fastjson JSONPath example 2019-07-15 09:26
We can use fastjson to serialize and deserialize json. When the json structure is complicated, it will be more troublesome to handle. The json is like following.
{
"name": "henryxi",
"address": "beijing",
"company": {
"companyName": "xiami",
"companyAddress": "beijing"
}
}
If you want to get the company name or company address. You can deserialize json to JSONObject and then get the value. Another way to handle it is to use JSONPath.
public class GetPathValueClient {
public static void main(String[] args) {
String json = "{\n" +
" \"name\": \"henryxi\",\n" +
" \"address\": \"beijing\",\n" +
" \"company\": {\n" +
" \"companyName\": \"xiami\",\n" +
" \"companyAddress\": \"beijing\"\n" +
" }\n" +
"}";
JSONObject jsonObject = JSON.parseObject(json);
String companyAddress = jsonObject.getJSONObject("company").getString("companyAddress");
System.out.println("company address:" + companyAddress);
String companyName = (String) JSONPath.extract(json, "$.company.companyName");
System.out.println("company name:" + companyName);
}
}
The advantage of using JSONPath is that there will be no NullPointerException.
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