我正在为遗留系统编写很多JUnit测试.
我常常提出这样的问题：断言复杂对象的最佳方法是什么？
这是我目前的代码
public class SomeParserTest {
@Test
public void testParse() throws Exception {
final SomeParser someParser = new SomeParser();
someParser.parse("string from some file");
final List<Result> listOfResults = someParser.getResults();
assertThat(listOfResults,hasSize(5));
assertResult(listOfResults.get(0),"20151223",2411189L,isEmptyOrNullString(),"2.71","16.99");
assertResult(listOfResults.get(1),"20151229",2411190L,"2.86","17.9");
assertResult(listOfResults.get(2),2411191L,is("1.26"),".75","23.95");
assertResult(listOfResults.get(3),2411192L,is("2.52"),"1.5","47.9");
assertResult(listOfResults.get(4),2411193L,"16.99");
final List<SubResult> listofSubResuls = someParser.getSubResultOf(listOfResults.get(0));
assertThat(listofSubResuls,hasSize(1));
assertSubResult(listofSubResuls.get(0),12.5D,1L,14.87D,16.99D,0L,null,67152L,"2","02411189",56744349L);
final List<SubResult> listofSubResuls1 = someParser.getListofBBBS(listOfResults.get(1));
assertThat(listofSubResuls1,hasSize(2));
assertSubResult(listofSubResuls1.get(0),30.0D,"20151228",12.53D,17.9D,67156L,"02411190",56777888L);
assertSubResult(listofSubResuls1.get(1),33.3D,4.66D,6.99D,"J",21L,56777889L);
//And 50 Lines more
}
// how to avoid so many parameters?
private void assertSubResult(final SubResult subResult,final double someDouble,final String bestellDatum,final long someLong,final double someDouble2,final double someDouble3,final long someLong3,final String someString,final long someLong1,final String someString4,final String someString3,final long someLong4,final String rechnungsNummer,final long someLong2) {
assertThat(subResult.getXXX(),is(nullValue()));
assertThat(subResult.getXYX().getTag(),is(someDouble2));
assertThat(subResult.getXYX(),is("some constant"));
// and much more
}
// how to avoid so many parameters?
private void assertResult(final Result result,final String string1234,final long abc,final String string1,final String string12,final String string134) {
assertThat(result.getXXX(),is(nullValue()));
assertThat(result.getXYX().getTag(),is(someDouble2));
assertThat(result.getXYX(),is("some constant"));
// and much more
}
}
没有简单的方法可以测试这样一个解析器的每一步,因为它是遗留代码,所以我无法改变这一点.
谢谢你的帮助！
解决方法
作为西西弗斯,我建议使用hamcrest matchers.
但我建议编程custom matcher.以下行
assertResult(listOfResults.get(0),"16.99");
可能看起来像：
assertThat(listOfResults,contains(
ResultMatcher.matchesResult().withFirstAttribute("20151223").andSecondAttribute(2411189L)...
... // here the matchers for the other elements of the list
));
您将需要自定义匹配器类ResultMatcher,它应具有以下形式：
class ResultMatcher extends TypeSafeMatcher<Result> {
Matcher<String> firstAttribute = Matchers.any(String.class);
Matcher<String> secondAttribute = Matchers.any(String.class);
...
ResultMatcher withFirstAttribute(String firstAttribute) {
this.firstAttribute = Matchers.equalTo(firstAttribute);
return this;
}
...
public boolean matchesSafely(Result result) {
if (!firstAttribute.matches(result.getFirstAttribute())) {
return false
}
...
return true;
}
}
这个设计有一些优点：
>您不需要定义equals方法>您可以为每个属性定义默认值,因此您可以仅测试您感兴趣的属性并默认匹配其他属性>测试不必检查对象的每个属性(匹配器执行此操作)
