How to handled error in onError ?

[lyquocnam]

Dear author, i handled error in onError, so i don't want use try catch it again in my code. How to do that ?
i tried return null; in onError but not work.

 dynamic _onError(DioError e) {
    // handled..
   // ....
    return null; // <---
  }

currently i must do this to ignore exception, because it was handled as above:

try {
var response = await dio.post(....);
} catch (e) {}

[wendux]

A request must correspond to a response, even if it failed.
you can handle successful response by only suppling callback to then instead of using async...await to ignore exception:

dio.get(...).then((r)=>print(r))

[jlcool]

@wendux 请求需要做单独的异常处理还必须得用await，不能用then。使用then，有异常catch不到，能不能不抛出异常而是返回一个异常状态，直接用then获取处理，比如这样
.then((response,error){ if(error!=null){ } })

[jlcool]

@wendux 请求需要做单独的异常处理还必须得用await，不能用then。使用then，有异常catch不到，能不能不抛出异常而是返回一个异常状态，直接用then获取处理，比如这样
.then((response,error){ if(error!=null){ } })

onError 返回Response 就可以