我正在开发一个非常简单的网站应用程序,只是一个基本的桌面应用程序。
所以我想出了如何获取我需要的所有JSON数据,如果可能的话,我试图避免使用外部库来解析JSON。
这就是我现在正在做的事情:
package me.thegreengamerhd.TTVPortable;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLConnection;
import me.thegreengamerhd.TTVPortable.Utils.Messenger;
public class Channel
{
URL url;
String data;
String[] dataArray;
String name;
boolean online;
int viewers;
int followers;
public Channel(String name)
{
this.name = name;
}
public void update() throws IOException
{
// grab all of the JSON data from selected channel, if channel exists
try
{
url = new URL("https://api.twitch.tv/kraken/channels/" + name);
URLConnection connection = url.openConnection();
BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
data = new String(in.readLine());
in.close();
// clean up data a little, into an array
dataArray = data.split(",");
}
// channel does not exist, throw exception and close client
catch (Exception e)
{
Messenger.sendErrorMessage("The channel you have specified is invalid or corrupted.", true);
e.printStackTrace();
return;
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dataArray.length; i++)
{
sb.append(dataArray[i] + "\n");
}
System.out.println(sb.toString());
}
}
所以这是我输入示例频道(正确抓取数据)时打印的内容
{"updated_at":"2013-05-24T11:00:26Z"
"created_at":"2011-06-28T07:50:25Z"
"status":"HD [XBOX] Call of Duty Black Ops 2 OPEN LOBBY"
"url":"http://www.twitch.tv/zetaspartan21"
"_id":23170407
"game":"Call of Duty: Black Ops II"
"logo":"http://static-cdn.jtvnw.net/jtv_user_pictures/zetaspartan21-profile_image-121d2cb317e8a91c-300x300.jpeg"
"banner":"http://static-cdn.jtvnw.net/jtv_user_pictures/zetaspartan21-channel_header_image-7c894f59f77ae0c1-640x125.png"
"_links":{"subscriptions":"https://api.twitch.tv/kraken/channels/zetaspartan21/subscriptions"
"editors":"https://api.twitch.tv/kraken/channels/zetaspartan21/editors"
"commercial":"https://api.twitch.tv/kraken/channels/zetaspartan21/commercial"
"teams":"https://api.twitch.tv/kraken/channels/zetaspartan21/teams"
"features":"https://api.twitch.tv/kraken/channels/zetaspartan21/features"
"videos":"https://api.twitch.tv/kraken/channels/zetaspartan21/videos"
"self":"https://api.twitch.tv/kraken/channels/zetaspartan21"
"follows":"https://api.twitch.tv/kraken/channels/zetaspartan21/follows"
"chat":"https://api.twitch.tv/kraken/chat/zetaspartan21"
"stream_key":"https://api.twitch.tv/kraken/channels/zetaspartan21/stream_key"}
"name":"zetaspartan21"
"delay":0
"display_name":"ZetaSpartan21"
"video_banner":"http://static-cdn.jtvnw.net/jtv_user_pictures/zetaspartan21-channel_offline_image-b20322d22543539a-640x360.jpeg"
"background":"http://static-cdn.jtvnw.net/jtv_user_pictures/zetaspartan21-channel_background_image-587bde3d4f90b293.jpeg"
"mature":true}
初始化用户界面 - 加入
所有这一切都是正确的。现在我想做的是,能够抓住,例如'成熟'标签,它的价值。因此,当我抓住它时,它就像:
// pseudo code
if(mature /*this is a boolean */ == true){ // do stuff}
因此,如果您不理解,我需要拆分值之间的引号和分号以检索Key,Value。
答案 0 :(得分:1)
可以使用以下代码:
public static Map<String, Object> parseJSON (String data) throws ParseException {
if (data==null)
return null;
final Map<String, Object> ret = new HashMap<String, Object>();
data = data.trim();
if (!data.startsWith("{") || !data.endsWith("}"))
throw new ParseException("Missing '{' or '}'.", 0);
data = data.substring(1, data.length()-1);
final String [] lines = data.split("[\r\n]");
for (int i=0; i<lines.length; i++) {
String line = lines[i];
if (line.isEmpty())
continue;
line = line.trim();
if (line.indexOf(":")<0)
throw new ParseException("Missing ':'.", 0);
String key = line.substring(0, line.indexOf(":"));
String value = line.substring(line.indexOf(":")+1);
if (key.startsWith("\"") && key.endsWith("\"") && key.length()>2)
key = key.substring(1, key.length()-1);
if (value.startsWith("{"))
while (i+1<line.length() && !value.endsWith("}"))
value = value + "\n" + lines[++i].trim();
if (value.startsWith("\"") && value.endsWith("\"") && value.length()>2)
value = value.substring(1, value.length()-1);
Object mapValue = value;
if (value.startsWith("{") && value.endsWith("}"))
mapValue = parseJSON(value);
else if (value.equalsIgnoreCase("true") || value.equalsIgnoreCase("false"))
mapValue = new Boolean (value);
else {
try {
mapValue = Integer.parseInt(value);
} catch (NumberFormatException nfe) {
try {
mapValue = Long.parseLong(value);
} catch (NumberFormatException nfe2) {}
}
}
ret.put(key, mapValue);
}
return ret;
}
您可以这样称呼它:
try {
Map<String, Object> ret = parseJSON(sb.toString());
if(((Boolean)ret.get("mature")) == true){
System.out.println("mature is true !");
}
} catch (ParseException e) {
}
但是,实际上,你不应该这样做,并且使用已经存在的JSON解析器,因为这段代码会破坏任何复杂或无效的JSON数据(比如&#34; :&#34;在键中),如果你想手工构建一个真正的JSON解析器,它将需要你更多的代码和调试!
答案 1 :(得分:0)
这是一个简单的json字符串的解析器:
public static HashMap<String, String> parseEasyJson(String json) {
final String regex = "([^{}: ]*?):(\\{.*?\\}|\".*?\"|[^:{}\" ]*)";
json = json.replaceAll("\n", "");
Matcher m = Pattern.compile(regex).matcher(json);
HashMap<String, String> map = new HashMap<>();
while (m.find())
map.put(m.group(1), m.group(2));
return map;
}