pull down/pull up refresh & UI Components
pull down/pull up refresh & UI Components
下拉/上拉刷新
https://mint-ui.github.io/docs/#/zh-cn2/loadmore
import { Loadmore } from 'mint-ui';
Vue.component(Loadmore.name, Loadmore);
<mt-loadmore :top-method="loadTop" :bottom-method="loadBottom" :bottom-all-loaded="allLoaded" ref="loadmore">
<ul>
<li v-for="item in list">{{ item }}</li>
</ul>
</mt-loadmore>
pull down/pull up refresh js
https://github.com/owenliang/pullToRefresh
https://github.com/owenliang/pullToRefresh/blob/master/pullToRefresh.js
https://github.com/owenliang/pullToRefresh/blob/master/iscroll.js
https://github.com/dwcares/pulltorefresh
http://dwcares.com/2013/10/06/pull-to-refresh-2/
pulltorefresh.js
https://www.boxfactura.com/pulltorefresh.js/
https://www.boxfactura.com/pulltorefresh.js/demos/basic.html
https://www.cssscript.com/tag/pull-to-refresh/
https://stackoverflow.com/questions/46190436/how-to-detect-pull-to-refresh
https://stackoverflow.com/questions/37005417/pull-up-to-refresh-in-android-recyclerview
https://github.com/WyrdNexus/js-minimal-android-swipe-detect
https://makitweb.com/pull-down-to-refresh-with-pulltorefresh-js/
https://www.npmjs.com/package/mobile-pull-to-refresh
https://www.cnblogs.com/xgqfrms/p/11071017.html
AB testing
https://github.com/jiangfengming/pull-to-refresh/blob/master/src/pullToRefresh.js
pull down/pull up refresh & UI Components的更多相关文章
- Angular 2 to Angular 4 with Angular Material UI Components
Download Source - 955.2 KB Content Part 1: Angular2 Setup in Visual Studio 2017, Basic CRUD applicat ...
- vue ui components
vue ui components h_ui https://www.npmjs.com/~hs_ui https://www.npmjs.com/package/h_ui_beta https:// ...
- Magento2 UI components概述
UI components 概述Magento UI components 是用来展示不同的UI元素,比如表,按钮,对话框等.他们被用于简单灵活的交互界面渲染.Components被用来渲染结果界面, ...
- how to share UI components
how to share UI components The shared component cloud · Bit https://bit.dev/ A better way to build w ...
- Vue & mobile UI components
Vue & mobile UI components https://github.com/vuejs/awesome-vue https://awesome-vue.js.org/ http ...
- Flutter & APP & UI Components
Flutter & APP & UI Components 下拉刷新或者上拉加载 https://github.com/OpenFlutter/flutter_screenutil h ...
- Chrome 的 Material Design Refresh UI初探
今天Chrome自动升级到69.0.3497.92, 发现UI已经变成了"Material Design Refresh". Chrome 浏览器的页面标签已经不再像以往那样倾斜和 ...
- Push pull, open drain circuit, pull up, pull down resistor
Push pull 就以下面這個 電路來說, 因為沒有 pull up resistor, 所以 output voltage 由 low 往 high 的速度會較快. 有兩個電晶體,一個on,一個 ...
- Force git to overwrite local files on pull 使用pull强制覆盖本地文件 转载自:http://snowdream.blog.51cto.com/3027865/1102441
How do I force an overwrite of local files on a git pull? I think this is the right way: $ git fetch ...
随机推荐
- Why should I avoid blocking the Event Loop and the Worker Pool?
Don't Block the Event Loop (or the Worker Pool) | Node.js https://nodejs.org/en/docs/guides/dont-blo ...
- Centos虚拟机上安装部署Tenginx,以及部署过程中遇到的问题
Tenginx下载网址: Tenginx 官网地址:http://tengine.taobao.org/ Tenginx的官方网址中可以阅读Nginx的文档,可以选择中文进行阅读.下载Tengine- ...
- loj10170
在 n×n 的棋盘上放 k 个国王,国王可攻击相邻的 8 个格子,求使它们无法互相攻击的方案总数. -------------------------------------------------- ...
- Codeforces 1220D 思维 数学 二分图基础
原题链接 题意 我们有一个含多个正整数的集合B,然后我们将所有的整数,也就是Z集合内所有元素,都当做顶点 两个整数 \(i , j\) 能建立无向边,当且仅当 \(|i - j|\) 这个数属于B集合 ...
- Vue3.0网页版聊天|Vue3.x+ElementPlus仿微信/QQ界面|vue3聊天实例
一.项目简介 基于vue3.x+vuex+vue-router+element-plus+v3layer+v3scroll等技术构建的仿微信web桌面端聊天实战项目Vue3-Webchat.基本上实现 ...
- ubuntu 安装新版的qq,可支持下载文件等常用功能
说明:此版本的QQ基本完美,但是有个缺点就是历史记录有些会显示乱码! 注意:此方法能完美解决这篇文章http://www.cnblogs.com/EasonJim/p/7118693.html的所有问 ...
- 动态规划TG.lv(1) (洛谷提高历练地)
动态规划TG.lv(1) P1005 矩阵取数游戏 分析:每行不超过80个数字,直接区间DP即可,\(dp[i][j]\)表示区间\([i,j]\)之间取数可以得到的答案,每次向右或者向左扩展即可.但 ...
- 2019 徐州网络赛 M Longest subsequence t
对于答案来说,一定是 前 i-1 个字符和 t的前 i 个一样,然后第 i 个字符比 t的 大 \(i\in [1,m]\) 前缀为t,然后长度比t长 对于第一种情况,枚举这个 i ,然后找最小的 p ...
- Codeforces Round #647 (Div. 2) D. Johnny and Contribution(BFS)
题目链接:https://codeforces.com/contest/1362/problem/D 题意 有一个 $n$ 点 $m$ 边的图,每个结点有一个从 $1 \sim n$ 的指定数字,每个 ...
- 【noi 2.6_6045】开餐馆(DP)
题意:有N个地址,从中选一些开餐馆,要保证相邻餐馆的距离大于k.问最大利润. 解法:f[i]表示在前 i 个地址中选的最大利润. 1 #include<cstdio> 2 #include ...