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问题描述

Future login() async { insert ( <String, dynamic> { "username": useremail.text, "password": password.text } );

          x = await  result.data.toString();
         if(x == null) {

           _showInSnackBar("somthing went wrong");


         }
         if(x != null)
         {
           _showInSnackBar("success");
         }

    }

标签: fluttergraphqlflutter-graphql

解决方案


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