ArrayList allows adding null values, which can easily cause java.lang.NullPointerException exceptions when converting objects in the list.
Scenario:
Database select min(id) as id,min(name) as name from user where 1=2;
Query It's not that there is no record, but a record of null, null, which causes the User object to be null
List
At this time, an error will be reported when operating the user object!
Solution
Method 1: Remove null elements
Example:
List<user> users = new ArrayList<user>(); users.add(null); users.add(null); users.add(null); System.out.println("size:"+users.size()); //size:3for(User user:users){try { System.out.println("id:" + user.getId() + ",name:" + user.getName()); }catch (Exception ex){ System.out.println(ex); //java.lang.NullPointerException } }users.remove(null); //移除第一个nullSystem.out.println("size:"+users.size()); //size:2 users.removeAll(Collections.singleton(null)); //移除所有的null元素System.out.println("size:"+users.size()); //size:0//不会进入循环for(User user:users){try { System.out.println("id:" + user.getId() + ",name:" + user.getName()); }catch (Exception ex){ System.out.println(ex); } }</user></user>
<span style="font-size: 14px">法二:保证数据库查询出来没有null值,即遇null值转换为默认值<br><br><br></span><span style="font-family: " microsoft yahei>mysql下使用ifnull/case when均可实现</span><br><span style="font-family: " microsoft yahei>select ifnull(min(id),-1) as id,ifnull(min(name),'defaultName') as name from user where 1=2;</span>
(有group by是分组统计,找不到记录就是没有记录)
<br>
<span style="font-size: 14px"><em><br><br></em></span>
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