map.entrySet().stream() .filter(entry -> entry.getValue()instanceofString)// 过滤出值为String类型的条目 .forEach(entry ->System.out.println("Key: " + entry.getKey() +", Value: " + entry.getValue())); 以上每种方法都可以有效地遍历Map<String, Object>。选择哪种方法取决于你的具体需求...
//第一种:普遍使用,二次取值 System.out.println("通过Map.keySet遍历key和value:"); for(String key : map.keySet()) { System.out.println("key= "+ key +" and value= "+ map.get(key)); } //第二种 System.out.println("通过Map.entrySet使用iterator遍历key和value:"); Iterator<Map.Entry<...
List<HashMap<String,Object>> DataSource; int i=0; for (Map<String, Object> item : DataSource) { for (String k : item.keySet()) { if (k.equals("input_date")) { //此句为判断条件可去掉 Object b = item.get(k); String data=b.toString(); boolean status = data.contains(".0")...
通过map.entrySet()转成Set 再遍历。上面的方法是加入list集合再遍历 Set<Map.Entry<String, Object>> entries = map.entrySet();for (Map.Entry<String, Object> entry:entries) { System.out.println("entry.getKey()---"+entry.getKey()+"===entry.getValue()---"+entry.getValue()...
1. 通过Map.entrySet()遍历key和value,推荐 Map<String,Object>map=newHashMap<String,Object>();map.put("aaa",111);map.put("bbb",222);map.put("ccc",333);for(Map.Entry<String,Object>m:map.entrySet()){System.out.println("key:"+m.getKey()+" value:"+m.getValue());} ...
public static void main(String[] args) throws Exception { List<Map<String, Object>> testList = new ArrayList<>();Map<String, Object> m1 = new HashMap<>();m1.put("a", "a1");m1.put("b", "b1");m1.put("x", "x1");Map<String, Object> m2 = new HashMap<>();...
//第一种方式foreach遍历,先遍历listmap,然后再遍历map,通过key拿value System.out.println("***foreach遍历***"); for(Map<String,Object> map : lstp){ for(String key : map.keySet()){ System.out.println(key+":"+map.get(key)); } } //普通for循环; ...
* 遍历List<Map<String,Object>>数据结构，根据指定的key，把所有的value取出来 * [{empid=397}, {empid=604}, {empid=473300}, {empid=598925}]* 数据库查询出来的结果，是上⾯这样的数据格式，取指定字段(empid)值得时候，⽤这种⽅法 */ public static void printList(List<Map<String,Object>> ...
Object>>();Map<String,Object>map1=newHashMap<String,Object>();map1.put("name","p");map1.put("cj","5");Map<String,Object>map2=newHashMap<String,Object>();map2.put("name","h");map2.put("cj","12");Map<String,Object>map3=newHashMap<String,Object>();map3.put("name","...