断言复杂对象时的Junit最佳实践

Question

我正在为遗留系统编写很多JUnit测试.

我经常会提出这样的问题: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, isEmptyOrNullString(), "2.86", "17.9");
        assertResult(listOfResults.get(2), "20151229", 2411191L, is("1.26"), ".75", "23.95");
        assertResult(listOfResults.get(3), "20151229", 2411192L, is("2.52"), "1.5", "47.9");
        assertResult(listOfResults.get(4), "20151229", 2411193L, isEmptyOrNullString(), "2.71", "16.99");

        final List<SubResult> listofSubResuls = someParser.getSubResultOf(listOfResults.get(0));
        assertThat(listofSubResuls, hasSize(1));
        assertSubResult(listofSubResuls.get(0), 12.5D, "20151223", 1L, 14.87D, 16.99D, 0L, null, 67152L, "20151223", "2", 0L, "02411189", 56744349L);

        final List<SubResult> listofSubResuls1 = someParser.getListofBBBS(listOfResults.get(1));
        assertThat(listofSubResuls1, hasSize(2));
        assertSubResult(listofSubResuls1.get(0), 30.0D, "20151228", 1L, 12.53D, 17.9D, 0L, null, 67156L, "20151229", "2", 0L, "02411190", 56777888L);
        assertSubResult(listofSubResuls1.get(1), 33.3D, "20151228", 1L, 4.66D, 6.99D, 1L, "J", 67156L, "20151229", "2", 21L, "02411190", 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
    }
}

没有简单的方法来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!

Answer 1

我会尝试使用assertj提取功能，例如：

// fellowshipOfTheRing is a List<TolkienCharacter>
assertThat(fellowshipOfTheRing).extracting("name", "age", "race.name")
                               .contains(tuple("Boromir", 37, "Man"),
                                         tuple("Sam", 38, "Hobbit"),
                                         tuple("Legolas", 1000, "Elf"));

该示例在这里详细描述：http://joel-costigliola.github.io/assertj/assertj-core-features-highlight.html#extracted-properties-assertion

您还可以使用特定的比较策略来比较实际结果和预期结果，最后支持逐个字段比较：isEqualToComparingFieldByField、isEqualToComparingOnlyGivenFields 和 isEqualToIgnoringGivenFields。

希望它能有所帮助

Answer 2

我会做类似的事情，而不是长方法签名。（伪代码，免责声明，以防无法编译）：

class AssertResult {
    private int xxx;
    private String yyy;
    public AssertResult setXXX(int xxx) {
        this.xxx = xxx;
        return this;
    }
    public AssertResult setYYY(String yyy) {
        this.yyy = yyy;
        return this;
    }
    public void check(Result result) {
        assertThat(result.getXXX(), is(xxx));
        assertThat(result.getYYY(), is(yyy));
    }
 }

然后我可以这样使用它：

new AssertResult().setXXX(123).setYYY("asdasd").check(result);