Laravel - 自定义处理程序中的Catch异常-Java 学习之路

我正在使用Laravel 5.5，我想处理来自自定义处理程序的自定义异常，而不是来自 app\Exceptions\Handler.php . 现在，我 grab 了一个异常，如果某些字段's empty in the form submitted by the user. It'完全像这样工作：

ProfileController.php ：

public function update(Request $request, $id){
    $this->guzzleService->put(
        $request,
        ApiEndPoints::UPDATE_PROFILE . $id,
        true
    );

    return back()->with('SavedCorrectly', 'Changes saved correctly');
}

app\Exceptions\Handler.php

public function render($request, Exception $exception)
{
    if($exception instanceof ClientException && $exception->getCode() == 422)
        return back()->withErrors(
            json_decode((string) $exception->getResponse()->getBody(), TRUE)["errors"]
        );

    return parent::render($request, $exception);
}

问题是我想重构它，所以它仍然是这样的：

ProfileController.php

public function update(Request $request, $id){
    try {
        $this->guzzleService->put(
            $request,
            ApiEndPoints::UPDATE_PROFILE . $id,
            true
        );

        return back()->with('SavedCorrectly', 'Cambios guardados correctamente');
    } catch(ClientException $exception) {
        if ($exception->getCode() == 500) throw new InternalServerErrorException;
        if ($exception->getCode() == 422) throw new UnprocessableEntityException;
    }
}

app\Exceptions\HttpExceptions\UnprocessableEntityException.php

<?php

namespace App\Exceptions\HttpExceptions;

use GuzzleHttp\Exception\ClientException;

class UnprocessableEntityException extends \Exception
{
    public function render($request, ClientException $exception)
    {
        return back()->withErrors(
            json_decode((string) $exception->getResponse()->getBody(), TRUE)["errors"]
        );
    }
}

但是我收到了这个错误：

类型错误：参数2传递给App \ Exceptions \ HttpExceptions \ UnprocessableEntityException :: render（）必须是GuzzleHttp \ Exception \ ClientException的实例，没有给出，在... \ vendor \ laravel \ framework \ src \ Illuminate \中调用第169行的Foundation \ Exceptions \ Handler.php

1 回答

这是因为您传递了一个新的异常

public function update(Request $request, $id){
    try {
        $this->guzzleService->put(
            $request,
            ApiEndPoints::UPDATE_PROFILE . $id,
            true
        );

        return back()->with('SavedCorrectly', 'Cambios guardados correctamente');
    } catch(ClientException $exception) {
        if ($exception->getCode() == 500) throw new InternalServerErrorException((string) $exception->getResponse()->getBody());
        if ($exception->getCode() == 422) throw new UnprocessableEntityException((string) $exception->getResponse()->getBody());
    }
}

和

<?php

namespace App\Exceptions\HttpExceptions;

use GuzzleHttp\Exception\ClientException;

class UnprocessableEntityException extends \Exception
{
    public function render($request)
    {
        return back()->withErrors(
            json_decode((string) $this->message, TRUE)["errors"]
        );
    }
}

回复于 2024-04-27T08:53:21+08:00

Laravel - 自定义处理程序中的Catch异常

1 回答

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