我在DB中查询两列,其中第一列是第二列的键。 如何将结果列表转换为单个地图? 它甚至可能吗?我刚看过豆子的例子。
List<Map<String, Object>> steps = jdbcTemplate.queryForList("SELECT key, value FROM table");
// well this doesn't work
Map<String, String> result = steps.stream().collect(Collectors.toMap(s -> s.get("key"), s -> s.get("value")));
答案 0 :(得分:8)
您忘记转换密钥和值映射以生成String
:
final Map<String, String> result = steps
.stream()
.collect(Collectors.toMap(s -> (String) s.get("key"), s -> (String) s.get("value")));
完整示例
public static void main(String[] args) {
final List<Map<String, Object>> steps = queryForList("SELECT key, value FROM table");
final Map<String, String> result = steps
.stream()
.collect(Collectors.toMap(s -> (String) s.get("key"), s -> (String) s.get("value")));
result.entrySet().forEach(e -> System.out.println(e.getKey() + " -> " + e.getValue()));
}
private static List<Map<String, Object>> queryForList(String s) {
final List<Map<String, Object>> result = new ArrayList<>();
for (int i = 0; i < 10; i++) {
final Map<String, Object> map = new HashMap<>();
map.put("key", "key" + i);
map.put("value", "value" + i);
result.add(map);
}
return result;
}
打印
key1 -> value1
key2 -> value2
key0 -> value0
key5 -> value5
key6 -> value6
key3 -> value3
key4 -> value4
key9 -> value9
key7 -> value7
key8 -> value8
答案 1 :(得分:0)
问题是你有一张地图清单。下面的代码应该有效:
Map<String, String> result = new HashMap<>();
steps.stream().forEach(map -> {
result.putAll(map.entrySet().stream()
.collect(Collectors.toMap(entry -> entry.getKey(), entry -> (String) entry.getValue())));
});
如果我们尝试运行此示例
Map<String, Object> steps1 = new HashMap<>();
steps1.put("key11", "value11");
steps1.put("key12", "value12");
Map<String, Object> steps2 = new HashMap<>();
steps2.put("key21", "value21");
steps2.put("key22", "value22");
List<Map<String, Object>> steps = new ArrayList<>();
steps.add(steps1);
steps.add(steps2);
Map<String, String> result = new HashMap<>();
steps.stream().forEach(map -> {
result.putAll(map.entrySet().stream()
.collect(Collectors.toMap(entry -> entry.getKey(), entry -> (String) entry.getValue())));
});
System.out.println(result);
很高兴为我们提供了这样的输出:
{key12=value12, key11=value11, key22=value22, key21=value21}
答案 2 :(得分:0)
加上上面的esin88所说的,如果有可能在映射列表中得到重复的键,我们希望得到一个包含键和值列表而不是键值对的映射,这可以像在下面。
public static void main(String[] args) {
final List<Map<String, Object>> steps = queryForList("SELECT key, value FROM table");
/*final Map<String, String> result = steps.stream()
.collect(Collectors.toMap(s -> (String) s.get("key"), s -> (String) s.get("value")));*/
final Map<String, List<String>> result1 = steps.stream()
.collect(Collectors.groupingBy(k -> String.valueOf(k.get("key")),
Collectors.mapping(l -> String.valueOf(l.get("value")), Collectors.toList())));
result1.entrySet().forEach(e -> System.out.println(e.getKey() + " -> " + e.getValue()));
}
private static List<Map<String, Object>> queryForList(String s) {
final List<Map<String, Object>> result = new ArrayList<>();
Map<String, Object> map = new HashMap<>();
for (int i = 0; i < 10; i++) {
map = new HashMap<>();
map.put("key", "key" + i);
map.put("value", "value" + i);
result.add(map);
}
map = new HashMap<>();
map.put("key", "key1");
map.put("value", "value20");
result.add(map);
return result;
}
结果看起来像
key1 -> [value1, value20]
key2 -> [value2]
key0 -> [value0]
key5 -> [value5]
key6 -> [value6]
key3 -> [value3]
key4 -> [value4]
key9 -> [value9]
key7 -> [value7]
key8 -> [value8]
答案 3 :(得分:0)
我们可以使用java流的reduce()将映射列表转换为Java中的单个映射。
请检查以下代码以了解其用法。
例如:
@Data
@AllArgsConstructor
public class Employee {
private String employeeId;
private String employeeName;
private Map<String,Object> employeeMap;
}
public class Test{
public static void main(String[] args) {
Map<String, Object> map1 = new HashMap<>();
Map<String, Object> map2 = new HashMap<>();
Map<String, Object> map3 = new HashMap<>();
map1.put("salary", 1000);
Employee e1 = new Employee("e1", "employee1", map1);
map2.put("department", "HR");
Employee e2 = new Employee("e2", "employee2", map2);
map3.put("leave balance", 14);
Employee e3 = new Employee("e3", "employee3", map3);
//now we create a employees list and add the employees e1,e2 and e3.
List<Employee> employeeList = Arrays.asList(e1,e2,e3);
//now we retreive employeeMap from all employee objects and therefore have a List of employee maps.
List<Map<String, Object>> employeeMaps = employeeList
.stream()
.map(Employee::getEmployeeMap)
.collect(Collectors.toList());
System.out.println("List of employee maps: " + employeeMaps);
// to reduce a list of maps to a single map, we use the reduce function of stream.
Map<String, Object> finalMap = employeeMaps
.stream()
.reduce((firstMap, secondMap) -> {
firstMap.putAll(secondMap);
return firstMap;
}).orElse(null);
System.out.println("final Map: "+ finalMap);
}
}
输出: 员工地图列表:[{salary = 1000},{department = HR},{leave balance = 14}]。
最终地图:{salary = 1000,部门= HR,请假余额= 14}
PS:为扩展答案表示歉意,这是我第一次参加stackoverflow。 谢谢:-)