Java 如何迭代 JSONObject (gson)
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How to iterate a JSONObject (gson)
提问by Floyd
I have a JsonObject e.g
我有一个 JsonObject 例如
JsonObject jsonObject = {"keyInt":2,"keyString":"val1","id":"0123456"}
Every JsonObjectcontains a "id"entry, but the number of other key/value pairs is NOT determined, so I want to create create an object with 2 attributes:
每个都JsonObject包含一个"id"条目，但其他键/值对的数量未确定，因此我想创建一个具有 2 个属性的对象：
class myGenericObject {
Map<String, Object> attributes;
String id;
}
So I want my attributes map to look like this:
所以我希望我的属性映射看起来像这样：
"keyInt" -> 4711
"keyStr" -> "val1"
I found this solution
我找到了这个解决方案
Map<String, Object> attributes = new HashMap<String, Object>();
Set<Entry<String, JsonElement>> entrySet = jsonObject.entrySet();
for(Map.Entry<String,JsonElement> entry : entrySet){
attributes.put(entry.getKey(), jsonObject.get(entry.getKey()));
}
but the values are enclosed by ""
但值被包围 ""
"keyInt" -> "4711"
"keyStr" -> ""val1""
How to get the plain values (4711and "val1")?
如何获得普通值（4711和"val1"）？
Input data:
输入数据：
{
"id": 0815,
"a": "a string",
"b": 123.4,
"c": {
"a": 1,
"b": true,
"c": ["a", "b", "c"]
}
}
or
或者
{
"id": 4711,
"x": false,
"y": "y?",
}
回答by atish shimpi
replace "" with blank.
将“”替换为空白。
Map<String, Object> attributes = new HashMap<String, Object>();
Set<Entry<String, JsonElement>> entrySet = jsonObject.entrySet();
for(Map.Entry<String,JsonElement> entry : entrySet){
if (! nonProperties.contains(entry.getKey())) {
properties.put(entry.getKey(), jsonObject.get(entry.getKey()).replace("\"",""));
}
}
回答by Ted Xu
I am not quite sure why you want to do manual manipulation in the first place. GSON decode will simply leave those absent key/value pairs as default value (zero,null). And then you can process as you want.
我不太确定你为什么要首先进行手动操作。GSON 解码只会将那些不存在的键/值对保留为默认值（零，空）。然后您可以根据需要进行处理。
回答by Syed Mauze Rehan
How are you creating your JsonObject? Your code works for me. Consider this
你是如何创建你的 JsonObject 的？你的代码对我有用。考虑这个
import com.google.gson.JsonElement;
import com.google.gson.JsonObject;
...
...
...
try{
JsonObject jsonObject = new JsonObject();
jsonObject.addProperty("keyInt", 2);
jsonObject.addProperty("keyString", "val1");
jsonObject.addProperty("id", "0123456");
System.out.println("json >>> "+jsonObject);
Map<String, Object> attributes = new HashMap<String, Object>();
Set<Entry<String, JsonElement>> entrySet = jsonObject.entrySet();
for(Map.Entry<String,JsonElement> entry : entrySet){
attributes.put(entry.getKey(), jsonObject.get(entry.getKey()));
}
for(Map.Entry<String,Object> att : attributes.entrySet()){
System.out.println("key >>> "+att.getKey());
System.out.println("val >>> "+att.getValue());
}
}
catch (Exception ex){
System.out.println(ex);
}
And it is working fine. Now I am interested in knowing how you created that JSON of yours?
它工作正常。现在我有兴趣知道您是如何创建您的 JSON 的？
You can also try this (JSONObject)
你也可以试试这个（JSONObject）
import org.json.JSONObject;
...
...
...
try{
JSONObject jsonObject = new JSONObject("{\"keyInt\":2,\"keyString\":\"val1\",\"id\":\"0123456\"}");
System.out.println("JSON :: "+jsonObject.toString());
Iterator<String> it = jsonObject.keys();
while( it.hasNext() ){
String key = it.next();
System.out.println("Key:: !!! >>> "+key);
Object value = jsonObject.get(key);
System.out.println("Value Type "+value.getClass().getName());
}
}
catch (Exception ex){
System.out.println(ex);
}
回答by Siddhesh Shirodkar
Just make following changes...
只需进行以下更改...
Map<String, Object> attributes = new HashMap<String, Object>();
Set<Entry<String, JsonElement>> entrySet = jsonObject.entrySet();
for(Map.Entry<String,JsonElement> entry : entrySet){
attributes.put(entry.getKey(), jsonObject.get(entry.getKey()).getAsString());
}
"getAsString" will do the magic for u
“getAsString” 会为你带来魔力
回答by Paul Benn
Your Map values are JsonElements. Whenever you print a JsonElement(e.g. using a debugger) its toString()method will be called - and since a JsonElementhas many implementing classes the default toStringimplementation wraps the value in quotes to ensure correct JSON. To get the value as a normal, unwrapped String, simply call getAsString():
您的 Map 值是JsonElements。每当您打印 a JsonElement（例如使用调试器）时，它的toString()方法将被调用 - 由于 aJsonElement有许多实现类，因此默认toString实现将值包装在引号中以确保正确的 JSON。要获取正常的未包装值String，只需调用getAsString()：
JsonElement elem;
// ...
String value = elem.getAsString();
With your example:
以你的例子：
Map<String, Object> attributes = new HashMap<String, Object>();
Set<Entry<String, JsonElement>> entrySet = jsonObject.entrySet();
for(Map.Entry<String,JsonElement> entry : entrySet){
attributes.put(entry.getKey(), jsonObject.get(entry.getKey()).getAsString());
}
Note there are many other methods you can call on a JsonElementto produce other types.
请注意，您可以调用许多其他方法JsonElement来生成其他类型。
