我通过遵循Laravel文档https://laravel.com/docs/5.6/validation#custom-validation-rules为Laravel应用程序创建了一些新的验证规则,但是,当我尝试将规则注册到我的自定义请求规则数组中时,会引发错误:
#message: "trim() expects parameter 1 to be string, object given"
#code: 0
#file: "/Users/ari/Projects/dps/sites/acg/vendor/laravel/framework/src/Illuminate/Validation/ValidationRuleParser.php"
#line: 217
#severity: E_WARNING
我的自定义规则-NotContainsEmail.php:
<?php
namespace App\Rules;
use Illuminate\Contracts\Validation\Rule;
class NotContainsEmail implements Rule
{
/**
* Determine if the validation rule passes.
*
* @param string $attribute
* @param mixed $value
* @return bool
*/
public function passes($attribute, $value)
{
return (strpos($value, '@') !== false);
}
/**
* Get the validation error message.
*
* @return string
*/
public function message()
{
return 'This field cannot contain an email address.';
}
}
我的请求-QuoteRespondRequest:
<?php namespace Client\Http\Requests\Quotes;
use App\Rules\NotContainsEmail;
use Client\Http\Requests\FormRequest;
class QuoteRespondRequest extends FormRequest
{
public function rules()
{
return [
'help' => ['string', 'nullable', new NotContainsEmail],
'description' => ['string', 'nullable', new NotContainsEmail],
'community' => ['string', 'nullable', new NotContainsEmail],
'funding' => ['string', new NotContainsEmail],
];
}
}
我对错误感到困惑,因为Laravel文档明确指出我可以通过一个对象,但是Illuminate\Validation\ValidationRuleParser.php@217
显然无法处理对象。
我哪里出错了?
答案 0 :(得分:1)
使用扩展 注册自定义验证规则的另一种方法是在Validator外观上使用extend方法。让我们在服务提供者中使用此方法来注册自定义验证规则: https://laravel.com/docs/5.7/validation
在AppServiceProvider中
public function boot()
{
Validator::extend('NotContainsEmail', function ($attribute, $value, $parameters, $validator) {
// code here
});
Validator::replacer('foo', function ($message, $attribute, $rule, $parameters) {
//return str_replace(...);
});
}