I want to send this List example :
我想发送这个List示例:
{"id":[1]}
To this controller :
到这个控制器:
public String addUsersToProject(@RequestBody List<String> usersIds, @PathVariable String projectTitle){..}
But I can't read the list :
但我无法阅读清单:
Could not read document: Can not deserialize instance of java.util.ArrayList
Any suggestion will be appreciated, Thank you.
任何建议将不胜感激,谢谢。
1 个解决方案
#1
0
The easiest way is create wrapper class with id field as List (userIds).
最简单的方法是创建包含类,其id字段为List(userIds)。
Example:
class IdsWrapper {
private List<Integer> id;
}
And use it in controller:
并在控制器中使用它:
public String addUsersToProject(@RequestBody IdsWrapper ids) {...}
Also you can check link Possible maping solutions
您也可以查看链接可能的maping解决方案
#1
0
The easiest way is create wrapper class with id field as List (userIds).
最简单的方法是创建包含类,其id字段为List(userIds)。
Example:
class IdsWrapper {
private List<Integer> id;
}
And use it in controller:
并在控制器中使用它:
public String addUsersToProject(@RequestBody IdsWrapper ids) {...}
Also you can check link Possible maping solutions
您也可以查看链接可能的maping解决方案