如何将列表变量传递到构造函数/方法中

Question

我是 Java 新手。当我创建类的对象时，我在传递作为构造函数定义一部分的列表变量时遇到问题    public class Patient {    private String patientfirstName;    private String patientLastName;    private List<String> allergyList;     public Patient(String patientfirstName, String patientLastName,      List<String> allergyList) {     this.patientfirstName = patientfirstName;     this.patientLastName = patientLastName;     this.allergyList = allergyList;      }     Patient patientobj = new Patient("sean","john","allegry1");给出错误：“构造函数“Str，str，str”未定义。”我需要如何消除过敏的帮助

Answer 1

您需要 aList<String>而不是单个String,Arrays.asList(T...)可能是最简单的解决方案：

Patient patientobj = new Patient("sean", "john", Arrays.asList("allergy1"));

如果你有更多的过敏症

Patient patientobj = new Patient("sean", "john", 
        Arrays.asList("allergy1", "allergy2"));

Answer 2

public class Patient {

private String patientfirstName;

private String patientLastName;

private List<String> allergyList;

 public Patient(String patientfirstName, String patientLastName, 

 List<String> allergyList) {

 this.patientfirstName = patientfirstName;

 this.patientLastName = patientLastName;

 this.allergyList = allergyList;

  }

 *Patient patientobj = new Patient("sean","john","allegry1");*// this is wrong you have to pass a list not the string. you should do something like this:

 // first create a list and add the value to it

 List<String> list = new ArrayList<>();

 list.add("allergy1");

 // now create a object and pass the list along with other variables

 Patient patientobj = new Patient("sean","john",list);

Answer 3

在我看来，你可以使用Varargs。感谢 varargs，您可以在参数中放入您想要的参数数量

public class Patient {

public String patientfirstName;

public String patientLastName;

public List<String> allergyList;

public Patient(String fName,String lName,String...aList) {

    this.patientfirstName = fName;

    this.patientLastName = lName;

    this.allergyList = Arrays.asList(aList);

}

public static void main(String[] args) {

    Patient firstPatient = new Patient("Foo", "Bar", "First Allergy","Second Allergy");

    Patient secondPatient = new Patient("Foo", "Baz", "First Allergy","Second Allergy","Third Allergy","Fourth Allergy");

    Patient ThirdPatient = new Patient("Foo", "Foo", "First Allergy");

}

参数“aList”就像一个数组，因为varargs就像一个没有特定长度的数组，你在输入参数时选择的长度，如你所见

allergyList 的类型是可以选择的。您也可以这样做：

在“患者”属性中：

 public String[] allergyList;

在构造函数中：

public Patient(String fName,String lName,String...aList) {

        this.patientfirstName = fName;

        this.patientLastName = lName;

        this.allergyList = allergyList;

    }

Answer 4

您还有一种解决方案，只需添加该类的一个构造函数即可Patient。

public Patient (String patientfirstName,String patientLastName,String allergeyList){

this.patientfirstName  = patientfirstName;

this.patientLastName = patientLastName;\

this.allergeyList = new ArrayList<>( Arrays.asList(allergeyList));

}