Question

我正在尝试解析以下JSON。但是获得例外请帮助。

{
  "method": "demo",
  "clazz": "storage",
  "params": [
    "",
    "LOGIN",
    "{"auth": {"tenantName": "AUTH_Tonny", "casauth": {"username": "Tonny", "tgt": "TGT-1876hkahaadcaweyfiowufssadsfsdf"}}}",
    "http://ipstorage.google.com"
  ]
}

这是Java代码：

        String tokenId = "";
        String message = "LOGIN";
        String url1 = "http://ipstorage.google.com";
        String authentication = "{\"auth\": {\"tenantName\": \"AUTH_" + test2.getUsername()
        + "\", \"casauth\": {\"username\": \""
        + test2.getUsername() + "\", \"tgt\": \""
        + test2.getPassword() + "\"}}}";
        String pp = "[\"" + tokenId + "\",\"" + message + "\",\""
                  + authentication + "\",\"" + url1 + "\"]";
        String msg1 = "{\"method\":\"demo\",\"clazz\":\"storage\",\"params\":" + pp + "}";
        System.out.println(msg1);
        JSONObject jo = (JSONObject) new JSONParser().parse(msg1);
        System.out.println("##");
        System.out.println(jo);

我得到的输出和例外是：

    {"method":"demo","clazz":"storage","params":["","LOGIN","{"auth":
   {"tenantName": "AUTH_Tonny", "casauth": {"username": "Tonny", "tgt": "TGT - 
   1876 hkahaadcaweyfiowufssadsfsdf"}}}","http://ipstorage.google.com"]}
   Exception in thread "main" Unexpected character (a) at position 59.
    at org.json.simple.parser.Yylex.yylex(Unknown Source)
    at org.json.simple.parser.JSONParser.nextToken(Unknown Source)
    at org.json.simple.parser.JSONParser.parse(Unknown Source)
    at org.json.simple.parser.JSONParser.parse(Unknown Source)
    at org.json.simple.parser.JSONParser.parse(Unknown Source)
    at com.t.g.i.e.utils.Test.main(Test.java:74)

请帮帮我。提前谢谢。

Answer 1

您可以使用http://www.jsoneditoronline.org/来验证您的json文件。

{
"method": "demo",
"clazz": "storage",
"params": [
    "",
    "LOGIN",
    {
        "auth": {
            "tenantName": "AUTH_Tonny",
            "casauth": {
                "username": "Tonny",
                "tgt": "TGT-1876hkahaadcaweyfiowufssadsfsdf"
            }
        }
    },
    "http://ipstorage.google.com"
]
}

Answer 2

您的JSON格式不对。也许你想要这样的JSON，

{
  "method": "demo",
  "clazz": "storage",
  "params": [
    "",
    "LOGIN",
    {"auth": {"tenantName": "AUTH_Tonny", "casauth": {"username": "Tonny", "tgt": "TGT-1876hkahaadcaweyfiowufssadsfsdf"}}},
    "http://ipstorage.google.com"
  ]
}

没有“JSONObject之前和之后。您可以使用此工具检查您的JSON格式。

https://jsonformatter.curiousconcept.com/

你的java代码应该是这样的，

...
String pp = "[\"" + tokenId + "\",\"" + message + "\","
                  + authentication + ",\"" + url1 + "\"]";
...

删除您的JSON。

Answer 3

使用Json验证器验证字符串并使用像Gson或Jackson这样的json解析器库，它可以自动完成你的很多工作而且你不需要编写很多代码来解析

JSON Java - 位置59处的意外字符（a）

3 个答案: