简体   繁体   English

我如何像这样解析此JSONArray / JSONObject?

[英]How can I parse this JSONArray/JSONObject like this?

I have JSON data like this: 我有这样的JSON数据:

  "books": [
        {
            "rating": {}, 
            "subtitle": "", 
            "author": [
                "The OReilly Java Authors"
            ], 
            "pubdate": "2003-9", 
            "tags": [], 
            "origin_title": "",
               …… …… …… …… … … 

and I want get the author data,some "author"data has more than one author,but I just confused weather I could parse it like this: 并且我想获取作者数据,一些“作者”数据有多个作者,但是我很困惑,我可以这样解析它:

JSONArray jsonbooks = mess.getJSONArray("books");
for(blabla){
   JSONObject obj = jsonbooks.getJSONObject(i);
   obj.getJSONArray("author").get(1):

or 要么

  JSONArray jsonbooks = mess.getJSONArray("books");
    for(blabla){
     JSONObject obj = jsonbooks.getJSONObject(i); 
     obj.getString("author");

sencond way,I could get the data like"["author,blabla "],I want get rid off [" "],and I need more string processing,if there a better way looks like the first way? 第二种方法,我可以得到像“ [” author,blabla“]这样的数据,我想摆脱[”“],并且我需要更多的字符串处理,如果有更好的方法看起来像第一种方法呢?

Since author is an array therefore first convert that into jsonArray 由于author是一个数组,因此首先将其转换为jsonArray

JSONArray jsonAuthor = mess.getJSONArray("author");

then String author = jsonAuthor.get(0);  //this will give you -> The OReilly Java Authors

First one is right just one change obj.getJSONArray("author").getString(0): 第一个是正确的,只是一个更改obj.getJSONArray("author").getString(0):

JSONArray jsonbooks = mess.getJSONArray("books");
for(blabla){
   JSONObject obj = jsonbooks.getJSONObject(i);
   obj.getJSONArray("author").getString(0):

Hope this helps: 希望这可以帮助:

                JSONArray jsonAuthorNode = booksJsonObject.optJSONArray("author");
                int jsonArrLen = jsonAuthorNode.length();

                for(int j=0;j<jsonArrLen;j++)
                {
                    String authorName = jsonAuthorNode.get(j);
                                    }

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM