android - 如何在 android fragment 中使用 SQLite？

Question

我是一名有抱负的 Android 开发者。我想做一个支持 SQLtite 的简单项目，但我遇到了 SQLite 和 Fragment 的问题。

预先感谢您的帮助

我尝试将 context: FragmentTwo 替换为 context: MainActivity。

结果又是一个错误。

据此，我得出结论，这不是解决问题的方法。😕😕😕

class sql(context: FragmentTwo):SQLiteOpenHelper(context, DATABASE_NAME, null, DATABASE_VER) {
    companion object{
        private val DATABASE_VER=1
        private val DATABASE_NAME="GUESTS.db"

        private val TABLE_NAME= "GUESTS"
        private val COL_ID= "Id"
        private val COL_NAME= "Name"
        private val COL_CITY= "City"

    }
    override fun onCreate(db: SQLiteDatabase?) {
        val CREATE_TABLE_QUERY = ("CREATE TABLE $TABLE_NAME +($COL_ID INTEGER PRIMARY KEY,$COL_NAME TEXT, $COL_CITY TEXT)")
        db!!.execSQL(CREATE_TABLE_QUERY)

    }

    override fun onUpgrade(db: SQLiteDatabase?, p1: Int, p2: Int) {
        db!!.execSQL("DROP TABLE IF EXISTS $TABLE_NAME")
        onCreate(db!!)
    }
    val allguest:List<guest>
    get(){
        val listguest = ArrayList<guest>()
        val selectQuery = "SELECT * FROM $TABLE_NAME"
        val db = this.writableDatabase
        val cursor = db.rawQuery(selectQuery, null)
        if (cursor.moveToFirst()){
            do {
                val guest= guest()
                guest.id=cursor.getInt(cursor.getColumnIndex(COL_ID))
                guest.name=cursor.getString(cursor.getColumnIndex(COL_NAME))
                guest.city=cursor.getString(cursor.getColumnIndex(COL_CITY))

                listguest.add(guest)

            }
                while (cursor.moveToNext())
        }
        db.close()
        return listguest
    }
    fun addGuest(guest: guest){

        val db = this.writableDatabase
        val values = ContentValues()
        values.put(COL_ID, guest.id)
        values.put(COL_NAME, guest.name)
        values.put(COL_CITY, guest.city)

        db.insert(TABLE_NAME, null, values)
        db.close()
    }

    fun updateGuest(guest: guest):Int{

        val db = this.writableDatabase
        val values = ContentValues()
        values.put(COL_ID, guest.id)
        values.put(COL_NAME, guest.name)
        values.put(COL_CITY, guest.city)

        return db.update(TABLE_NAME, values, "$COL_ID=?", arrayOf(guest.id.toString()))

    }

    fun deleteGuest(guest: guest){

        val db = this.writableDatabase



        db.delete(TABLE_NAME, "$COL_ID=?", arrayOf(guest.id.toString()))
        db.close()

    }
}

不能使用提供的参数调用以下函数:

public constructor SQLiteOpenHelper(@RecentlyNullable p0: Context!, @RecentlyNullable p1: String!, @RecentlyNullable p2: ((SQLiteDatabase!, SQLiteCursorDriver!, String!, SQLiteQuery!) -> Cursor!)!, p3: Int) defined in android.database.sqlite.SQLiteOpenHelper
public constructor SQLiteOpenHelper(@RecentlyNullable p0: Context!, @RecentlyNullable p1: String!, @RecentlyNullable p2: SQLiteDatabase.CursorFactory!, p3: Int) defined in android.database.sqlite.SQLiteOpenHelper
public constructor SQLiteOpenHelper(@RecentlyNullable p0: Context!, @RecentlyNullable p1: String!, p2: Int, @RecentlyNonNull p3: SQLiteDatabase.OpenParams!) defined in android.database.sqlite.SQLiteOpenHelper

Answer 1

SQLiteOpenHelper 构造函数的第一个参数是 Context。 Fragment 不是 Context。

战术上，替换:

class sql(context: FragmentTwo):SQLiteOpenHelper(context, DATABASE_NAME, null, DATABASE_VER) {

与:

class sql(fragment: FragmentTwo):SQLiteOpenHelper(fragment.requireContext(), DATABASE_NAME, null, DATABASE_VER) {

但是，在现代 Android 应用程序开发中， fragment 不应直接与 SQLiteOpenHelper 一起使用。至少，该 fragment 应该与 ViewModel 一起工作，而后者又与单例 SQLiteOpenHelper 一起工作。确保您用于学习 Android 应用程序开发的任何书籍或类(class)都向您展示了如何使用 Jetpack/AndroidX 库，包括 ViewModel 和 LiveData 等内容。