由於在運行時出現語法錯誤而在顯示列表視圖時出錯

Question

如何解決SQLiteLog（3507）：（1）在「=」附近：在android中語法錯誤sqlite。我的數據庫在錯誤活動中正常工作。該logcat的顯示在第1行

public void openAndQueryDatabase() 
{ 
    try 
{   
db = openOrCreateDatabase("mydatabase.db", SQLiteDatabase.CREATE_IF_NECESSARY , null); 

     Cursor cursor = db.rawQuery("select * from "+ table + "where name='"+ name + "'", null); //line 1 

     int count = cursor.getColumnCount(); 


     if (cursor != null) 
     { 
      if (cursor.moveToFirst()) 
      { 
       do 
       { 
        for (int i =0 ; i< count; i++) 
        { 
        String data = cursor.getString(i); 
        details.add(data); 
        } 
       } 
       while (cursor.moveToNext()); 
      } 
     }  

    } 
    catch (SQLiteException se) 
    { 

     Log.e(getClass().getSimpleName(), "Error in the code"); 
    } 

    finally 
    { 

      db.close(); 
    } 

} 

Answer 1

你需要和where關鍵字的表名之間的空間中的語法錯誤：

Cursor cursor = db.rawQuery("select * from "+ table + " where name='"+ name + "'", null); 

還要考慮使用?綁定args作爲文字，例如

Cursor cursor = db.rawQuery("select * from "+ table + " where name=?", new String[] { name });