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我正在编写一个android应用程序,它与Web服务进行交互以获取数据。 Web服务使用PHP编写,我写了一个使用AsyncTask获取数据的库,问题在于类只接受JSONObject。我的大多数服务都只返回一个JSONObject,有一个返回一个json数组。返回JSON对象而不是JSONArray
$array = array();
while ($row = mysql_fetch_array($query))
{
$array[] = $row;
}
echo json_encode($array);
返回是这样的:
[{ "0":"10","id":"10","1":"17.9409915","lat":"17.9409915","2":"-77.1003625","lon":"-77.1003625"},{"0":"9","id":"9","1":"17.9410143","lat":"17.9410143","2":"-77.1003672","lon":"-77.1003672"}]
我想回报
{result:[{"0":"10","id":"10","1":"17.9409915","lat":"17.9409915","2":"-77.1003625","lon":"-77.1003625"},{"0":"9","id":"9","1":"17.9410143","lat":"17.9410143","2":"-77.1003672","lon":"-77.1003672"}]}
我试图这样做来实现:
echo json_encode("{result: " .$array. "}");
但是,不起作用。它返回。
"{result: Array}"
我该如何做到这一点?