我正在开发一个用于联系人的应用程序,为此,我使用sqlite数据库存储姓名和电话号码,因此我想过滤姓名和电话号码以搜索 我尝试了该查询,但 这对我不起作用 我使用过像“ =”和“ =”一样的问题的任何解决方案吗?
public boolean onQueryTextChange(String newText) {
try {
final String temp = newText.toLowerCase();
final ArrayList<Map> map1 = new ArrayList();
try {
dbhelper =new FeedReaderDbHelper(getActivity());
SQLiteDatabase db=dbhelper.getReadableDatabase();
String projection[]={FeedReaderContract.FeedEntry.COLUMNS_TITLE,FeedReaderContract.FeedEntry.COLUMN_SUB_TITLE};
Cursor cursor =db.query(FeedReaderContract.FeedEntry.TABLE_NAME,projection ,FeedReaderContract.FeedEntry.COLUMNS_TITLE+"= ?",new String[]{newText},null,null,null);
while(cursor.moveToFirst()){
String name = cursor.getString(cursor.getColumnIndex(FeedReaderContract.FeedEntry.COLUMNS_TITLE));
String number = cursor.getString(cursor.getColumnIndex(FeedReaderContract.FeedEntry.COLUMN_SUB_TITLE));
Map<String, Object> map = new HashMap();
map.put(name, number);
if (name.toLowerCase().contains(temp)) {
map1.add(map);
Phone.this.adapter.Filter(map1);
}
} }
catch(Exception e){
System.out.print(Retriving_list);
}
}catch(Exception e){
Toast.makeText(getActivity(), First_Retrive_Data, Toast.LENGTH_SHORT).show();
}
// DbHelperClass
公共类FeedReaderDbHelper扩展了SQLiteOpenHelper {
public static final int DB_VERSION =1;
public static String getDataBASE_NAME() {
return DataBASE_NAME;
}
public static void setDataBASE_NAME(String dataBASE_NAME) {
DataBASE_NAME = dataBASE_NAME;
}
public static String DataBASE_NAME;
Context context;
private static final String SQL_CREATE_ENTRIES =
"CREATE TABLE " + FeedReaderContract.FeedEntry.TABLE_NAME + "(" +
FeedReaderContract.FeedEntry._ID + "INTEGER PRIMARY KEY," +
FeedReaderContract.FeedEntry.COLUMNS_TITLE + " TEXT ," +
FeedReaderContract.FeedEntry.COLUMN_SUB_TITLE + " TEXT UNIQUE)";
private static final String SQL_DELETE_ENTRIES =
"DROP TABLE IF EXISTS " + FeedReaderContract.FeedEntry.TABLE_NAME;
public FeedReaderDbHelper(Context context) {
super(context, FeedReaderDbHelper.getDataBASE_NAME(), null, DB_VERSION);
this.context=context;
}
@Override
public void onCreate(SQLiteDatabase sqLiteDatabase) {
sqLiteDatabase.execSQL(SQL_CREATE_ENTRIES);
}
void addContactValues(FeedReaderDbHelper dbhelper ,String name ,String number){
SQLiteDatabase db= dbhelper.getWritableDatabase();
ContentValues values =new ContentValues();
values.put(FeedReaderContract.FeedEntry.COLUMNS_TITLE ,name);
values.put(FeedReaderContract.FeedEntry.COLUMN_SUB_TITLE,number);
long rows= db.insertWithOnConflict(FeedReaderContract.FeedEntry.TABLE_NAME,null,values,SQLiteDatabase.CONFLICT_IGNORE);
if (rows==-1){
// Toast.makeText(context, String.valueOf(rows)+"not interested", Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(context, String.valueOf(rows)+"inserted", Toast.LENGTH_SHORT).show();
}
}
@Override
public void onUpgrade(SQLiteDatabase sqLiteDatabase, int i, int i1) {
sqLiteDatabase.execSQL(SQL_DELETE_ENTRIES);
onCreate(sqLiteDatabase);
}
}
答案 0 :(得分:0)
目前尚不清楚您想要的是什么,但是如果您只想获取具有给定值的行,它应该像这样:
String mSearchString = "what do you want to search";
String selection = setupSelectionString();
cursor = database.query(NoteContract.NoteEntry.NOTES_TABLE_NAME, projection, selection, selectionArgs,
null, null, sortOrder);
private String setupSelectionString() {
String selection = null;
if (mSearchString != null) {
selection = FeedReaderContract.FeedEntry.COLUMNS_TITLE + " LIKE '%"
+ mSearchString + "%'"
+ " OR " + FeedReaderContract.FeedEntry.COLUMN_SUB_TITLE + " LIKE '%"
+ mSearchString + "%'";
}
return selection;
它可以在我的应用中使用