java json-simple无法解析有效的json字符串,但代码构建正常吗?

时间:2012-03-01 21:56:03

标签: java json json-simple

我正在尝试使用json-simple 1.1.1(http://code.google.com/p/json-simple/)解析json,并且它无法在OSX上运行。它在AWS实例上运行正常,但我想在本地运行我的代码来调试它。

我的代码在两个环境中构建都很好,而且我传入了有效的json:

{"name":"Penus","loc":"","timezone":"-8"}

如果我使用JSONValue.parse,我会得到一个空对象:

logger.info("Received data " + msg);
Object obj=JSONValue.parse(msg); // null

如果我使用

Object obj=parser.parse(msg); // blows up

我得到了这个例外:

INFO: Received data {"name":"Penus","loc":"","timezone":"-8"}
SEVERE: exception
java.lang.ClassCastException: org.json.simple.JSONArray cannot be cast to java.util.List
    at org.json.simple.parser.JSONParser.parse(JSONParser.java:220)
    at org.json.simple.parser.JSONParser.parse(JSONParser.java:81)
    at org.json.simple.parser.JSONParser.parse(JSONParser.java:75)
    at Reactor$Parser.parse(Reactor.java:621)
    at Reactor$Handler.process(Reactor.java:317)
    at Reactor$Handler.read(Reactor.java:362)
    at Reactor$Handler.run(Reactor.java:329)
    at Reactor.dispatch(Reactor.java:207)
    at Reactor.run(Reactor.java:86

0 个答案:

没有答案