Php json-JSONarray无法转换为JSONObject

我正在尝试解析从服务器获取的JSON响应，并收到错误消息：JSONArray无法转换为JSONObject

org.json_JSONEXCEPTION: Value [{}] at 0 of type org.json.JSONArray cannot be converted to JSONObject 





 try {

           JSONArray jArray = new JSONArray(result);
           for(int i=0;i<jArray.length();i++){
               JSONObject json_data = jArray.getJSONObject(i);

               DataFood foodData = new DataFood();

               foodData.foodImage= json_data.getString("gambar");
               foodData.foodName= json_data.getString("nama");
               foodData.foodId= json_data.getInt("id");
               foodData.price= json_data.getInt("harga");
               data.add(foodData);
           }

换行

JSONObject json_data = jArray.getJSONObject(i);

到

然后再次循环通过JSONArray

for (int i = 0; i < innerArray.length(); i++) {
    JSONObject json_data = innerArray.getJSONObject(i);
    DataFood foodData = new DataFood();

    foodData.foodImage= json_data.getString("gambar");
    foodData.foodName= json_data.getString("name");
    foodData.foodId= json_data.getInt("id");
    foodData.price= json_data.getInt("harga");
    data.add(foodData);
}

for（int i=0；i

外部JSONArray包含JSONArray，而不是可以在PHP代码中看到的JSONObject

$result = array(); <-- your main array
while($row = mysqli_fetch_array($res)){
array_push($result, <-- you add a new array to your main array
array('id'=>$row[0],
'name'=>$row[1],
'harga'=>$row[2],
'gambar'=>$row[3]
));
}

$result=array（）$第[1]行，
“harga'=>美元行[2]，
“gambar”=>美元行[3]
));
}

您的结果字符串是什么？请提供您的响应字符串。我必须更改此行吗？JSONArray jArray=新JSONArray（结果）；不，这行代码仍然可以遵循您的建议，但它说“无法解析方法大小（）”对不起，先生，我是新手。如果问题正在减少，响应字符串是“org.JSONException No value for nama”，应该是

name

，我刚刚复制了您的JSON

JSONArray innerArray = jArray.getJSONArray(i);

for (int i = 0; i < innerArray.length(); i++) {
    JSONObject json_data = innerArray.getJSONObject(i);
    DataFood foodData = new DataFood();

    foodData.foodImage= json_data.getString("gambar");
    foodData.foodName= json_data.getString("name");
    foodData.foodId= json_data.getInt("id");
    foodData.price= json_data.getInt("harga");
    data.add(foodData);
}

$result = array(); <-- your main array
while($row = mysqli_fetch_array($res)){
array_push($result, <-- you add a new array to your main array
array('id'=>$row[0],
'name'=>$row[1],
'harga'=>$row[2],
'gambar'=>$row[3]
));
}