异常的消息数据中，我能不能只要response这部分，其他多余的信息不要

[leeboo]

Client error: POST https://api.mch.weixin.qq.com/v3/applyment4sub/applyment/ resulted in a 400 Bad Request response: {"code":"PARAM_ERROR","message":"开户银行省市编码(/bank_account_info/bank_address_code)取值有误，请检查

我只想要 {"code":"PARAM_ERROR","message":"开户银行省市编码(/bank_account_info/bank_address_code)取值有误，请检查这部分json

怎么弄？

[TheNorthMemory]

两种方式:

1. 同步模式：

try {
    $r = $instance->chain('v3/applyment4sub/applyment/')->post([]);
    $m = json_decode((string)$r->getBody());
    print_r($m);
} catch(Exception $e) {
    if ($e instanceof \GuzzleHttp\Exception\RequestException && $e->hasResponse()) {
        $r = $e->getResponse();
        $m = json_decode((string)$r->getBody());
        print_r($m);
    }
}

2. 异步链式：

$instance->chain('v3/applyment4sub/applyment/')
->postAsync([])
->then(static function(\Psr\Http\Message\ResponseInterface $r) {
    $m = json_decode((string)$r->getBody());
    print_r($m);
})
->otherwise(static function (\GuzzleHttp\Exception\RequestException $e) {
    $r = $e->getResponse();
    $m = json_decode((string)$r->getBody());
    print_r($m);
})
->wait();