我是stackoverflow的新手。
我正在创建一个Java应用程序,它将从Web服务器获取数据。数据采用json格式。实施例“
[
{
"item_name": "Adame",
"item_type": "Special",
"item": "Chestplate",
"item_min_lvl": "50",
"enchantment": {
"health": "0.3",
"dam": "24%",
"life": "0.1",
"xp": "24%",
"loot": "22%"
},
"def": "73"
},
{
"item_name": "Sticks'",
"item_type": "Unique",
"item": "Stick",
"item_min_lvl": "4",
"enchantment": {
"health": "0.6",
"mana": "1",
"dam": "12%",
"life": "0.3",
"xp": "17%",
"loot": "17%"
},
"min_dam": "39",
"max_dam": "34"
}
]
我知道如何使用Gson反序列化json。如您所见,它始于[
。我之前从未反复将此案件反序列化。此外,json数据不相同(例如enchantment
)。我也在谷歌搜索,但我找不到任何类似的情况。任何人都可以帮我代码吗?谢谢你提前。
答案 0 :(得分:0)
尝试使用此代码。你会得到你的问题的答案。它是一个包含2个项目的列表。
StringBuilder builder = new StringBuilder();
BufferedReader reader = new BufferedReader(new FileReader(new File("resources/json1.txt")));
String line = null;
while ((line = reader.readLine()) != null) {
builder.append(line);
}
reader.close();
Gson gson = new Gson();
Type listType = new TypeToken<ArrayList<MyJSON>>() {
}.getType();
List<MyJSON> list = gson.fromJson(builder.toString(), listType);
// you can try this form as well
// MyJSON[] list = gson.fromJson(builder.toString(), MyJSON[].class);
for (MyJSON json : list) {
System.out.println(json.toString());
}
...
class MyJSON {
String item_name;
String item_type;
String item;
String item_min_lvl;
Enchantment enchantment;
@Override
public String toString() {
StringBuilder builder = new StringBuilder();
builder.append("\nitem_name:").append(item_name);
builder.append("\nitem_type:").append(item_type);
builder.append("\nitem:").append(item);
builder.append("\nitem_min_lvl:").append(item_min_lvl);
builder.append("\n\nEnchantment Details:");
builder.append("\nhealth:").append(enchantment.health);
builder.append("\ndam:").append(enchantment.dam);
builder.append("\nlife:").append(enchantment.life);
builder.append("\nxp:").append(enchantment.xp);
builder.append("\nloot:").append(enchantment.loot);
return builder.toString();
}
}
class Enchantment {
String health;
String dam;
String life;
String xp;
String loot;
}
输出:
item_name:Adame
item_type:Special
item:Chestplate
item_min_lvl:50
Enchantment Details:
health:0.3
dam:24%
life:0.1
xp:24%
loot:22%
item_name:Sticks'
item_type:Unique
item:Stick
item_min_lvl:4
Enchantment Details:
health:0.6
dam:12%
life:0.3
xp:17%
loot:17%
修改强>
每个条目的结构都不相同,因此您不能将POJO用于此类JSON。
只需使用ArrayList<Map<String, Object>>
并根据地图中的密钥访问该值。
Gson gson = new Gson();
Type listType = new TypeToken<ArrayList<Map<String, Object>>>() {
}.getType();
ArrayList<Map<String, Object>> list = gson.fromJson(builder.toString(), listType);
for (Map<String, Object> json : list) {
for (String key : json.keySet()) {
System.out.println(key + ":" + json.get(key));
}
System.out.println("===========");
}
输出:
item_name:Adame
item_type:Special
item:Chestplate
item_min_lvl:50
enchantment:{health=0.3, dam=24%, life=0.1, xp=24%, loot=22%}
def:73
===========
item_name:Sticks'
item_type:Unique
item:Stick
item_min_lvl:4
enchantment:{health=0.6, mana=1, dam=12%, life=0.3, xp=17%, loot=17%}
min_dam:39
max_dam:34
===========
答案 1 :(得分:0)
这实际上在Java和GSON中有效:
YourObject[] locs = gson.fromJson (someJsonString, YourObject[].class);
它将解析并返回YourObject
的数组。只需创建代表JSON对象的Java类,并根据需要替换占位符。
编辑: 正如Braj之前所说,你可以创建一个完全形成的POJO,包括其他(非对称)属性(我在这里借用Braj的答案代码):
//... snip ...
class MyJSON
{
String item_name;
String item_type;
String item;
String item_min_lvl;
Enchantment enchantment;
// Heres the other attributes
String min_dam;
String max_dam;
}
//... snip ...
如果原始JSON中没有提供值,GSON会解析它并将值设置为null
。
然而,从另一个问题来看,似乎JSON(Java - JSON Parser Error)的结界不一致,所以这会引起问题。我建议发送JSON作为一个数组以获得一致性,然后你可以将你的POJO结构化为:
//... snip ...
class MyJSON
{
String item_name;
String item_type;
String item;
String item_min_lvl;
Enchantment[] enchantment;
// Heres the other attributes
String min_dam;
String max_dam;
}
//... snip ...