Question

我目前正在调整Android应用程序，但是有一个我无法解决的相当简单的问题。

在“ onClick”之后，我想在我的数据库中检索特殊数据（在这种情况下，是一种加密货币的名称），但是该数据存在并且名称很好，问题出在请求中，但不是解决方法。

由以下原因引起：android.database.sqlite.SQLiteException：无此类列：名称（代码1 SQLITE_ERROR）：，而在编译时：选择* FROM COINS WHERE cryptoName = name

这是我的sqliteDatabase.java的源代码

package com.example.cryptoapp;

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

import java.util.ArrayList;

public class SqliteDatabase extends SQLiteOpenHelper {

    private static final int DATABASE_VERSION = 5;
    private static final String DATABASE_NAME = "cryptoBase";
    private static final String TABLE_COINS = "COINS";

    private static final String COLUMN_ID = "_id";
    private static final String COLUMN_CRYPTONAME = "cryptoName";
    private static final String COLUMN_QUANTITY = "quantity";
    private static final String COLUMN_TOTALVALUE = "totalValue";

    public SqliteDatabase(Context context) {
        super(context, DATABASE_NAME, null, DATABASE_VERSION);
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        String  CREATE_COINS_TABLE = "CREATE    TABLE " + TABLE_COINS + "(" + COLUMN_ID + " INTEGER PRIMARY KEY," + COLUMN_CRYPTONAME + " TEXT," + COLUMN_QUANTITY + " TEXT," + COLUMN_TOTALVALUE + " TEXT" + ")";
        db.execSQL(CREATE_COINS_TABLE);
    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        db.execSQL("DROP TABLE IF EXISTS " + TABLE_COINS);
        onCreate(db);
    }

    public ArrayList<Coins> listCoins(){
        String sql = "select * from " + TABLE_COINS;
        SQLiteDatabase db = this.getReadableDatabase();
        ArrayList<Coins> storeCoins = new ArrayList<>();
        Cursor cursor = db.rawQuery(sql, null);
        if(cursor.moveToFirst()){
            do{
                int id = Integer.parseInt(cursor.getString(0));
                String cryptoname = cursor.getString(1);
                String quantity = cursor.getString(2);
                String totalvalue = cursor.getString(3);
                storeCoins.add(new Coins(id, cryptoname, quantity, totalvalue));
            }while (cursor.moveToNext());
        }
        cursor.close();
        return storeCoins;
    }

    public void addCoins(Coins coins){
        Log.d("TESTINADD", coins.getCryptoName());
        ContentValues values = new ContentValues();
        values.put(COLUMN_CRYPTONAME, coins.getCryptoName());
        values.put(COLUMN_QUANTITY, coins.getQuantity());
        values.put(COLUMN_TOTALVALUE, coins.getTotalValue());
        SQLiteDatabase db = this.getWritableDatabase();
        db.insert(TABLE_COINS, null, values);
    }

    public void updateCoins(Coins coins){
        ContentValues values = new ContentValues();
        values.put(COLUMN_CRYPTONAME, coins.getCryptoName());
        values.put(COLUMN_QUANTITY, coins.getQuantity());
        values.put(COLUMN_TOTALVALUE, coins.getTotalValue());
        SQLiteDatabase db = this.getWritableDatabase();
        db.update(TABLE_COINS, values, COLUMN_ID    + " = ?", new String[] { String.valueOf(coins.getId())});
    }

    public Coins findCrypto(String name){
        String query = "Select * FROM " + TABLE_COINS + " WHERE " + COLUMN_CRYPTONAME + " = " + "name";
        SQLiteDatabase db = this.getWritableDatabase();
        Coins coins = null;
        Log.d("TESTquery", query);
        Cursor cursor = db.rawQuery(query,  null);
        if  (cursor.moveToFirst()){
            int id = Integer.parseInt(cursor.getString(0));
            String mName = cursor.getString(1);
            String mQuantity = cursor.getString(2);
            String mTotalValue = cursor.getString(3);
            coins = new Coins(id, mName, mQuantity, mTotalValue);
        }
        cursor.close();
        Log.d("TEST", name);
        Log.d("TEST1", coins.getCryptoName());
        return coins;
    }

    public void deleteCoins(int id){
        SQLiteDatabase db = this.getWritableDatabase();
        db.delete(TABLE_COINS, COLUMN_ID    + " = ?", new String[] { String.valueOf(id)});
    }
}

提前谢谢！

Answer 1

String query = "Select * FROM " + TABLE_COINS + " WHERE " + COLUMN_CRYPTONAME + " = " + "name";

这里应该是：

String query = "Select * FROM " + TABLE_COINS + " WHERE " + COLUMN_CRYPTONAME + " LIKE " + "'" + name + "'";

Answer 2

您将名称作为字符串而不是变量进行传递，因此它经过了硬编码，因此数据库中没有名为名称的表，因此您遇到了例外情况，可以解决此问题，使查询像这样

String query = "Select * FROM " + TABLE_COINS + " WHERE " + COLUMN_CRYPTONAME + " = " + name;

android.database.sqlite.SQLiteException：无此类列：名称（代码1 SQLITE_ERROR）

2 个答案: