这是一个简单的!但是,我缺少一些东西。请帮帮我。
在这里,我试图通过id来获取值,但无法这样做。即使更改id的值,它也会返回相同的值。
db = openOrCreateDatabase("DBSOURCE", 0, null);
Cursor cursorc = db.rawQuery("SELECT * FROM LIST WHERE ID="+id+"", null);
cursorc.moveToFirst();
int NameID = cursorc.getColumnIndex("Name");
int mobilenumberID = cursorc.getColumnIndex("MoblieNumber");
edName.setText(cursorc.getString(NameID));
edMobNum.setText(cursorc.getString(mobilenumberID));
cursorc.close();
db.close();
最佳答案
1-
或更好地使用参数化语句
String query = "SELECT COUNT(*) FROM " + tableName + " WHERE columnName = ?";
cursor = db.rawQuery(query, new String[] {comment});
2-与条件c.moveToFirst()或c.getCount()> 0或(!c.isAfterLast())if (c.moveToFirst()){
do{
//if you not need the loop you can remove that
id = c.getInt(c.getColumnIndex("_id"));
}
while(cursor.moveToNext());
}c.close();
一起使用