Java List<Map<String,Object>> 到 org.json.JSONObject？
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List<Map<String,Object>> to org.json.JSONObject?
提问by yogman
List<Map<String, Object>> list = new ArrayList<Map<String, Object>>();
Map<String, Object> map = new HashMap<String, Object>();
map.put("abc", "123456");
map.put("def", "hmm");
list.add(map);
JSONObject json = new JSONObject(list);
try {
System.err.println(json.toString(2));
} catch (JSONException e) {
e.printStackTrace();
}
What's wrong with this code?
这段代码有什么问题？
The output is:
输出是：
{"empty": false}
回答by WolfmanDragon
You have a map nested inside a list. you are trying to call the Map without ever iterating through the list first. JSON sometimes feels like magic but in fact it is not.
I'll post some code in a moment.
It would be more consistent with JSON to make a Map of Maps instead of a List of Maps.
您有一个嵌套在列表中的地图。您试图调用 Map 而不先遍历列表。JSON 有时感觉很神奇，但实际上并非如此。稍后我会发布一些代码。
制作地图的地图而不是地图的列表会更符合 JSON。
JSONObject json = new JSONObject(list);
Iterator<?> it = json.keys();
while (keyed.hasNext()) {
String x = (String) it.next();
JSONObject jo2 = new JSONObject(jo.optString(x));
}
回答by gavinandresen
You need to end up with a JSONArray (corresponding to the List) of JSONObjects (the Map).
您需要以 JSONObjects（地图）的 JSONArray（对应于列表）结束。
Try declaring the json variable as a JSONArray instead of a JSONObject (I believe the JSONArray constructor will do the right thing).
尝试将 json 变量声明为 JSONArray 而不是 JSONObject （我相信 JSONArray 构造函数会做正确的事情）。
回答by StaxMan
Also: you could consider using one of other parsers from json.org's list: most of them allow your Json "objects" and "arrays" to map natively to java.util.Maps and java.util.Lists; or in some cases to real Java objects too.
另外：您可以考虑使用 json.org 列表中的其他解析器之一：它们中的大多数允许您的 Json“对象”和“数组”本地映射到 java.util.Maps 和 java.util.Lists；或者在某些情况下也是真正的 Java 对象。
My recommendation would be Hymanson, http://Hymanson.codehaus.org/Tutorialwhich allows for mapping to List/Map/Integer/String/Boolean/null, as well as to real Beans/POJOs. Just give it the type and it maps data to you, or writes Java objects as Json. Others like "json-tools" from berlios, or google-gson also expose similar functionality.
我的建议是 Hymanson，http://Hymanson.codehaus.org/Tutorial，它允许映射到 List/Map/Integer/String/Boolean/null，以及真正的 Beans/POJO。只需给它类型，它就会将数据映射给您，或者将 Java 对象编写为 Json。其他像 berlios 的“json-tools”或 google-gson 也公开了类似的功能。
回答by Jose
This worked for me:
这对我有用：
List<JSONObject> jsonCategories = new ArrayList<JSONObject>();
JSONObject jsonCategory = null;
for (ICategory category : categories) {
jsonCategory = new JSONObject();
jsonCategory.put("categoryID", category.getCategoryID());
jsonCategory.put("desc", category.getDesc());
jsonCategories.add(jsonCategory);
}
try {
PrintWriter out = response.getWriter();
response.setContentType("text/xml");
response.setHeader("Cache-Control", "no-cache");
_log.info(jsonCategories.toString());
out.write(jsonCategories.toString());
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
回答by sivanza
List<Map<String, Object>> list = new ArrayList<Map<String, Object>>();
Map<String, Object> map = new HashMap<String, Object>();
map.put("abc", "123456");
map.put("def", "hmm");
list.add(map);
// it's wrong JSONObject json = new JSONObject(list);
// if u use list to add data u must be use JSONArray
JSONArray json = JSONArray.fromObject(list);
try {
System.err.println(json.toString(2));
} catch (JSONException e) {
e.printStackTrace();
}
回答by ShadowJohn
public String listmap_to_json_string(List<Map<String, Object>> list)
{
JSONArray json_arr=new JSONArray();
for (Map<String, Object> map : list) {
JSONObject json_obj=new JSONObject();
for (Map.Entry<String, Object> entry : map.entrySet()) {
String key = entry.getKey();
Object value = entry.getValue();
try {
json_obj.put(key,value);
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
json_arr.put(json_obj);
}
return json_arr.toString();
}
alright, try this~ This worked for me :D
好吧，试试这个~这对我有用：D
回答by Flávio Ferreira
You can do it using both:
您可以使用两者来做到这一点：
JSONArraydirectly as,
JSONArray直接作为，
String toJson(Collection<Map<String, Object>> list)
{
return new JSONArray(list).toString();
}
Or by iterating the list with Java8 (like @ShadowJohn solution):
或者通过使用 Java8 迭代列表（如@ShadowJohn 解决方案）：
String toJson(Collection<Map<String, Object>> list)
{
return new JSONArray(
list.stream()
.map((map) -> new JSONObject(map))
.collect(Collectors.toList()))
.toString();
}
回答by Balaji Dubey
If you are using org.json.simple.JSONArray
如果您使用org.json.simple.JSONArray
(https://mvnrepository.com/artifact/com.googlecode.json-simple/json-simple/1.1.1)
（https://mvnrepository.com/artifact/com.googlecode.json-simple/json-simple/1.1.1）
List<Map<String, Object>> list = new ArrayList<Map<String, Object>>();
Map<String, Object> map = new HashMap<String, Object>();
map.put("abc", "123456");
map.put("def", "hmm");
list.add(map);
JSONArray jsonArray = new JSONArray();
jsonArray.addAll(listOfMaps);
