android - Java后台把数据用json发到手机端，里面是map存放的，必须要说明数据的类型？

Question

               List<HashMap> reList = new ArrayList<HashMap>();
                      for(int i=0;i<bjdList.size();i++){
                          Zqwl_yd_jdxx bjd = bjdList.get(i);
                          HashMap temp = new HashMap();
                          temp.put("id",bjd.getId());
                          temp.put("yd_number",bjd.getYd_number());
                          temp.put("status",bjd.getStatus());
                          reList.add(temp);
                      }
                
                 str = JSONValue.toJSONString(reList);
                 response.getWriter().write(str);

我写后台的接口，ios的程序员说必须标注每一个值得类型（string int 。。。）Android的程序员就不用？转正json返回之后不都是字符串了吗？为什么还要标注出int类型？

Answer 1

Since it is an agreement, it must be agreed upon by both parties. It is best to have an interface document. Although json is just a string, the values ​​​​in it have different types. Android said that he does not need to mark it, maybe it is just that he thinks he can understand your json data. type (but not necessarily the data type that your backend originally wanted to give, for example, if you give it a float type, it uses double to get it. There is no difference for values ​​​​that do not require high precision), and iOS may want to be more clear and rigorous. , he wants to determine the exact value type. It must be unified at this time, otherwise you will not be able to explain clearly if there is a problem in the future.

Answer 2

json is just a string.
It’s not necessary. I didn’t have this kind of problem when connecting with ios programmers.

Answer 3

No need to mark ~

Answer 4

I think it’s right for you to mark it.
I have encountered problems with NSNumber and NSString that caused the program to crash

Answer 5

Actually, JSON has a type. The expanded "" is a string. "value":"1" and "value":1 are different.
Of course, IOS says it must be marked, but he doesn't want to Processing, I hope the background can handle it and use it directly, android just handles it for you.

It doesn’t matter who is right or wrong, but who is willing to do more.

Answer 6

The core of this problem lies in accurately understanding the definition of the JSON format
Control characters in JSON
{The beginning of the key-value pair container (A collection of name/value pairs)
} The end of the key-value pair container
[Array The beginning of the container
] The end of the array container
: key-value pair delimiter
" The beginning or end of the string
, the delimiter inside the container
The type of value in JSON
string The part between two double quotes, official website The original text is a collection of any number of Unicode characters surrounded by double quotes, using backslash escape. A character (character), that is, a separate string, is represented in OC as an NSString object
number with the character '0' or '9. ' at the beginning, the original text of the official website removes the unused octal and hexadecimal formats, and is represented in OC as an NSNumber object.
object object type, and in OC is represented as an ordered list of NSDictionary
values ​​(An ordered list of values ), commonly known as array, the performance in OC is NSArray
null null object, empty object, the performance in OC is NSNull
bool true/false Since there is no existence of Boolean objects in OC, the performance in OC is NSNumber

You can watch my podcast
http://cocoa1024.com/2016/03/31/stackjson/

Answer 7

Json is just a format. You can treat it as a string when receiving it, but when parsing, the value is typed

Answer 8

As for the oia document, if you don’t have oia, at least it must have json schema

Answer 9

No need, map can also be parsed directly

{"aaa,bbb";"aaa,bbb"}

   ArrayMap arrayMap = new ArrayMap();
      try {
           org.json.JSONObject obj = new org.json.JSONObject(content);
           Iterator<String> iterator = obj.keys();
           while (iterator.hasNext()) {
               String key = iterator.next();
               String value = (String) obj.get(key);
               arrayMap.put(key, value);
            }
        } catch (JSONException e) {
            e.printStackTrace();
      }