Appium checking if element is displayed
up vote
2
down vote
favorite
I'm using Appium for Android
the following works for clicking on the element
driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).click();
But I am trying to check if an element is on the screen and I tried the following
if (driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed()) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
but it returns an Exception saying
INFO: HTTP Status: '405' -> incorrect JSON status mapping for 'unknown error' (500 expected)
org.openqa.selenium.WebDriverException: Method is not implemented
How can I check to see if an element is on the screen?
appium
asked Nov 7 at 2:23
david
18610
add a comment |
up vote
2
down vote
favorite
I'm using Appium for Android
the following works for clicking on the element
driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).click();
But I am trying to check if an element is on the screen and I tried the following
if (driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed()) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
but it returns an Exception saying
INFO: HTTP Status: '405' -> incorrect JSON status mapping for 'unknown error' (500 expected)
org.openqa.selenium.WebDriverException: Method is not implemented
How can I check to see if an element is on the screen?
appium
asked Nov 7 at 2:23
david
18610
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm using Appium for Android
the following works for clicking on the element
driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).click();
But I am trying to check if an element is on the screen and I tried the following
if (driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed()) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
but it returns an Exception saying
INFO: HTTP Status: '405' -> incorrect JSON status mapping for 'unknown error' (500 expected)
org.openqa.selenium.WebDriverException: Method is not implemented
How can I check to see if an element is on the screen?
appium
asked Nov 7 at 2:23
david
18610
I'm using Appium for Android
the following works for clicking on the element
driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).click();
But I am trying to check if an element is on the screen and I tried the following
if (driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed()) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
but it returns an Exception saying
INFO: HTTP Status: '405' -> incorrect JSON status mapping for 'unknown error' (500 expected)
org.openqa.selenium.WebDriverException: Method is not implemented
How can I check to see if an element is on the screen?
appium
appium
asked Nov 7 at 2:23
david
18610
asked Nov 7 at 2:23
david
18610
asked Nov 7 at 2:23
david
18610
asked Nov 7 at 2:23
david
18610
asked Nov 7 at 2:23
david
18610
18610
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You can try this, hope it helps
//If the element found, do as you want
if (driver.findElements(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).size() > 0) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
answered Nov 8 at 5:01
Al Imran
531415
add a comment |
up vote
0
down vote
you can surround your code with try catch block.
public boolean isElementDisplayed(){
try{
return driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
you can also make the generic function to check if element is displayed.
public boolean isElementDisplayed(MobileElement element){
try{
return element.isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
answered Nov 7 at 13:20
Suban Dhyako
549115
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can try this, hope it helps
//If the element found, do as you want
if (driver.findElements(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).size() > 0) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
answered Nov 8 at 5:01
Al Imran
531415
add a comment |
up vote
1
down vote
accepted
You can try this, hope it helps
//If the element found, do as you want
if (driver.findElements(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).size() > 0) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
answered Nov 8 at 5:01
Al Imran
531415
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can try this, hope it helps
//If the element found, do as you want
if (driver.findElements(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).size() > 0) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
answered Nov 8 at 5:01
Al Imran
531415
You can try this, hope it helps
//If the element found, do as you want
if (driver.findElements(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).size() > 0) {
System.out.println("FOUND");
} else {
System.out.println("NOT FOUND!");
}
answered Nov 8 at 5:01
Al Imran
531415
edited Nov 9 at 3:59
answered Nov 8 at 5:01
Al Imran
531415
answered Nov 8 at 5:01
Al Imran
531415
answered Nov 8 at 5:01
Al Imran
531415
531415
add a comment |
add a comment |
up vote
0
down vote
you can surround your code with try catch block.
public boolean isElementDisplayed(){
try{
return driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
you can also make the generic function to check if element is displayed.
public boolean isElementDisplayed(MobileElement element){
try{
return element.isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
answered Nov 7 at 13:20
Suban Dhyako
549115
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
add a comment |
up vote
0
down vote
you can surround your code with try catch block.
public boolean isElementDisplayed(){
try{
return driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
you can also make the generic function to check if element is displayed.
public boolean isElementDisplayed(MobileElement element){
try{
return element.isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
answered Nov 7 at 13:20
Suban Dhyako
549115
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
add a comment |
up vote
0
down vote
up vote
0
down vote
you can surround your code with try catch block.
public boolean isElementDisplayed(){
try{
return driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
you can also make the generic function to check if element is displayed.
public boolean isElementDisplayed(MobileElement element){
try{
return element.isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
answered Nov 7 at 13:20
Suban Dhyako
549115
you can surround your code with try catch block.
public boolean isElementDisplayed(){
try{
return driver.findElement(By.xpath("//*[@resource-id='com.app.android:id/prelogin_signup']")).isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
you can also make the generic function to check if element is displayed.
public boolean isElementDisplayed(MobileElement element){
try{
return element.isDisplayed();
}
}catch(Exception e){
//System.out.println(e);
return false;
}
}
answered Nov 7 at 13:20
Suban Dhyako
549115
edited Nov 12 at 6:07
answered Nov 7 at 13:20
Suban Dhyako
549115
answered Nov 7 at 13:20
Suban Dhyako
549115
answered Nov 7 at 13:20
Suban Dhyako
549115
549115
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
add a comment |
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
You should use the original poster's example locator in your example. You should also consider that while there should be a try/catch block, you're not checking for specific exceptions, so any exception might be misleading and cause debugging problems under certain circumstances, especially since your example doesn't show any exception handling at all.
– Bill Hileman
Nov 7 at 20:52
add a comment |
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