. How to create dynamic grids in flutter app Skip to main content

How to create dynamic grids in flutter app

Grid is a layout design system to show items in a list.
Today we will learn that how can we create grid in any flutter app.
So Grid is just like a box that can be same size or different.
Forgot the static grid, Now we will create dynamic grid in flutter app -
Let get the example -
import 'package:flutter/material.dart';

void main() => runApp(MyApp());

class MyApp extends StatelessWidget {
  // This widget is the root of your application.
  @override
  Widget build(BuildContext context) {
    return MaterialApp(
      title: 'Flutter Grid Demo',
      theme: ThemeData(
        primarySwatch: Colors.blue,
      ),
      home: MyHomePage(title: 'FLutter Grid'),
    );
  }
}

class MyHomePage extends StatefulWidget {
  MyHomePage({Key key, this.title}) : super(key: key);
  final String title;

  @override
  _MyHomePageState createState() => _MyHomePageState();
}

class _MyHomePageState extends State {

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text(widget.title),
      ),
      body: new GridView.count(
        crossAxisCount: 4,
        children: new List.generate(16, (index) {
          return new GridTile(
            child: new Card(
                color: Colors.teal.shade200,
                child: new Center(
                  child: new Text('Tile $index'),
                )
            ),
          );
        }),
      ),
    );
  }
} 
 
 
 
 
  Example will look like this. In this we don't need to write the code for each and every tile. In this all titles will automatically generate.
I have already write for change the color of tile, today I'll tell you about  change the no of tiles in a row.
In above example 4 tiles are in a row. Suppose we only want three grid in in a row then !!!

So it's very easy in flutter, just change crossAxisCount: 4, to the crossAxisCount: 3 and you will get 3 tiles in a row. You can also change it to at your requirement.





Comments

Popular posts from this blog

Day 9 - Challenge 1 - Product of Array Except Self

Solving the "Product of Array Except Self" Problem in JavaScript Are you ready to dive into a common coding challenge that not only tests your programming skills but also sharpens your problem-solving mindset? If you're up for the challenge, let's tackle the "Product of Array Except Self" problem together using JavaScript. This problem requires us to return an array where each element at index i is the product of all the elements in the original array except the one at index i . Understanding the Problem:   Imagine you're given an array of integers, let's call it nums . Your task is to create a new array where the value at index i in this new array is the product of all the elements in nums , except the one at index i . In other words, you're calculating the product of all the elements to the left of nums[i] and the product of all the elements to the right of nums[i] , and then multiplying these two products to get the final value at index i ...

Day 9 - Challenge 2 - Reverse Linked List

Reversing a Singly Linked List in JavaScript: An In-Place Approach Introduction:   Singly linked lists are fundamental data structures in computer science that consist of a sequence of nodes, each containing data and a reference to the next node in the list. Reversing a singly linked list is a classic problem that challenges programmers to manipulate pointers effectively to achieve the desired outcome. In this blog post, we'll explore the problem of reversing a singly linked list using an in-place approach and provide a step-by-step solution in JavaScript. Problem Statement:   Given the head of a singly linked list, our task is to reverse the list in-place and return its new head. In other words, we need to modify the pointers of the nodes in such a way that the direction of the linked list is reversed. Solution Approach:   To solve this problem, we will iterate through the linked list while maintaining three pointers: previous , current , and next . The previous pointer...

Day 8 - Challenge 2 - Move Zeroes to the End

Moving Zeroes to the End: A JavaScript Solution Introduction:   When working with arrays, there are often times when we need to manipulate their elements to achieve a specific goal. One common problem is moving all zeroes to the end of an array while keeping the order of non-zero elements unchanged. In this blog post, we will explore an elegant solution to this problem using JavaScript. We'll discuss the problem statement, the approach we'll take, and provide a step-by-step guide to implementing the solution. The Problem:   Given an array of integers, the task is to move all zeroes to the end of the array while maintaining the relative order of the non-zero elements. This means that after rearranging the array, all the zeroes should be at the end, and the order of the non-zero elements should remain the same. The Approach:   To solve this problem, we can utilize a two-pointer approach. We'll maintain two pointers, one for iterating through the array and another for keepin...