allowedIncludes` here.
    ->allowedFilters([ // <-- I believe that `withTrashed` is a scope query,
                       // so you can use this. You cannot use `allowedIncludes`
                       // because it works with relations.
        Filter::scope('withTrashed')
    ])
    ->get();

Answer 2

请注意，现在包中原生支持此功能。

https://spatie.be/docs/laravel-query-builder/v5/features/filtering

QueryBuilder::for(Booking::class)
    ->allowedFilters([
        AllowedFilter::trashed(),
    ]);

// GET /bookings?filter[trashed]=only will only return soft deleted models

Laravel 空间查询生成器

Laravel Spatie Query Builder

php

laravel

laravel-5

laravel-query-builder