Question

我有一个成功运行并返回json数据的服务器的soap请求。

{"GetAccountResult":{"Header":{"Rquid":"D560BC95-24F2-F705-0585-7CCB38E37ECE","Status":{"StatusCode":"-4","Message":"Account not found","Details":null}},"Account":null}}

我想知道如何在php中访问它并只打印消息？不是整个json的结果？

我试过了

$account   = $client->GetAccount($arrParams);
            $results = json_encode($account);

            echo $results->Message;

但是这仍然打印完整的json对象

Answer 1

以下是解决方案：

$account   = '{"GetAccountResult":{"Header":{"Rquid":"D560BC95-24F2-F705-0585-7CCB38E37ECE","Status":{"StatusCode":"-4","Message":"Account not found","Details":null}},"Account":null}}';
$results = json_decode($account,true);    // 2nd argument `true` returns array from JSON
echo $results['GetAccountResult']['Header']['Status']['Message'];

Answer 2

$result = json_decode($your_json, true);

$message = $result["GetAccountResult"]["Header"]["Status"]["Message"];

我强烈建议使用isset()检查索引是否存在。否则你可能会出错。

Answer 3

您可以通过json_decode检索数据：

$json = '{"GetAccountResult":{"Header":{"Rquid":"D560BC95-24F2-F705-0585-7CCB38E37ECE","Status":{"StatusCode":"-4","Message":"Account not found","Details":null}},"Account":null}}';
$t =  json_decode($json,true);
print_r($t['GetAccountResult']['Header']['Status']['Message']);

Answer 4

检查此解决方案：

$account   = $client->GetAccount($arrParams);
$results   = json_decode($account);

echo $results->GetAccountResult->Header->Status->Message;

在PHP中阅读Json结果

4 个答案: