Question

<soapenv:envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/">
    <soapenv:body>
        <ns1:mbillcommandresponse xmlns:ns1="http://www.mysoap.com/SoapService/">
            <ReturnValues>
                <name>status</name>
                <value>TEHNICAL_ERROR</value>
            </ReturnValues>
            <ReturnValues>
                <name>description</name>
                <value>Please contact your administrator</value>
            </ReturnValues>
        </ns1:mbillcommandresponse>
    </soapenv:body>
</soapenv:envelope>

在CURL响应中，我得到了响应。这是我的PHP代码：

    $response = curl_exec($ch);
    $xml = simplexml_load_string($response);
    //$result = $xml->xpath('//name'); //echo "<pre>"; print_r($result); exit;
    $xml->registerXPathNamespace('ns1', 'http://www.mysoap.com/SoapService/');
    foreach ($xml->xpath('//returnvalues') as $item) {
        $json = json_encode($item);
        $convrt_arr = json_decode($json, true);
        break;
    }
    print_r($json); exit;

在上面的代码中，我得到了空的json。你能帮我吗？

Answer 1

您的问题编辑是一个重要线索。您已将XML从小写改为CamelCase元素。更改：

CurrentTemperatureAddonm

收件人：

foreach ($xml->xpath('//returnvalues') as $item) {

，它将起作用。 Xpath查询区分大小写。

Answer 2

希望这会有所帮助

$response = '<soapenv:envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/">
    <soapenv:body>
        <ns1:mbillcommandresponse xmlns:ns1="http://www.mysoap.com/SoapService/">
            <returnvalues>
                <name>status</name>
                <value>TEHNICAL_ERROR</value>
            </returnvalues>
            <returnvalues>
                <name>description</name>
                <value>Please contact your administrator</value>
            </returnvalues>
        </ns1:mbillcommandresponse>
    </soapenv:body>
</soapenv:envelope>';
$response = preg_replace("/(<\/?)(\w+):([^>]*>)/", "$1$2$3", $response);

$xml = new SimpleXMLElement($response);
$body = $xml->xpath('//returnvalues');
$json = json_encode((array)$body); 
print_r($json);

如何将SOAP响应转换为XML或JSON格式？

2 个答案: