Symfony5 ChoiceType TransformationFailedException

Question

使用Sf表单，但无法通过ChoiceType验证值

表格：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $countryChoiceArr = [];
    
    foreach ($options['countries'] as $country) {
        if (array_key_exists('fullName', $country) && array_key_exists('countryCode', $country)) {
            $countryChoiceArr[$country['fullName']] = $country['countryCode'];
        }
    }

    $builder
        ->add('choice', ChoiceType::class, [
            'choices'  => $countryChoiceArr, 
        ])
}

public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults([
            'countries' => [],
        ]);

        $resolver->setAllowedTypes('countries', 'array');
    }

数组是$ countryChoiceArr = ['一个'=>'一个'，'两个'=>'两个'] 我发现问题是当我在变量中传递$ countryChoiceArr时，如果在括号中传递相同的值，则表单会出错，例如

 'choices'  => ['one' => 'one', 'two' => 'two']

没有错误。

表单错误本身：

#message: "Unable to reverse value for property path "[country]": The choice "ago" does not exist or is not unique."
      #code: 0
      #file: "\vendor\symfony\form\Form.php"
#line: 1141
      -previous: Symfony\Component\Form\Exception\TransformationFailedException {#558 ▼
        -invalidMessage: null
        -invalidMessageParameters: []
        #message: "The choice "ago" does not exist or is not unique."
        #code: 0
        #file: "ChoiceToValueTransformer.php"

我已经看到一些涉及EventListener和DataType更改的解决方案，但是，这应该更简单，因为我传递的是同时具有键和值的两个字符串数组，只需要从下拉列表中进行选择并以字符串形式获取值即可。

将感谢您的建议。