Question

我将字符串传递给下面的代码并且它给出了错误：

Error: 08-24 12:57:55.583: E/AndroidRuntime(11640): android.database.sqlite.SQLiteException:    near "RAMDEV": syntax error: , while compiling: SELECT chnno FROM EPG WHERE title = BABA RAMDEV KA YOGA

我的代码在这里：

public String SearchChnNo(String title){

    String selectQuery = "SELECT " + chn_no + " FROM " + EPG + " WHERE " + pgm_title + " = " + title ; //error

    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);
    String chnno = cursor.getString(1);
    return chnno;

}

Answer 1

Lalit Poptani的回答是正确的，但我建议您使用原始查询方法的选择参数，方法如下：

每次您需要在查询中写入值时，请改为编写问号。

然后，您必须在字符串数组中添加值本身。

在您的情况下，它将如下所示：

public String SearchChnNo(String title){

    String selectQuery = "SELECT " + chn_no + " FROM " + EPG + " WHERE " + pgm_title + " = ? " ;
    String selectionArgs [] = new String [] { title };

    SQLiteDatabase db = this.getReadableDatabase ();
    Cursor cursor = db.rawQuery ( selectQuery , selectionArgs );
    // The default value, if none is found
    String chnno = "N/A";
    if ( cursor.moveToFirst () )
        chnno = cursor.getString ( 1 );
    // Do not forget to close the cursor
    cursor.close ();
    cursor = null;
    return chnno;

}

Answer 2

期待pgm_title是String，您必须将title放入单引号中，

pgm_title + " = '" + title+"'"

因此，您的最终查询将是，

String selectQuery = "SELECT " + chn_no + " FROM " + EPG + " WHERE " + 
                                             pgm_title + " = '" + title+"'" ;

Answer 3

要添加Leeeeelo和Lalit的答案，你应该做的是：

SQLiteDatabse db = this.getReadableDatabase();
Cursor cursor = db.query(EPG, new String[] {chn_no}, ""+pgm_title+"=?", new String [] { title}, null, null, null);
cursor.moveToFirst(); // that assumes there are results, you should probably make sure this returns true
String chnno = cursor.getString(1);
cursor.close();
return chnno;

Answer 4

   public String SearchChnNo(String title){

    String chnno = null;

   String selectQuery = "SELECT " + chn_no + " FROM " + EPG + " WHERE " + pgm_title + "   = '" + title + "'" ;
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);


    if (cursor.moveToFirst()) {
        do {
            chnno = cursor.getString(cursor.getColumnIndex(chn_no));
            //chnno = cursor.getString(chn_no);
        } while (cursor.moveToNext());
    }

    // closing connection
    cursor.close();
    db.close();

    // returning
    return chnno;

  }

SQLite数据库错误在android中选择查询

4 个答案: