在一个按钮上实现两个方法

问题描述 投票:-1回答:2

我有两种方法来添加和删除sqlite表中的数据。这两种方法都可以单独使用。我想做的是在一个按钮上实现两种方法。在按钮上单击如果表中存在数据则删除数据,否则添加数据(如果不存在)并更改图像按钮中的可绘制数据。我尝试了很多东西,但我无法弄清楚如何做到这一点。

这是我的代码

public boolean addBookmark(String id, String word, String definition) {

    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(COL_ID, id);
    values.put(COL_WORD, word);
    values.put(COL_DEFINITION, definition);
    db.insert(TABLE_BOOKMARK, null, values);

    return true;
}

public Integer deleteBookmark(String id){

    SQLiteDatabase db = this.getWritableDatabase();
    return db.delete(TABLE_BOOKMARK, "id = ?",new String[]{id});
}

我做了类似的事情,但它不起作用我在logcat中得到错误。

  public  boolean isBookmark(String id) {
        SQLiteDatabase db = this.getWritableDatabase();
        String Query = "Select * from " + TABLE_BOOKMARK + " where " + id + " = ?" ;
        Cursor cursor = db.rawQuery(Query, null);
        if(cursor.getCount() <= 0){
            cursor.close();
            return false;
        }
        cursor.close();
        return true;
    }

并在onclick

  public void onClick(View v) {

            boolean isBookmark = mDBHelper.isBookmark(id);

            if (isBookmark){

                mDBHelper.deleteBookmark(id);
                btnBookmark.setImageResource(R.drawable.ic_bookmark_border);
            } else {
                mDBHelper.addBookmark(id,word,definition);
                btnBookmark.setImageResource(R.drawable.ic_bookmark_fill);
            }

这是我的logcat错误

2019-02-22 19:36:53.391 28210-28210/com.elytelabs.testnav E/SQLiteDatabase: Error inserting definition=Create a new folder by going to New > Directory . In the dialog box that opens name the directory fragment or any name of your choice.
Create a new Fragment file inside the created directory and name it dictionary or any word of your choice.
  id=7 word=rvvg
android.database.sqlite.SQLiteConstraintException: UNIQUE constraint failed: bookmark.id (code 1555 SQLITE_CONSTRAINT_PRIMARYKEY)
    at android.database.sqlite.SQLiteConnection.nativeExecuteForLastInsertedRowId(Native Method)
    at android.database.sqlite.SQLiteConnection.executeForLastInsertedRowId(SQLiteConnection.java:796)
    at android.database.sqlite.SQLiteSession.executeForLastInsertedRowId(SQLiteSession.java:788)
    at android.database.sqlite.SQLiteStatement.executeInsert(SQLiteStatement.java:86)
    at android.database.sqlite.SQLiteDatabase.insertWithOnConflict(SQLiteDatabase.java:1564)
    at android.database.sqlite.SQLiteDatabase.insert(SQLiteDatabase.java:1433)
    at com.elytelabs.testnav.database.DatabaseHelper.addBookmark(DatabaseHelper.java:125)
    at com.elytelabs.testnav.DetailActivity$1.onClick(DetailActivity.java:68)
    at android.view.View.performClick(View.java:6597)
    at android.view.View.performClickInternal(View.java:6574)
    at android.view.View.access$3100(View.java:778)
    at android.view.View$PerformClick.run(View.java:25885)
    at android.os.Handler.handleCallback(Handler.java:873)
    at android.os.Handler.dispatchMessage(Handler.java:99)
    at android.os.Looper.loop(Looper.java:193)
    at android.app.ActivityThread.main(ActivityThread.java:6669)
    at java.lang.reflect.Method.invoke(Native Method)
    at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:493)
    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:858)
android android-sqlite buttonclick android-imagebutton
2个回答
0
投票

我认为isBookmark中的以下代码行不正确:

String Query = "Select * from " + TABLE_BOOKMARK + " where " + id + " = ?" ;

我相信会创建一个如下所示的select语句:

"Select * from TABLE where 5 = ?"

*当然我不知道你的TABLE_BOOKMARK const值究竟是什么,所以我将表名放在TABLE中。并且,我随意使用了id值为5。

声明应该形成如下:

String Query = "Select * from " + TABLE_BOOKMARK + " where id = " + id;

这似乎也有意义,因为这意味着你的isBookmark永远不会删除你希望它删除的记录值(因为它永远不会成功找到与查询匹配的记录)。这意味着插入总是失败,因为记录实际上仍然存在。


0
投票

首先需要检查数据是否存在,然后可以添加或删除数据。

例如,使用这样的方法。

public static boolean CheckIsDataAlreadyInDBorNot(String TableName,
        String dbfield, String fieldValue) {
    SQLiteDatabase sqldb = EGLifeStyleApplication.sqLiteDatabase;
    String Query = "Select * from " + TableName + " where " + dbfield + " = " + fieldValue;
    Cursor cursor = sqldb.rawQuery(Query, null);
        if(cursor.getCount() <= 0){
            cursor.close();
            return false;
        }
    cursor.close();
    return true;
}

取自https://stackoverflow.com/a/20416004/3106174

然后,你可以做到。

if( CheckIsDataAlreadyInDBorNot( ... ) ) {
    deleteBookmark( ... );
} else {
    addBookmark( ... );
}
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