Spring History

The first version of Spring was written by Rod Johnson, who released the framework with the publication of his book Expert One-on-One J2EE Design and Development in October 2002. The framework was first released under the Apache 2.0 license in June 2003.

As of Spring Framework 5.1, Spring requires JDK 8+ (Java SE 8+) and provides out-of-the-box support for JDK 11 LTS. Spring is open source. It has a large and active community that provides continuous feedback based on a diverse range of real-world use cases. This has helped Spring to successfully evolve over a very long time.

Spring continues to innovate and to evolve. Beyond the Spring Framework, there are other projects, such as Spring Boot, Spring Security, Spring Data, Spring Cloud, Spring Batch, among others. It’s important to remember that each project has its own source code repository, issue tracker, and release cadence. See spring.io/projects for the complete list of Spring projects.



Below are the version history of Spring 

  • The framework was first released in June 2003 under the Apache license version 2.0. 

  • The first milestone release of Spring framework (1.0) was released in March 2004. 

  • Spring 2.0, which came in 2006, simplified the XML config files.
  • Spring 2.5, which came in 2007, introduced annotation configurations.

  • Spring 3.2, which came in 2012, introduced Java configuration, had support for Java 7, Hibernate 4, Servlet 3.0, and also required a minimum of Java 1.5. 

  • Spring 4.0, which came in 2014, had support for Java 8. 

  • Spring Boot also was introduced in 2014.

  • Spring 5.0 came out in 2017. Spring Boot 2.x has support for Spring 5.


    Next -> Spring Architecture


    Final Overview of Spring Versions

    Developer(s)Pivotal Software
    Initial release1 October 2002; 17 years ago
    Stable release
    5.2.5.RELEASE / March 24, 2020; 
    Preview release
    5.2.0-RC1 / August 5, 2019;
    RepositorySpring Repository
    Written inJava
    PlatformJava EE
    TypeApplication framework
    LicenseApache License 2.0
    Websitespring.io

The intention behind this blog is to help all my fellow programmers. Also to provide solutions for various coding platform problems with solution and explanation. I will add many tutorials in future. I am not a professional blogger but whenever time permits I will post. I hope you’ll enjoy the content and find it useful! Check out my Writer blog for a more valuable quotes. For any queries or suggestions, please feel free to reach out to me. Happy coding!

Comments

Post a Comment

Popular posts from this blog

First Unique Character in a String

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase letters. class Solution {     public int firstUniqChar(String s) {         int length = s.length();         String str;         for(int i=0;i<length;i++){             if(i==length-1){                 str = s.substring(0, i);             }             else{                 str = s.substring(0,i)+s.sub...
Read more

Balanced Binary Tree

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of  every  node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true   Constraints: The number of nodes in the tree is in the range  [0, 5000] . -10 4  <= Node.val <= 10 4 Solution: /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int ...
Read more

Majority Element

Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int ...
Read more

Sequential Digits

An integer has sequential digits if and only if each digit in the number is one more than the previous digit. Return a sorted list of all the integers in the range [low, high]  inclusive that have sequential digits. Example 1: Input: low = 100, high = 300 Output: [123,234] Example 2: Input: low = 1000, high = 13000 Output: [1234,2345,3456,4567,5678,6789,12345] Constraints: 10 <= low <= high <= 10^9 class Solution {     public List<Integer> sequentialDigits(int low, int high) {         List<Integer> result = new ArrayList();         String str = "123456789";         int i, j, length = str.length(), num;         for( i = 2; i <= length; i++) {             for( j = 0; j <= length - i; j++) {      ...
Read more