0

如果我想使用 java 在下面的 sql 数据库中搜索一个字符串，我该怎么做？老实说，我不知道从哪里开始他们的任何在线材料都涵盖了这一点？...谢谢。我虽然可以有某种 strstr() 函数或收集表中字符串的东西。

package f.s.l;

import android.content.ContentValues;
import android.content.Context;
import android.database.Cursor;
import android.database.SQLException;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteDatabase.CursorFactory;
import android.database.sqlite.SQLiteOpenHelper;

public class HotOrNot {
public static final String KEY_ROWID ="_id";
public static final String KEY_NAME ="persons_name";
public static final String KEY_HOTNESS ="persons_hotness";

private static final String DATABASE_NAME ="HotOrNotdb";
private static final String DATABASE_TABLE ="peopleTable";
private static final int DATABASE_VERSION =1;

private DbHelper ourHelper;
private final Context ourContext;
private SQLiteDatabase ourDatabase;

private static class DbHelper extends SQLiteOpenHelper{

public DbHelper(Context context) {
super(context, DATABASE_NAME, null, DATABASE_VERSION);
// TODO Auto-generated constructor stub
}

@Override
public void onCreate(SQLiteDatabase db) {
// TODO Auto-generated method stub
db.execSQL("CREATE TABLE " + DATABASE_TABLE + " (" +
        KEY_ROWID + " INTEGER PRIMARY KEY AUTOINCREMENT, " +
        KEY_NAME + " TEXT NOT NULL, " +
        KEY_HOTNESS + " TEXT NOT NULL);"


        );

}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
// TODO Auto-generated method stub
db.execSQL("DROP TABLE IF EXISTS " + DATABASE_TABLE);
onCreate(db);

}

}
public HotOrNot(Context c){
ourContext =c;
}
public HotOrNot open() throws SQLException{ 
ourHelper = new DbHelper(ourContext);
ourDatabase = ourHelper.getWritableDatabase();
return this;
}
public void close(){

 ourHelper.close();
 }
 public long createEntry(String name, String hotness) {
 // TODO Auto-generated method stub
 ContentValues cv = new ContentValues();
 cv.put(KEY_NAME, name);
 cv.put(KEY_HOTNESS, hotness);
 return ourDatabase.insert(DATABASE_TABLE, null, cv);

 }
 public String getData() {
 // TODO Auto-generated method stub
String [] columns = new String[]{ KEY_ROWID, KEY_NAME, KEY_HOTNESS};
Cursor c = ourDatabase.query(DATABASE_TABLE, columns, null, null, null, null, null);
String result = "";

int iRow = c.getColumnIndex(KEY_ROWID);
int iName = c.getColumnIndex(KEY_NAME);
int iHotness = c.getColumnIndex(KEY_HOTNESS);

for(c.moveToFirst(); !c.isAfterLast(); c.moveToNext()) {


result = result + c.getString(iRow) + " "+ c.getString(iName) + " " +                c.getString(iHotness) + "\n";
}



return result;
}
}
4

4 回答 4

5

在query方法中，您需要声明 WHERE 子句。例如：

Cursor c = ourDatabase.query(DATABASE_TABLE, columns, "some_col like " + "'%Somevalue%'", null, null, null, null);
于 2012-04-29T23:47:57.007 回答
2

LIKE 是一个很好的 SQL 关键字供您查找...

于 2012-04-30T03:08:52.387 回答
1

尝试以下代码进行搜索：-

public Cursor getROUTINENAME(String search_str) {
        String query = "SELECT * FROM TBNAME where COL_NAME LIKE '%"+search_str+"%'";

        Cursor mCursor = db.rawQuery(query, null);
        if (mCursor != null) {
            mCursor.moveToFirst();
        }
        return mCursor;
    }

用 string 调用上面的函数。

于 2013-09-03T14:02:31.870 回答
0

字符串查询 = " SELECT * FROM " + DB_NAME + " WHERE " + COLUMN_dars + " LIKE " + "'"+select+"'";

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(query,null);
于 2016-07-17T04:00:08.773 回答