Espresso: return boolean if view exists

问题:

I am trying to check to see if a view is displayed with Espresso. Here is some pseudo code to show what I am trying:

if (!Espresso.onView(withId(R.id.someID)).check(doesNotExist()){
   // then do something
 } else {
   // do nothing, or what have you
 }

But my problem is .check(doesNotExist()) does not return boolean. It is just an assertion. With UiAutomator I was able to just do something like so:

 if (UiAutomator.getbyId(SomeId).exists()){
      .....
   }

回答1:

We need that functionality and I ended up implementing it below:

https://github.com/marcosdiez/espresso_clone

if(onView(withText("click OK to Continue")).exists()){ 
    doSomething(); 
} else { 
   doSomethingElse(); 
}

I hope it is useful for you.

回答2:

Conditional logic in tests is undesirable. With that in mind, Espresso's API was designed to guide the test author away from it (by being explicit with test actions and assertions).

Having said that, you can still achieve the above by implementing your own ViewAction and capturing the isDisplayed check (inside the perform method) into an AtomicBoolean.

Another less elegant option - catch the exception that gets thrown by failed check:

    try {
        onView(withText("my button")).check(matches(isDisplayed()));
        //view is displayed logic
    } catch (NoMatchingViewException e) {
        //view not displayed logic
    }

回答3:

I think to mimic UIAutomator you can do this:
(Though, I suggest rethinking your approach to have no conditions.)

ViewInteraction view = onView(withBlah(...)); // supports .inRoot(...) as well
if (exists(view)) {
     view.perform(...);
}

@CheckResult
public static boolean exists(ViewInteraction interaction) {
    try {
        interaction.perform(new ViewAction() {
            @Override public Matcher<View> getConstraints() {
                return any(View.class);
            }
            @Override public String getDescription() {
                return "check for existence";
            }
            @Override public void perform(UiController uiController, View view) {
                // no op, if this is run, then the execution will continue after .perform(...)
            }
        });
        return true;
    } catch (AmbiguousViewMatcherException ex) {
        // if there's any interaction later with the same matcher, that'll fail anyway
        return true; // we found more than one
    } catch (NoMatchingViewException ex) {
        return false;
    } catch (NoMatchingRootException ex) {
        // optional depending on what you think "exists" means
        return false;
    }
}

---

Also exists without branching can be implemented really simple:

onView(withBlah()).check(exists()); // the opposite of doesNotExist()

public static ViewAssertion exists() {
    return matches(anything());
}

Though most of the time it's worth checking for matches(isDisplayed()) anyway.

回答4:

You check with the below code also. If view is displayed it will click else it will pass on.

onView(withText("OK")).withFailureHandler(new FailureHandler() {
        @Override
        public void handle(Throwable error, Matcher<View> viewMatcher){
        }
    }).check(matches(isDisplayed())).perform(customClick());

回答5:

Based on the answer by Dhiren Mudgil, I ended up writing the following method:

public static boolean viewIsDisplayed(int viewId)
{
    final boolean[] isDisplayed = {true};
    onView(withId(viewId)).withFailureHandler(new FailureHandler()
    {
        @Override
        public void handle(Throwable error, Matcher<View> viewMatcher)
        {
            isDisplayed[0] = false;
        }
    }).check(matches(isDisplayed()));
    return isDisplayed[0];
}

I'm using this to help determine which View in a ViewFlipper is currently displayed.