最近我正在为遗留系统编写大量 JUnit 测试。

我经常遇到这样的问题:断言复杂对象的最佳方式是什么？

这是我当前的代码

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个步骤，我不能 cahnge 太多，因为它是遗留代码......

感谢您的帮助!

最佳答案

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"));

此处详细描述了示例: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，最后支持逐字段比较:isEqualToComparingFieldByField、isEqualToComparingOnlyGivenFields 和 isEqualToIgnoringGivenFields。

希望对你有帮助

关于java - 断言复杂对象时的 Junit 最佳实践，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34531301/