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?
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);
}
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);
}
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]}