Home > Java > javaTutorial > How to Efficiently Iterate Through a JSON Array in Android/Java?

How to Efficiently Iterate Through a JSON Array in Android/Java?

Linda Hamilton
Release: 2024-12-09 02:50:10
Original
719 people have browsed it

How to Efficiently Iterate Through a JSON Array in Android/Java?

JSON Array Iteration in Android/Java

Accessing a JSON object from an Android app requires decoding the JSON array. To iterate through the array and extract the desired data, several approaches can be considered.

Looping through JSONArray

A straightforward method is to loop through the JSONArray using a for loop:

int id;
String name;
JSONArray array = new JSONArray(jsonString);
for (int i = 0; i < array.length(); i++) {
    JSONObject row = array.getJSONObject(i);
    id = row.getInt("id");
    name = row.getString("name");
}
Copy after login

Using a Library

Alternatively, utilizing a third-party library like Gson or Jackson simplifies the JSON parsing process. For instance, using Gson:

Gson gson = new Gson();
Type type = new TypeToken<List<JsonObject>>() {}.getType();
List<JsonObject> rows = gson.fromJson(jsonString, type);
Copy after login

Iterable Approach

Another option is to convert the JSONArray to an iterable using the JSONPointer class:

JSONArray array = new JSONArray(jsonString);
JSONPointer pointer = new JSONPointer("/0");
int id = (int) pointer.getValue(array);
String name = (String) pointer.getValue(array, "/1");
Copy after login

Legacy Solution

The code snippet mentioned in the question required an implicit conversion from the JSONArray to an iterable array. Unfortunately, this approach may encounter compatibility issues and is not recommended.

Optimization

For large datasets, consider using asynchronous tasks or background threads to avoid blocking the main UI. Additionally, caching the parsed data can enhance performance.

The above is the detailed content of How to Efficiently Iterate Through a JSON Array in Android/Java?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
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
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template