json - 我要用java解析一段字符串，该怎么做？

Question

字符串比如是这样 {代码...} 是不是要把这个字符串先变成json？我最终要解析出里面BTC多少LTC多少等 具体代码如何写，求指教。

高洛峰 · Answer

It’s nothing more than nesting json one by one. Just follow the json format from simple to complex and figure it out one by one.

Reuse the third-party jar package for analysis. .
The ones I often use include gson, fastjson, etc. It is recommended to use fastjson. I feel that the efficiency is good.

PHPz · Answer

https://github.com/alibaba/fastjson

wiki : https://github.com/Alibaba/fastjson/wiki/%E9%A6%96%E9%A1%B5

大家讲道理 · Answer

You can also use the gson package provided by Google, or Alibaba's fastjson. You can write a bean based on what you want to parse, and then use the above package to parse it

PHPz · Answer

Look at this http://json.org/
The class libraries available for java are:

org.json.
org.json.me.
Jackson JSON Processor.
Json-lib.
JSON Tools.
Stringtree.
SOJO.
Jettison.
json-taglib.
XStream.
Flexjson.
JON tools.
Argo.
jsonij.
fastjson.
mjson.
jjson.
json-simple.
json-io.
JsonMarshaller.
google-gson.
Json-smart.
FOSS Nova JSON.
Corn CONVERTER.

怪我咯 · Answer

The native code may look like this;

String jsonString = '{"info":{"funds":{"free":{"btc":"1.042309","cny":"510.021","ltc":"0.01468"},"freezed":{"btc":"0","cny":"7490.319","ltc":"0"}}},"result":true}';  //当然咯不能使用单引号“'”

JSONObject json = new JSONObject(jsonString);
JSONObject info= json.getJSONObject("info");
JSONObject funds = info.getJSONObect("funds");
JSONObject free = funds.getJSONObject("free");
int btc = funds.getInt("btc");

Hmm, it’s so annoying! ! !

json - 我要用java解析一段字符串，该怎么做？

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