插入SQLite数据库-Android

问题描述 投票:0回答:1

我目前正在从事Android类项目。我对数据库中的插入语句有一些麻烦;我正在使用SQLite。当我运行该应用程序时,我的数据库已正确创建,但未插入值。我试图将自己介绍给在线提案和解决方案,并尝试实施它,但无济于事。

请我需要帮助。谢谢。

DBHandler.java

package ck.edu.com.birch;

import android.content.ContentValues;
import android.content.Context;
import android.database.Cursor;
import android.database.SQLException;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteOpenHelper;
import android.util.Log;

import java.text.SimpleDateFormat;
import java.util.Date;

public class DBHandler {

    String TAG = "Tennis Tracker";

    String DB_NAME = "TennisEFREI.db";
    String DB_TABLE = "statistics";
    int DB_VERSION = 1;

    String SQL_CREATE = "CREATE TABLE " + DB_TABLE +
            " (_id INTEGER PRIMARY KEY AUTOINCREMENT, " +
            "playerOnename TEXT NOT NULL, " +
            "playerTwoname TEXT NOT NULL, " +
            "winner TEXT NOT NULL, " +
            "wonPointsPlayer1 INTEGER NOT NULL, " +
            "wonPointsPlayer2 INTEGER NOT NULL, " +
            "wonSetsPlayer1 INTEGER NOT NULL, " +
            "wonSetsPlayer2 INTEGER NOT NULL, " +
            "acePlayer1 INTEGER, " +
            "acePlayer2 INTEGER, " +
            "winPlayer1 INTEGER, " +
            "winPlayer2 INTEGER, " +
            "errorPlayer1 INTEGER, " +
            "errorPlayer2 INTEGER, " +
            "date TIMESTAMP DEFAULT CURRENT_TIMESTAMP)";
    String SQL_DROP = "DROP TABLE IF EXISTS " + DB_TABLE;

    DatabaseHelper databaseHelper;

    private SQLiteDatabase db;

    public DBHandler(Context context) {
        databaseHelper = new DatabaseHelper(context);
    }

    class DatabaseHelper extends SQLiteOpenHelper {
        DatabaseHelper(Context context) {
            super(context, DB_NAME, null, DB_VERSION);
        }


        @Override
        public void onCreate(SQLiteDatabase db) {
            db.execSQL(SQL_CREATE);

        }

        @Override
        public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
            db.execSQL(SQL_DROP);
            onCreate(db);
        }
    }

    public void open() throws SQLException {
        db = databaseHelper.getWritableDatabase();
    }

    public void close() {
        databaseHelper.close();
    }

    public long insertStatistics(String playerOnename, String playerTwoname, String winner, int wonPointsPlayer1, int wonPointsPlayer2, int wonSetsPlayer1, int wonSetsPlayer2, int acePlayer1, int acePlayer2, int winPlayer1, int winPlayer2, int errorPlayer1, int errorPlayer2) {
        SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd/MM/yyyy");
        String date = simpleDateFormat.format(new Date());
        ContentValues contentValues = new ContentValues();
        contentValues.put("playerOnename", playerOnename);
        contentValues.put("playerTwoname", playerTwoname);
        contentValues.put("winner", winner);
        contentValues.put("wonPointsPlayer1", wonPointsPlayer1);
        contentValues.put("wonPointsPlayer2", wonPointsPlayer2);
        contentValues.put("wonSetsPlayer1", wonSetsPlayer1);
        contentValues.put("wonSetsPlayer2", wonPointsPlayer2);
        contentValues.put("acePlayer1", acePlayer1);
        contentValues.put("acePlayer2", acePlayer2);
        contentValues.put("winPlayer1", winPlayer1);
        contentValues.put("winPlayer2", winPlayer2);
        contentValues.put("errorPlayer1", errorPlayer1);
        contentValues.put("errorPlayer2", errorPlayer2);
        contentValues.put("date", date);
        Log.d(TAG, "VALUES : " + contentValues);
        return db.insert(DB_TABLE, null, contentValues);
    }
}


java android sqlite android-studio
1个回答
0
投票

documentation说:

SQLite没有为存储日期和/或时间预留存储类。相反,内置的SQLite Date And Time Functions能够将日期和时间存储为TEXT,REAL或INTEGER值:

  • TEXT作为ISO8601字符串(“ YYYY-MM-DD HH:MM:SS.SSS”)。
  • REAL作为儒略日数,是公元前4714年11月24日格林威治中午以来的天数。根据多久的公历。
  • INTEGER作为Unix时间,自1970-01-01 00:00:00 UTC以来的秒数。

因此,请以这种格式转换日期,然后尝试将表的日期列数据类型插入或设置为文本(我不推荐)。

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