Question

我有一个像下面这样的json数组，想要从spinner中选择一个选项时选择相应的id，它也是动态的，也就是在Spinner中显示的json数组

{
   "DoctorName": ["0001 DR. Sameer", "0001 DR.Krishna murti", "110 Mr. Ram", "4 Mr. Yash Pathak.", "99 Dr. Varma"],
    "DoctorId": [3,2,110,4,99]
};

我必须在Android中执行此操作。任何帮助将不胜感激。

Answer 1

1.首先创建一个类

 public class DoctorName
  {

public String id = "";
public String name = "";

public void setId(String id)
{
    this.id = id;
}

public void setName(String name)
{
    this.name = name;
}


public String getName()
{
    return name;
}

public String getId()
{
    return id;
}

// A simple constructor for populating our member variables for this tutorial.
public DoctorName( String _id, String _name)
{
    id = _id;
    name = _name;

}

// The toString method is extremely important to making this class work with a Spinner
// (or ListView) object because this is the method called when it is trying to represent
// this object within the control.  If you do not have a toString() method, you WILL
// get an exception.
public String toString()
{
    return( name );
}

}

2.创建另一个班级的 MainClass.java

ArrayList<DoctorName> doctList = new ArrayList<DoctorName>() ; for(int i=0;i<arr_name.length;i++) { doctList.add(new DoctorName(arr_id[i],arr_name[i])); } //fill data in spinner //ArrayAdapter<DoctorName> adapter = new ArrayAdapter<DoctorName>(getApplicationContext(), android.R.layout.simple_spinner_dropdown_item, answers); ArrayAdapter <DoctorName>adapter= new ArrayAdapter<DoctorName> (getApplicationContext(), android.R.layout.simple_spinner_dropdown_item,doctList ); Doctor_selection.setAdapter(adapter); Doctor_selection.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() { @Override public void onItemSelected(AdapterView<?> parent, View view, int position, long id) { DoctorName doctorName = (DoctorName) parent.getSelectedItem(); Log.i("SliderDemo", "getSelectedItemId" +doctorName.getId()); } @Override public void onNothingSelected(AdapterView<?> parent) { } });

Answer 2

您必须使用ArrayAdapter将json数组值显示到微调器

Spinner spinner = new Spinner(this);
ArrayAdapter<String> spinnerArrayAdapter = new ArrayAdapter<String>(this, android.R.layout.simple_spinner_item, list_values); //selected item will look like a spinner set from XML
spinnerArrayAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
spinner.setAdapter(spinnerArrayAdapter);

//Set on item select Listener
        spinner.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
            @Override
            public void onItemSelected(AdapterView<?> parent, View view, int position, long id) {

             // here you can get your selected id

            }

            @Override
            public void onNothingSelected(AdapterView<?> parent) {

            }
        });

了解更多check here.

Answer 3

创建两个数组，即DoctorName和DoctorId

使用上面的数组创建动态HashMap，使用for循环将所有值置于键值形式。但是对于这个，上面两个数组的长度应该是相同的。

HashMap<String, String> hash;
for(int i = 0; i < DoctorName.size() ; i++) {

    hash = new HashMap<String, String>();
    hash.put(DoctorId.get(i), DoctorName.get(i));
}

对于微调器，只从Map（散列）发送医生名单，并且微调器的onclick获取其为doctorId的Id。在Spinner onclick中写下以下代码
```
String name = spinner.getSelectedItem().toString();
String id = hash.get(name);
```
在id中，您将获得所选名称的相应ID。

希望它有所帮助：）

如何将json数组元素映射到Spinner中选择的选项？

3 个答案: