Question

我需要执行大量的帐户关闭。可以在屏幕上一次显示200个帐户的列表，这些帐户可以作为下拉列表使用。仅6个帐户，其他人需要向下滚动。如何执行它使用Appium脚本。我试过的流程和我试过的示例源代码如下所示

package ******;

import java.util.List;
import io.appium.java_client.MobileBy;
import io.appium.java_client.android.AndroidElement;
import org.openqa.selenium.By;
import org.openqa.selenium.Dimension;
import org.openqa.selenium.WebElement;
import org.openqa.selenium.support.ui.ExpectedConditions;
import org.openqa.selenium.support.ui.WebDriverWait;
import org.testng.annotations.Test;


public class QuickPay extend ********{  



     @Test
     public void T1a_Landingpage() {

        public void  scroll(int i){
            do{
                size = driver.manage().window().getSize(); //Get the size of screen.
                System.out.println(size);                     
                int starty = (int) (size.height * 0.80);//Find starty point which is at bottom side of screen.
                int endy = (int) (size.height * 0.20);  //Find endy point which is at top side of screen.            
                int startx = size.width / 2; //Find horizontal point where you wants to swipe. It is in middle of screen width.     
                System.out.println("starty = " + starty + " ,endy = " + endy + " , startx = " + startx);     // int startx = size.width;
                driver.swipe(startx, starty, startx, endy, 3000);//Swipe from Bottom to Top.
                //Thread.sleep(2000);   
                }                                    

                 //AAD Closing starts
                     driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-1')]")).sendKeys("1111");
                     driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-tab-5')]")).click();                   

                     for(int i=1;i<10;i++){
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-18')]")).click();                    
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-input-5')]")).click();
                         do{
                             if (driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-simplelistitem-"+i+"')]")) != null){

                                 driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-simplelistitem-"+i+"')]")).click();
                                }
                             else{
                                    scroll(i);
                                }i++;
                             }
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-simplelistitem-"+i+"')]")).click();                         
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-input-6')]")).click();
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-simplelistitem-3')]")).click();                     
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-30')]")).click();                    
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-31')]")).click();                    
                                       driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-37') and @index='1']")).click();  
                         driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-37') and @index='1']")).click();
                                     driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-37') and @index='1']")).click();
                                    driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-37') and @index='1']")).click();
                                   driver.findElement(By.xpath("//android.view.View[contains(@resource-id,'ext-button-49')]")).click();

             }

Answer 1

如果您使用的是1.5以上的appium版本，则可以执行以下操作

Dimension size = driver.manage().window().getSize();
    logger.info("size of screen" + size);

    // Find swipe start and end point from screen's with and height.
    // Find startx point which is at right side of screen.
    int startx = (int) (size.width * 0.70);
    // Find endx point which is at left side of screen.
    int endx = (int) (size.width * 0.30);
    // Find vertical point where you wants to swipe. It is in middle of screen
    // height.
    int starty = size.height / 2;
    logger.info("Scrolling down with the coordinates " + "startx = " + startx + " ,endx = " + endx + " , starty = "
            + starty);

    // Scroll down
    driver.swipe(startx, starty, endx, starty, 3000);

注意：屏幕尺寸尺寸适用于iPad

否则如果使用1.6以下，则可以使用JavaScript滚动

Answer 2

通常我会从屏幕上的第一个元素位置滑动到屏幕上第6个元素的位置，直到屏幕上的第一个元素与上一个迭代相似，如下面的代码所示

public boolean removeGroupByName() {
    nameStart = "";
    while (true) {
        doYourLogicOnThe6Elements();
        if (!keepSwippingGroup()) {
            break;
        }
    }
    return true;
}

private static String nameStart = "";

private boolean keepSwippingList() {
    if (isElementPresent(By.xpath(FIRST_ELEMENT_XPATH)) && isElementPresent(By.xpath(LAST_ELEMENT_XPATH))) {
        AndroidElement firstElement = (AndroidElement) driver.findElementByXPath(FIRST_ELEMENT_XPATH);
        AndroidElement lastElement= (AndroidElement) driver.findElementByXPath(LAST_ELEMENT_XPATH);
        if (nameStart.equals(firstElement.getText())) {
            return false;
        } else {
            nameStart = firstElement.getText();
            swipe(lastElement.getLocation().x, lastElement.getLocation().y, firstElement.getLocation().x,
                    firstElement.getLocation().y, 2000);
        }
        return true;
    }
    return false;
}

如何滚动列表以使用appium执行应用程序的迭代流程？

2 个答案: