如何在 guzzlehttp/guzzle 错误中访问返回错误的标题、状态、详细信息字段？

Question

在 laravel 10 应用程序中使用 guzzlehttp/guzzle ^7.8.0 捕获 https://github.com/mailchimp/mailchimp-marketing-php (v3.0) 的 mailchimpListId 方法的运行时错误 库 getContents 方法出现错误：

try {
    $response = $this->getApiClient()->lists->addListMember($this->mailchimpListId, [
        "email_address" => $email,
        "status" => "subscribed",
        'merge_fields' => $userMergeFields,
    ]);

} catch (ClientException $e) {
    \Log::info($e->getMessage());
    \Log::info(varDump($e->getResponse()->getBody()->getContents());

返回包含所有 3 个字段的字符串：

[2024-02-18 07:18:45] local.INFO: scalar => (string) :{"title":"Member Exists","status":400,"detail":"[email protected] is already a list member. Use PUT to insert or update list members.","instance":"4d85bc54-136f-e637-92e2-3c1bf9f1048c"}

I try to get access to all 3 fields title", status, detail, but I did not find how can I do it. I tried like :

\Log::info(varDump($e->getResponse()->getBody()->getTitle(), ' -10 $e->getResponse()->getTitle()::'));

但出现错误：

Call to undefined method GuzzleHttp\Psr7\Response::getTitle()

我怎样才能访问这3个字段？

Answer 1

您收到的错误消息是 JSON 字符串。您可以使用 PHP 中的

json_decode

函数将此 JSON 字符串转换为关联数组。之后，您可以将字段

title

、

status

和

detail

作为数组元素访问。

您可以这样做：

try {
    $response = $this->getApiClient()->lists->addListMember($this->mailchimpListId, [
        "email_address" => $email,
        "status" => "subscribed",
        'merge_fields' => $userMergeFields,
    ]);

} catch (ClientException $e) {
    $errorResponse = json_decode($e->getResponse()->getBody()->getContents(), true);

    \Log::info('Title: ' . $errorResponse['title']);
    \Log::info('Status: ' . $errorResponse['status']);
    \Log::info('Detail: ' . $errorResponse['detail']);
}