将Resultset转换为Map

问题描述 投票:0回答:3

我有查询将返回两个值(列A和列B),如下所示

A        B
------------
a   aaa
a   aaa
a   aaa
a   aaa
b   bbb
c   ccc
c   ccc
b   bbb
c   ccc
b   bbb

我正在尝试创建一个java方法(Java 7),它将一次性获取所有这些值并将其存储在集合变量(Map)中,就像下面格式中的所有值一样

(a -> (aaa,aaa,aaa,aaa,aaa),
 b -> (bbb,bbb,bbb),
 c -> (ccc,ccc,ccc))

下面是我正在尝试的方法,但我甚至无法在第一时间获取所有数据:

import java.sql.*;
import java.util.ArrayList;

public class CollectionFrame {


    public static void main(String[] args) {

        try {
            // step1 load the driver class
            Class.forName("oracle.jdbc.driver.OracleDriver");

            // step2 create the connection object
            Connection con = DriverManager.getConnection("jdbc:oracle:thin:@localhost:1521:xe", "hr", "hr");

            // step3 create the statement object
            Statement stmt = con.createStatement();

            // step4 execute query

            // Lists of Lists to store the values
            ArrayList<ArrayList<String>> listOLists = new ArrayList<ArrayList<String>>();
            ArrayList<String> obj = new ArrayList<String>();

            ResultSet rs = stmt.executeQuery("select * from t");
            while (rs.next()) {
                // System.out.println(rs.getString(1) + " " + rs.getString(2));
                obj.add(rs.getString(1));
                // obj.add(rs.getString(2));
                listOLists.add(obj);
                obj.removeAll(obj);

            }

            // step5 close the connection object
            con.close();

            System.out.println(listOLists.toString());

        } catch (Exception e) {
            System.out.println(e);
        }

    }

}

上面的代码给出了以下结果

[[a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b], [a, a, a, a, b, c, c, b, c, b]]

如果我取消注释行obj.removeAll(obj);我得到以下内容:

[[], [], [], [], [], [], [], [], [], []]

我被困在这里。有人可以帮助我如何继续或建议更好的解决方案吗?

collections java-7 resultset
3个回答
2
投票

你应该使用一张地图。

ResultSet rs = stmt.executeQuery("select * from t");
Map<String, List<String>> valueMap = new HashMap<>();

while (rs.next()) {
    String columnAstring = rs.getString(1);
    String columnBstring = rs.getString(2);

    valueMap.putIfAbsent(columnAstring, new ArrrayList<>());
    valueMap.get(columnAstring).add(columnBstring);

}

编辑:因为我将创造和抛弃许多arraylists,所以putifabsent可能是非常低效的。正如@Andreas指出的那样。所以这将是一个不那么清洁,但更有效的方式来做到这一点!

与JAVA 7兼容

ResultSet rs = stmt.executeQuery("select * from t");
Map<String, List<String>> valueMap = new HashMap<>();

while (rs.next()) {
    String columnAstring = rs.getString(1);
    String columnBstring = rs.getString(2);

    if(!valueMap.containsKey(columnAstring)){
        valueMap.put(columnAstring, new ArrayList());
    }
    valueMao.get(columnA).add(columnBstring);

}

使用Java 8 Lambdas

@ Mureinik的answer指出了使用computeIfAbsent更清晰的方法。

while (rs.next()) {
    String columnAstring = rs.getString(1);
    String columnBstring = rs.getString(2);

    valueMap.computeIfAbsent(columnAstring, k -> new ArrayList<>())
    valueMap.get(columnAstring).add(columnBstring);
}

1
投票

我将迭代ResultSet并将更改应用于地图Map<String, List<String>>。在每次迭代中,如果键(列A)不存在,则需要使用空列表添加它,然后一旦确定有该键的列表,则附加B列中的值。幸运的是, Java 8对Map界面的改进使其非常优雅:

Map<String, List<String>> result = new HashMap<>();
while (rs.next()) {
    String a = rs.getString("a");
    String b = rs.getString("b");

    result.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
}

1
投票

也许你可以试试这个。这使用HashMap。

public class Main {
public static void main(String[] args) {

    //sample CSV strings...pretend they came from a file
    String[] csvStrings = new String[] {
      "a aaa","a aaa","a aaa","a aaa","b bbb","b bbb","b bbb",
      "b bbb","c ccc","c ccc","c ccc"
    };

    List<List<String>> csvList = new ArrayList<List<String>>();
    Map<String,ArrayList<String>> outVal 
        = new HashMap<String, ArrayList<String>>();

    for(String val:csvStrings){
        String outList[] = val.split(" ");
        if (outVal.containsKey(outList[0])){
            outVal.get(outList[0]).add(outList[1]);
        } else {
            ArrayList<String> inputList = new ArrayList<String>();
            inputList.add(outList[1]);
            outVal.put(outList[0],inputList);
        }
    }

   System.out.println(outVal.toString());  
}

}

这是输出:{a = [aaa,aaa,aaa,aaa],b = [bbb,bbb,bbb,bbb],c = [ccc,ccc,ccc]}

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