如何在android volley上解析此JSON
这是我需要解析的输出。如何在android volley上解析此JSON,android,json,Android,Json,这是我需要解析的输出。usertype\u name的值应存储到字符串数组中。请帮帮我 { 0: { usertype_name: "Member-Individual" }, 1: { usertype_name: "Member-Institutional" }, 2: { usertype_name: "Officer-Member" }, 3: { usertype_name: "Officer-Ad
usertype\u name
的值应存储到字符串数组中。请帮帮我
{
0: {
usertype_name: "Member-Individual"
},
1: {
usertype_name: "Member-Institutional"
},
2: {
usertype_name: "Officer-Member"
},
3: {
usertype_name: "Officer-Admin"
},
4: {
usertype_name: "Officer-President"
},
current_position: "Member-Individual"
}
在截取上解析JSON时,通常我会这样做:
public void onResponse(String s) {
JSONArray array= new JSONArray(s.toString());
for(int i = 0; i < array.length(); i++) {
JSONObject object = array.getJSONObject(i);
String text = object.getString("usertype_name");
}
}
像这样更改json字符串
{
"array": [
"Member-Individual",
"Member-Institutional",
"Officer-Member"
],
"current_position": "6"
}
php代码示例:
<?php
class User {
public $array = array("Member-Individual", "Member-Institutional", "Officer-Member");
public $current_position = "6";
public function __construct() {
}
}
$user = new User();
echo json_encode($user);
?>
是的。。。您将不得不添加更多数据。你的密码在哪里?你已经用截击了吗?展示出来,事实上。。。这不是有效的JSON。整数和纯文本不能是键。请检查我的编辑。希望它真的有用,因为您没有有效的JSON如何使它成为有效的JSON,同时保持当前位置
值?您能帮助我在php上如何做到这一点吗?这一点我不熟悉:(参考我已经编辑了答案。请看一看。
<?php
class User {
public $array = array("Member-Individual", "Member-Institutional", "Officer-Member");
public $current_position = "6";
public function __construct() {
}
}
$user = new User();
echo json_encode($user);
?>