Question

我有一个包含姓名，年龄和汽车的人员列表。我可以打印出一组精确的字符串值。但是，当我向列表中添加另一个名称，年龄，汽车时，仅显示最后的项目/列表。如何打印两组人员数据？

    List<People> peopleInfo = getInfo(ppl);
    String peopleResults = mapper.writeValueAsString(peopleInfo);
    for(People model : peopleInfo) {          
        People people = new People(model.getName(), model.getAge(), model.getCar());
        ToJSON peop = new ToJSON(people);

         ObjectMapper mapper2 = new ObjectMapper();
         ObjectWriter writer = mapper2.writer(new DefaultPrettyPrinter());
         writer.writeValue(new File("file.json"), peop);

getInfo方法

private List<People> getInfo(String ppl) {
    List<People> peopleList = new ArrayList<People>();
    People people = new People();

    //1st Person
    people.setName("Tim");
    people.setAge("22");
    people.setCar("Ford");

    //Second Person
    people.setName("Susan");
    people.setAge("42");
    people.setCar("Toyota");

    peopleList.add(people);

    return peopleList;
}

Answer 1

您在创建第二个人时重新定义了人，因此只有第二个人被添加到列表中

//1st Person
people.setName("Tim");
people.setAge("22");
people.setCar("Ford");

//Second Person
people.setName("Susan");
people.setAge("42");
people.setCar("Toyota");

peopleList.add(people);

修复方法是创建多个人物对象或在定义后立即添加人物

People people = new People();
People people2 = new People();

People.setName("Time");
People2.setName("Susan");

peopleList.add(people);
peopleList.add(people2);

或后者

people.setName("Tim");
people.setAge("22");
people.setCar("Ford");

peopleList.add(people);

people.setName("Susan");
people.setAge("42");
people.setCar("Toyota");

peopleList.add(people);

Answer 2

因为people对象将包含最新值。你需要每次都像这样传递新对象，

People people = new People();

一个简单的例子就是你可以这样做，

People people = new People();

//1st Person
people.setName("Tim");
people.setAge("22");
people.setCar("Ford");

//Second Person
 People people2 = new People();

people2.setName("Susan");
people2.setAge("42");
people2.setCar("Toyota");

peopleList.add(people);
peopleList.add(people2);

这样您的peopleList将会容纳2个对象。

Answer 3

你真的忘记了一步。让我们一步一步走：

<plugin>
    <artifactId>maven-ear-plugin</artifactId>
    <version>2.9.1</version>
    <configuration>
        <defaultLibBundleDir>lib</defaultLibBundleDir>
        <goals>
            <goal>generate-application-xml</goal>
            <initializeInOrder>true</initializeInOrder>
        </goals>
    </configuration>
</plugin>

您创建了空列表

List<People> peopleList = new ArrayList<People>();

您创建了一个People people = new People(); //1st Person people.setName("Tim"); people.setAge("22"); people.setCar("Ford");对象实例并在其上设置了一些值

People

在这里，您不会创建新人，但会覆盖第一个人的值

//Second Person
people.setName("Susan");
people.setAge("42");
people.setCar("Toyota");

最后，将第一个人（包含第二个人的值）添加到列表中。

您必须为第二个人创建一个新的peopleList.add(people);个实例，在该个实例上设置值并将其添加到列表中，就像第一个人一样。

Answer 4

 Private Sub Worksheet_Change(ByVal Target As Excel.Range)

If Target.Column = 1 Then

    ThisRow = Target.Row

    If Target.Value = "Final Recon" Then

    With Selection.Validation
        .Range ("N" & ThisRow)


.Delete
.Add Type:=xlValidateList, Formula1:="=Final", Formula2:="=Under Review"
 End With

    Else
        Range("N" & ThisRow) = ""
    End If
  End If
 End Sub

您不是在创建另一个人对象并添加上一个对象，而只是修改它。

Answer 5

你创造了一个＆＃34; People＆＃34;对象并设置名称，年龄和汽车。。。然后再次设置人物对象的名称，年龄和汽车。您需要两个不同的People对象：

People people1 = new People();
People people2 = new People();

//1st Person
people1.setName("Tim");
people1.setAge("22");
people1.setCar("Ford");

//Second Person
people2.setName("Susan");
people2.setAge("42");
people2.setCar("Toyota");

peopleList.add(people1);
peopleList.add(people2);

另外，您将参数传递给方法getInfo(String ppl)，但不要使用它。该参数是什么？

如何显示ArrayList中包含重复值的所有项？

5 个答案: