Home  >  Article  >  Java  >  Java method to determine whether it is json

Java method to determine whether it is json

尚
Original
2019-11-30 10:09:464542browse

Java method to determine whether it is json

Java method to determine whether it is in json format: (Recommended: java video tutorial)

json format: {"name":"Michael" ,"age":24,"birthday":"2018-09-09"};

private boolean isjson(String str){
    try {
        JSONObject jsonStr= JSONObject.parseObject(string);
        return  true;
    } catch (Exception e) {
        return false;
    }
}

JSONObject.parseObject(String str) is a method in fastjson.

fastjson is a high-performance JSON parser and generator implemented in Java language, developed by engineers from Alibaba. Its main features are:

① Fast: fastjson uses an original algorithm to increase the speed of parse to the extreme, surpassing all Java-based json libraries, including jackson, which was once known as the fastest;

② Powerful: Fastjson fully supports the standard of http://json.org (also one of the reference implementations included on Google’s official website); supports various JDK types; including basic types, JavaBean, Collection, Map, Enum, generics, etc.;

③Zero dependency: Does not depend on any other class library except JDK, can run directly on Java SE 5.0 or above; supports Android; open source (Apache 2.0).

For more java knowledge, please pay attention to the java basic tutorial column.

The above is the detailed content of Java method to determine whether it is json. For more information, please follow other related articles on the PHP Chinese website!

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn