[英]How to get full coverage for Java Code Coverage? Junit test cases
[英]How to get full coverage for Java Code Coverage? Junit 5 Eclipse IDE
我正在为一门课程做作业,我需要全面了解这种方法
public void ourcompanyname(String companyName) {
if (companyName == ("")) {
throw new AirlineException("Warning. The company must have a name !");}
---->这是构造函数
public String getCompanyName() {
return companyName;
}
/**
* @param companyName the companyName to set
*/
public void setCompanyName(String companyName) {
this.companyName = companyName;
----->这是我用的测试
public void testSet() throws AirlineException {
airlinecompany.setCompanyName((""));
Assertions.assertEquals((""),airlinecompany.getCompanyName());
Assertions.assertThrows(AirlineException.class,()->airlinecompany.ourcompanyname(("")));
----->问题是我在测试这段代码时得到了 50% 的覆盖率。 ("") 字符串代表一个空白字段,因此当您不填写名称时,您会得到一个 AirlineException。 但我也想测试它是否像“FLYAIR”这样的字符串,它不应该抛出异常。 我有哪些选择? 更好地为空白字段 ("") 编码还是仅更改此字段?
提前多谢!
尝试这样的事情:---
@Test
public void testUsernameIsNull() {
Throwable exception = Assertions.assertThrows(
AirlineException.class, () -> {
Airlinecompany airlinecompany = new Airlinecompany();
airlinecompany.setName("");
}
);
Assertions.assertEquals("Warning. The company must have a name !", exception.getMessage());
if(true){
Airlinecompany airlinecompany = new Airlinecompany();
airlinecompany.setName("FLYAIR");
Assertions.assertEquals(("FLYAIR"),airlinecompany.getCompanyName());
}
}
希望它可以解决您的问题。
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