GuzzleHttp 处理xml数据格式的请求及返回

1. 数组转成xml的方法private function arrayToXml($arr) {$xml = "<xml>";foreach ($arr as $key => $val) {if (is_array($val)) {$xml .= "<" . $key . ">" . $this->arrayToXml($val) . "</" . $ke

qq_39688201

1192人浏览 · 2021-10-26 14:02:37

qq_39688201 · 2021-10-26 14:02:37 发布

1. 数组转成xml的方法

private function arrayToXml($arr) {
        $xml = "<xml>";
        foreach ($arr as $key => $val) {
            if (is_array($val)) {
                $xml .= "<" . $key . ">" . $this->arrayToXml($val) . "</" . $key . ">";
            } else {
                $xml .= "<" . $key . ">" . $val . "</" . $key . ">";
            }
        }
        $xml .= "</xml>";
        return $xml;
    }

2.请求及处理返回结果

// $body 提交的数据
$xmlData = $this->arrayToXml($body);
$response = $this->httpClient()->post('pay/unifiedorder', [
            'body' => $xmlData
        ]);
        if ($response->getStatusCode() === 200) {
            $response = simplexml_load_string($response->getBody()->getContents(), 'SimpleXMLElement', LIBXML_NOCDATA);
            $response = json_decode(json_encode($response), true);
            dump($response);
        } else {
            throw new BaseException('请求失败');
        }