我米试图删除通过Android应用程序一个SQLiteDB的记录，这是我使用的code删除记录的方式

// code的片段在我DBAdapter类公共布尔DeleteRecord（字符串联系人姓名）{    Log.i（TAGDeleteRecord（字符串联系人姓名））;    Log.i（TAG名称，联系人姓名）;    返回db.delete（TABLE_SIMPLETABLE_CLIENT1，KEY_EMPLOYEE_NAME +=             +联系人姓名，NULL）;    }

和从另一个类，我试图打开数据库，并调用此方法，然后再次关闭数据库调用这个方法，这是我怎么也得codeD，

//我DBApplication类code的段

公共无效的onClick（DialogInterface对话框，    其中INT）{DBAdapter.open（）;DBAdapter.DeleteRecord（DeleteRecord）;// DeletRecord是字符串值，在DB（Sivaram）DBAdapter.close（）;}

当我部署它，我一收到以下错误，

08-25 14：52：20.602：ERROR / AndroidRuntime（231）：android.database.sqlite.SQLiteException：没有这样的列：Sivaram：，在编译：DELETE FROM WHERE SimpleTable1 = Employee_Name Sivaram

请帮助我，找到了解决办法

在此先感谢...

 编辑 -  

试试这个，CHK如果它工作第一

 返回db.delete（TABLE_SIMPLETABLE_CLIENT1，KEY_EMPLOYEE_NAME +='+联系人姓名+'，NULL）; 

在这里你的字符串应该是你的where子句（第3位），你只需要提及你的表名作为你的第二个参数，我相信

 删除（字符串tableName值，字符串whereClause，字符串[] whereArgs）{}返回db.delete（TABLE_SIMPLETABLE_CLIENT1，KEY_EMPLOYEE_NAME，新的String [] {联系人姓名}）; 

I m trying to delete a record from a SQLiteDB via Android Application,This is the way that i used the code to delete the record

//Snippet of code in my DBAdapter Class

public boolean DeleteRecord(String ContactName) {

    Log.i(TAG, "DeleteRecord(String ContactName)");
    Log.i(TAG, ContactName);
    return db.delete(TABLE_SIMPLETABLE_CLIENT1,KEY_EMPLOYEE_NAME + "="
             +ContactName,null);
    }

and from the another class, i trying to call this method by opening the DB and calling this method and again closing the DB, this is how i have coded,

//Snippet of code in my DBApplication Class

public void onClick(DialogInterface dialog,
    int which) {
DBAdapter.open();
DBAdapter.DeleteRecord(DeleteRecord);  
//DeletRecord is string value that in DB(Sivaram)                       
DBAdapter.close();
}

When i deploy it , i an getting the following error,

08-25 14:52:20.602: ERROR/AndroidRuntime(231): android.database.sqlite.SQLiteException: no such column: Sivaram: , while compiling: DELETE FROM SimpleTable1 WHERE Employee_Name=Sivaram

please help me, in finding out the solution

Thanks in advance...

Edit-

Try this and chk if it works first

return db.delete(TABLE_SIMPLETABLE_CLIENT1,KEY_EMPLOYEE_NAME + "='"+ContactName+"'",null);

here your string should be in your where clause(3rd position) and you just require to mention your table name as your second parameter i believe

delete(String tableName, String whereClause, String[] whereArgs) {}
return db.delete(TABLE_SIMPLETABLE_CLIENT1,KEY_EMPLOYEE_NAME,new String[]{ContactName});