我更喜欢tddavies的答案（只要验证mock被调用了--因为你知道调用它意味着它会抛出）。但这里有另一个解决方案，它可能使您试图Assert的内容更加明确。
doAnswer是更通用的Mockito原语，它可以返回值或抛出。在那里你可以只保留一个计数器。

AtomicInteger blinkIOExceptions = new AtomicInteger();
doAnswer(invocation -> {
    blinkIOExceptions.incrementAndGet();
    throw new IOException();
}).when(mockOptionalService).blink();
assertEquals(1, blinkIOExceptions.get());

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我不认为您可以检查异常是否真的被抛出，但是您可以验证您为抛出异常而被删除的方法是否真的被调用了。参见下面的示例，我们控制是否调用该方法，当它没有被调用时，测试失败：

package org.example;
import org.junit.jupiter.api.Test;
import org.junit.jupiter.api.extension.ExtendWith;
import org.mockito.InjectMocks;
import org.mockito.Mock;
import org.mockito.junit.jupiter.MockitoExtension;
import java.io.IOException;
import static org.mockito.Mockito.doThrow;
import static org.mockito.Mockito.verify;
class ToTest {
    private final InjectedClass injectedClass;
    ToTest(InjectedClass injectedClass) {
        this.injectedClass = injectedClass;
    }
    public void doIt(boolean callMethod) {
        try {
            if (callMethod) {
                injectedClass.mightThrow();
            }
        } catch (IOException e) {
            // ignore (this seems like a bad idea...
        }
    }
}
interface InjectedClass {
    void mightThrow() throws IOException;
}
@ExtendWith(MockitoExtension.class)
public class AppTest {
    @Mock
    InjectedClass injectedClass;
    @InjectMocks
    ToTest toTest;
    @Test
    public void passingTest() throws IOException {
        doThrow(IOException.class).when(injectedClass).mightThrow();
        toTest.doIt(true);
        verify(injectedClass).mightThrow();
    }
    @Test
    public void failingTest() throws IOException {
        doThrow(IOException.class).when(injectedClass).mightThrow();
        toTest.doIt(false);
        verify(injectedClass).mightThrow();
    }
}

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