如何在 ListView 中显示 SQLite 查询的输出

Posted 2023-05-07

【中文标题】如何在 ListView 中显示 SQLite 查询的输出

【英文标题】：How to display output of SQLite query in ListView

【发布时间】：2013-01-10 05:55:40

【问题描述】：

我正在学习安卓。我正在尝试学习 SQLite 数据库，但我遇到了问题。 我无法读取 Cursor 的输出并将其显示在 ListView 中。这是我的 MainActivity.java 代码-

package com.example.addname;

import android.app.Activity;
import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;
import android.os.Bundle;
import android.view.Menu;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.ArrayAdapter;
import android.widget.Button;
import android.widget.EditText;
import android.widget.ListView;
import android.widget.TextView;

public class MainActivity extends Activity 

    Button b1, b2;
    EditText t1;
    ListView lv1;
    SQLiteDatabase db;
    Cursor c;

    @Override
    protected void onCreate(Bundle savedInstanceState) 
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        b1 = (Button) findViewById(R.id.button1);
        b2 = (Button) findViewById(R.id.button2);
        t1 = (EditText) findViewById(R.id.editText1);
        lv1 = (ListView) findViewById(R.id.listView1);


        db = openOrCreateDatabase("MyDB", MODE_PRIVATE, null);
        db.execSQL("CREATE TABLE IF NOT EXISTS Names(Name VARCHAR)");
        b1.setOnClickListener(new OnClickListener() 

            @Override
            public void onClick(View arg0) 
                // TODO Auto-generated method stub

                if (t1.getText().toString().equals("")) 
                    t1.setText("Enter Name");
                

                else 
                    String input = t1.getText().toString();
                    db.execSQL("INSERT INTO Names VALUES(" + "'" + input + "'"
                            + ")");
                

            
        );

        b2.setOnClickListener(new OnClickListener() 


            @Override
            public void onClick(View arg0) 
                // TODO Auto-generated method stub
                c = db.rawQuery("SELECT * FROM Names",null);
                int count = c.getCount();
                String values[] = new String[count+1];
                int i = 0;

                do
                 
                    values[i]= c.getString(c.getColumnIndex("Name"));
                    i++;
                


                while(c.moveToNext());

                ArrayAdapter<String> adapter = new ArrayAdapter<String>(getBaseContext(),
                          android.R.layout.activity_list_item, android.R.id.text1,values);

                lv1.setAdapter(adapter);
            
        );
    

    @Override
    public boolean onCreateOptionsMenu(Menu menu) 
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.activity_main, menu);
        return true;
    

这里是 activity_main.xml 代码 -

<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_
    android:layout_
    tools:context=".MainActivity" >

    <EditText
        android:id="@+id/editText1"
        android:layout_
        android:layout_
        android:layout_alignParentLeft="true"
        android:layout_alignParentRight="true"
        android:layout_alignParentTop="true"
        android:layout_marginTop="25dp"
        android:ems="10" >

        <requestFocus />
    </EditText>

    <Button
        android:id="@+id/button1"
        android:layout_
        android:layout_
        android:layout_below="@+id/editText1"
        android:layout_centerHorizontal="true"
        android:text="@string/button1" />

    <ListView
        android:id="@+id/listView1"
        android:layout_
        android:layout_
        android:layout_alignParentLeft="true"
        android:layout_below="@+id/button2" >
    </ListView>

    <Button
        android:id="@+id/button2"
        android:layout_
        android:layout_
        android:layout_alignLeft="@+id/button1"
        android:layout_below="@+id/button1"
        android:layout_marginTop="24dp"
        android:text="@string/button2" />

</RelativeLayout>

当我点击“显示”按钮时，我收到错误“不幸的是 addname 已停止”（addname 是我的项目名称）。请帮助。

【问题讨论】：

您的问题到底是什么？更具体！

【参考方案1】：

使用SimpleCursorAdapter 代替 ArrayAdapter。

c = db.rawQuery("SELECT * FROM Names",null);

while(c.isAfterLast() == false)
        c.moveToNext();


ListAdapter adapter = new SimpleCursorAdapter(this,
                android.R.layout.activity_list_item,
                c,
                new String[] "Name",
                new int[] android.R.id.text1,0);

lv1.setAdapter(adapter);

【讨论】：

得到错误 - 构造函数 SimpleCursorAdapter(new View.OnClickListener(), int, Cursor, String[], int[], int) 未定义

不要传递 OnClickListener()... 传递当前活动上下文。即this

【参考方案2】：

用这个替换 do...while 部分

while(c.moveToNext())
 
      values[i]= c.getString(c.getColumnIndex("Name"));
       i++;
  

【参考方案3】：

            c = db.rawQuery("SELECT * FROM Names",null);
            int count = c.getCount();
            String values[] = new String[count+1];
            int i = 0;

   while(c.moveToNext())

             
                values[i]= c.getString(c.getColumnIndex("Name"));
                i++;
            




            ArrayAdapter<String> adapter = new ArrayAdapter<String>(getBaseContext(),
                      android.R.layout.activity_list_item, android.R.id.text1,values);

            lv1.setAdapter(adapter);