Iterating elements using NightWatchJS
1) used the following strategy to iterate over DOM elements using Nightwatch:
// Executing a function in the application context.
client.execute(function () { // Get elements by CSS selector.
var elements = document.querySelectorAll('.elements'); // Iterate over them.
[].forEach.call(elements, function (element) { // Manipulate each element.
element.click();
});
});
2) the number returned by allitems.value[0].ELEMENT will work as the id.
3) jQuery 定位
$('#padt_lic_numb > ul > li > label').each(function(){
$(this).click();
});
4) nightwatchjs
client.elements("css selector","#padt_lic_numb > ul > li > label",function(results)
{
results.value.forEach(function(index){
// console.log('Element value is '+index.ELEMENT);
client.elementIdText(index.ELEMENT,function(result){
//console.log(result.value);
if(result.value == plate_number)
{
console.log(result.value);
console.log(plate_number);
client.elementIdClick(index.ELEMENT);
}
});
})
});
Iterating elements using NightWatchJS的更多相关文章
- js Form.elements[i]的使用实例
function pdf(){ //一个html里面可能存在多个form,所以document.form[0]指的是第一个form,document.form[1]返回就是第二个form,如果没 ...
- View and Data API Tips: Hide elements in viewer completely
By Daniel Du With View and Data API, you can hide some elements in viewer by calling "viewer.hi ...
- [LeetCode] Minimum Moves to Equal Array Elements II 最少移动次数使数组元素相等之二
Given a non-empty integer array, find the minimum number of moves required to make all array element ...
- [LeetCode] Minimum Moves to Equal Array Elements 最少移动次数使数组元素相等
Given a non-empty integer array of size n, find the minimum number of moves required to make all arr ...
- [LeetCode] Top K Frequent Elements 前K个高频元素
Given a non-empty array of integers, return the k most frequent elements. For example,Given [1,1,1,2 ...
- [LeetCode] Remove Linked List Elements 移除链表元素
Remove all elements from a linked list of integers that have value val. Example Given: 1 --> 2 -- ...
- Chrome 开发工具之Elements
友情提示:全文图片高能,如使用手机阅读,请确保在wifi情况下或者流量充足.图片有点渣,也算辛苦做出来的,请别嫌弃- Elements面板主要展示当前页面的组织结构,在如今的应用程序中,HTML页面初 ...
- T-SQL Recipes之Separating elements
Separating elements Separating elements is a classic T-SQL challenge. It involves a table called Arr ...
- POJ2167Irrelevant Elements[唯一分解定理 组合数 杨辉三角]
Irrelevant Elements Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 2407 Accepted: 59 ...
随机推荐
- Appium+python自动化8-Appium Python API【转载】
前言: Appium Python API全集,不知道哪个大神整理的,这里贴出来分享给大家. 1.contexts contexts(self): Returns the contexts withi ...
- VirtualBox安装部署的Ubuntu16.04的步骤
1.下载ubuntu16.04镜像 http://cn.ubuntu.com/download/ 以及虚拟机软件VirtualBox https://www.virtualbox.org/wiki/D ...
- [BZOJ1072][SCOI2007]排列perm 状压dp
1072: [SCOI2007]排列perm Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 2488 Solved: 1546[Submit][St ...
- Mex函数与C++数据矩阵交互方法
#include "mex.h" #include <iostream> #include <windows.h> #include "math. ...
- DB Link
oracle中DB Link select * from TB_APP_HEADER@SSDPPORTAL
- Ubuntu 16.04/CentOS 6.9安装Apache压力(并发)测试工具ab
说明: ab工具已经在Apache中包含,如果不想安装Apache,那么可以使用下面方法单独安装. 安装: Ubuntu: sudo apt-get install apache2-utils Cen ...
- 教程:基于Spring快速开发电子邮件发送功能
在Spring框架的spring-context-support.jar中有对电子邮件发送功能的封装: 基于Spring开发简单省事,而且更稳定.需要mail.jar包支持 @Component pu ...
- Delphi CRC16校验算法实现(转)
循环冗余码校验英文名称为Cyclical Redundancy Check,简称CRC.它是利用除法及余数的原理来作错误侦测(Error Detecting)的.实际应用时,发送装置计算出CRC值并随 ...
- Understanding Memory Technology Devices in Embedded Linux
转: NAND Chip Drivers NAND technology users such as USB pen drives, DOMs, Compact Flash memory, and S ...
- vertex buffer 数据结构 如何读vb的memory pool
vertex attribute (declaration) vertex stream (memory pool) 这两部分 通过attribute 里面对memory的描述把两部分 vbo ...