Php 是否可以提取fzaninotto伪造格式器的单个组件？

是否有可能访问和提取伪造提供商（如地址）的组件。例如，我只需要stateAbr来填充数据库中的一列，而不是整个地址

我尝试了类似于

$faker->address->stateAbr

的方法，但它当然会抛出一个错误

// Faker\Provider\en_US\Address
cityPrefix                          // 'Lake'
secondaryAddress                    // 'Suite 961'
state                               // 'NewMexico'
stateAbbr                           // 'OH'
citySuffix                          // 'borough'
streetSuffix                        // 'Keys'
buildingNumber                      // '484'
city                                // 'West Judge'
streetName                          // 'Keegan Trail'
streetAddress                       // '439 Karley Loaf Suite 897'
postcode                            // '17916'
address                             // '8888 Cummings Vista Apt. 101, Susanbury, NY 95473'
country                             // 'Falkland Islands (Malvinas)'
latitude($min = -90, $max = 90)     // 77.147489
longitude($min = -180, $max = 180)  // 86.211205 

//This is my code

use Faker\Generator as Faker;    
$factory->define(App\Company::class, function (Faker $faker) {
    return [
        'city'=>$faker->city,
        'state'=>$faker->address->stateAbbr,//Getiing an error here
    ];

});

---

直接访问stateAbbr格式化程序

$factory->define(App\Company::class, function (Faker $faker) {
    return [
        'city'=>$faker->city,
        'state'=>$faker->stateAbbr
    ];
});

---

直接访问stateAbbr格式化程序

$factory->define(App\Company::class, function (Faker $faker) {
    return [
        'city'=>$faker->city,
        'state'=>$faker->stateAbbr
    ];
});