Java 如何在没有名字的android中解析json数组？

如何在android中解析json数组，如下所示：

["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

我尝试的是：

     try {
            JSONObject json = new JSONObject(s);

            JSONArray jsonArray = json.names();

            for ( int i = 0 ; i < jsonArray.length();i++)
            {
                Toast.makeText(getApplicationContext(),jsonArray.getInt(i),Toast.LENGTH_LONG).show();
            }

        } catch (JSONException e) {
            e.printStackTrace();
        }

试试看{
JSONObject json=新的JSONObject；
JSONArray JSONArray=json.names（）；
for（int i=0；i

但是我没有得到任何输出。

试试这个

try {
        JSONArray jsonArray = new JSONArray(s);

        for ( int i = 0 ; i < jsonArray.length();i++) {

            Toast.makeText(getApplicationContext(),jsonArray.getString(i),Toast.LENGTH_LONG).show();

        }

} catch (JSONException e) {
     e.printStackTrace();
}

试试看{
JSONArray JSONArray=新JSONArray；
for（int i=0；i

试试这个

try {
        JSONArray jsonArray = new JSONArray(s);

        for ( int i = 0 ; i < jsonArray.length();i++) {

            Toast.makeText(getApplicationContext(),jsonArray.getString(i),Toast.LENGTH_LONG).show();

        }

} catch (JSONException e) {
     e.printStackTrace();
}

试试看{
JSONArray JSONArray=新JSONArray；
for（int i=0；i

uday的答案是正确的，让我提供一些更详细的解释

JSON提供了两种基本数据结构：对象和数组

对象的格式如下：

"name":{value(s)}

[{value}, {value}, {value}...]

"myArray":["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

值可以是另一个对象、数组或只是一些数据

另一方面，数组的格式如下所示：

"name":{value(s)}

[{value}, {value}, {value}...]

"myArray":["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

同样，值可以是一个对象、另一个数组，或者，如您的示例所示，一个简单的数据段

发生错误的原因是您假设接收的数据是一个包含数组的对象。但是，它实际上是一个匿名数组，这意味着您不能解析成

JSONObject

，首先，您需要从一开始就使用

JSONArray

这是一个匿名数组：

["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

如果它是一个对象（例如，称为myArray），它将如下所示：

"name":{value(s)}

[{value}, {value}, {value}...]

"myArray":["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

---

差别很小（只是“myArray”：部分），但必须以正确的方式解析，否则解析器将失败。

uday的答案是正确的，让我提供一些更详细的解释

JSON提供了两种基本数据结构：对象和数组

对象的格式如下：

"name":{value(s)}

[{value}, {value}, {value}...]

"myArray":["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

值可以是另一个对象、数组或只是一些数据

另一方面，数组的格式如下所示：

"name":{value(s)}

[{value}, {value}, {value}...]

"myArray":["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

同样，值可以是一个对象、另一个数组，或者，如您的示例所示，一个简单的数据段

发生错误的原因是您假设接收的数据是一个包含数组的对象。但是，它实际上是一个匿名数组，这意味着您不能解析成

JSONObject

，首先，您需要从一开始就使用

JSONArray

这是一个匿名数组：

["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

如果它是一个对象（例如，称为myArray），它将如下所示：

"name":{value(s)}

[{value}, {value}, {value}...]

"myArray":["BHUBANESWAR","BANGALORE CANT","BRAHMAPUR","VISAKHAPATNAM",
"VIJAYAWADA ROAD","ASOKHAR","CHAURAKHERI","BANIHAL","SADURA","ANANTNAG",
"PANJGAM","AWANTIPURA","KAKAPORA","PAMPORE"]

---

差别很小（只是

“myArray”：

部分），但必须以正确的方式对其进行解析，否则解析器将失败。

您确定不会因为试图将字符串强制为int而使每个项都出现异常吗？您确定不会因为试图将字符串强制为int而使每个项都出现异常吗？请参阅此部分以了解JsonArray和JsonObject请参考本文了解JsonArray和JsonObject之间的区别