假设以下简单测试:
sashelp.cars
测试不会编译,因为" moreThan"只能应用于\
类型的实例。但我想断言@Test
public void test() throws Exception {
Object value = 1;
assertThat(value, greaterThan(0));
}
是一个大于零的整数。我怎样才能用Hamcrest来表达呢?
简单的解决方案是通过强制转换匹配器来删除泛型:
Comparable
可能,但会生成编译器警告并感觉不对。
一个冗长的选择是:
value
感觉"整洁"并且"纠正",但实际上很多代码看起来很简单。我试图在hamcrest找到类似内置的东西,但我没有成功,但也许我错过了什么?
在我的实际测试用例中代码是这样的:
assertThat(value, (Matcher)greaterThan(0));
在问题的简化案例中,我也可以写@Test
public void testName() throws Exception {
Object value = 1;
assertThat(value, instanceOfAnd(Integer.class, greaterThan(0)));
}
private static<T> Matcher<Object> instanceOfAnd(final Class<T> clazz, final Matcher<? extends T> submatcher) {
return new BaseMatcher<Object>() {
@Override
public boolean matches(final Object item) {
return clazz.isInstance(item) && submatcher.matches(clazz.cast(item));
}
@Override
public void describeTo(final Description description) {
description
.appendText("is instanceof ")
.appendValue(clazz)
.appendText(" and ")
.appendDescriptionOf(submatcher);
}
@Override
public void describeMismatch(final Object item, final Description description) {
if (clazz.isInstance(item)) {
submatcher.describeMismatch(item, description);
} else {
description
.appendText("instanceof ")
.appendValue(item == null ? null : item.getClass());
}
}
};
}
。在我的实际情况中,我可以写Map<String, Object> map = executeMethodUnderTest();
assertThat(map, hasEntry(equalTo("the number"), greaterThan(0)));
,但如果出现问题,这当然会使错误信息恶化。
答案 0 :(得分:5)
这个答案将不显示如何使用Hamcrest执行此操作,我不知道是否有比建议更好的方法。
但是,如果您可以包含其他测试库,AssertJ正好支持:
import org.junit.Test;
import static org.assertj.core.api.Assertions.assertThat;
public class TestClass {
@Test
public void test() throws Exception {
Object value = 1;
assertThat(value).isInstanceOfSatisfying(Integer.class, integer -> assertThat(integer).isGreaterThan(0));
}
}
不需要任何演员,AssertJ会为你做这件事。
此外,如果断言失败,它会打印一条漂亮的错误消息,value
太小:
java.lang.AssertionError:
Expecting:
<0>
to be greater than:
<0>
或者如果value
的类型不正确:
java.lang.AssertionError:
Expecting:
<"not an integer">
to be an instance of:
<java.lang.Integer>
but was instance of:
<java.lang.String>
可以找到isInstanceOfSatisfying(Class<T> type, Consumer<T> requirements)
的Javadoc here,其中还包含一些更复杂的断言示例:
// second constructor parameter is the light saber color
Object yoda = new Jedi("Yoda", "Green");
Object luke = new Jedi("Luke Skywalker", "Green");
Consumer<Jedi> jediRequirements = jedi -> {
assertThat(jedi.getLightSaberColor()).isEqualTo("Green");
assertThat(jedi.getName()).doesNotContain("Dark");
};
// assertions succeed:
assertThat(yoda).isInstanceOfSatisfying(Jedi.class, jediRequirements);
assertThat(luke).isInstanceOfSatisfying(Jedi.class, jediRequirements);
// assertions fail:
Jedi vader = new Jedi("Vader", "Red");
assertThat(vader).isInstanceOfSatisfying(Jedi.class, jediRequirements);
// not a Jedi !
assertThat("foo").isInstanceOfSatisfying(Jedi.class, jediRequirements);
答案 1 :(得分:4)
问题是你在这里丢失了类型信息:
sqlite3_exec(dbHandle, "PRAGMA journal_mode=WAL;", 0, 0, 0);
如果你想一想,这是一个非常奇怪的线。这里 Object value = 1;
是最通用的东西,没有可以被合理地告知它,除了可能检查它是value
还是检查它的字符串表示是否&# 39;不是。我试图想象一下现代Java中上述行的合法用例,我有点茫然。
明显的解决方法是null
更好的解决方案是投射到assertThat((Comparable)value, greaterThan(0));
,因为你要比较一个整数常量;字符串也是可比较的,但仅限于它们之间。
如果您不能认为Integer
是value
,那么将它与任何内容进行比较都是毫无意义的。如果您的广告投放到Comparable
的测试失败,则表示您从其他内容动态投放到Comparable
的报告失败了。
答案 2 :(得分:3)
包含Object值的map的问题在于您必须假设要比较的特定类。
缺少的hamcrest是一种将匹配器从给定类型转换为另一种类型的方法,例如这个要点中的一个: https://gist.github.com/dmcg/8ddf275688fd450e977e
public class TransformingMatcher<U, T> extends TypeSafeMatcher<U> {
private final Matcher<T> base;
private final Function<? super U, ? extends T> function;
public TransformingMatcher(Matcher<T> base, Function<? super U, ? extends T> function) {
this.base = base;
this.function = function;
}
@Override
public void describeTo(Description description) {
description.appendText("transformed version of ");
base.describeTo(description);
}
@Override
protected boolean matchesSafely(U item) {
return base.matches(function.apply(item));
}
}
有了这个,你可以这样写你的断言:
@Test
public void testSomething() {
Map<String, Object> map = new HashMap<>();
map.put("greater", 5);
assertThat(map, hasEntry(equalTo("greater"), allOf(instanceOf(Number.class),
new TransformingMatcher<>(greaterThan((Comparable)0), new Function<Object, Comparable>(){
@Override
public Comparable apply(Object input) {
return Integer.valueOf(input.toString());
}
}))));
}
但问题是,您需要指定一个给定的Comparable数字类(在本例中为Integer)。
如果出现断言错误,则消息为:
java.lang.AssertionError
Expected: map containing ["string"->(an instance of java.lang.Number and transformed version of a value greater than <0>)]
but: map was [<string=NaN>]
答案 3 :(得分:2)
原始尝试的轻微修改版本如何:
@Test
public void testName() throws Exception {
Map<String, Object> map = executeMethodUnderTest();
assertThat(map, hasEntry(equalTo("the number"),
allOf(instanceOf(Integer.class), integerValue(greaterThan(0)))));
}
private static<T> Matcher<Object> integerValue(final Matcher<T> subMatcher) {
return new BaseMatcher<Object>() {
@Override
public boolean matches(Object item) {
return subMatcher.matches(Integer.class.cast(item));
}
@Override
public void describeTo(Description description) {
description.appendDescriptionOf(subMatcher);
}
@Override
public void describeMismatch(Object item, Description description) {
subMatcher.describeMismatch(item, description);
}
};
}
现在自定义匹配器不那么冗长,你仍然可以达到你想要的效果。
如果值太小:
java.lang.AssertionError:
Expected: map containing ["the number"->(an instance of java.lang.Integer and a value greater than <0>)]
but: map was [<the number=0>]
如果值错误,请输入:
java.lang.AssertionError:
Expected: map containing ["the number"->(an instance of java.lang.Integer and a value greater than <0>)]
but: map was [<the number=something>]