如何处理错误以停止函数执行

Question

它会转到突变中的 onError，但是在 try catch 中，下一个响应的状态为 200，因此它不会执行 catch 方法。我应该怎么做，这样如果它转到 onError，它就会捕获。 我也愿意接受任何可以正确返回/结束函数的解决方案

const useValidatePassword = () =>{  
    return useMutation({
      mutationFn: async (data: any) => {
          const body: validatePasswordPayload = 
            {
               UserName: data.username,
               Password: data.currentPassword,
            };
         await apiAccountSettings.validatePassword(tenantName, body);
        }
      ,
      onSuccess: () => {
        return true
      },
      onError: (data) => {
        toast.error(t('Current password is incorrect'));
      },
      mutationKey: ['ValidatePassword', { tenant: tenantName }],
    });
}
       const {mutate: validatePassword, error} = useValidatePassword();

      Somefn()=>{
        try {
            await validatePassword(validatePayload);
        } catch (error) {
            toast.error(t("Something went wrong"));
            return
        }
      }

Answer 1

通过在

onError

回调中抛出错误，可以确保错误传播到

try-catch

块。

new Promise

validatePassword

调用的包装器确保

try-catch

块可以等待突变结果并捕获引发的任何错误。

const useValidatePassword = () => {
  return useMutation({
    mutationFn: async (data: any) => {
      const body: validatePasswordPayload = {
        UserName: data.username,
        Password: data.currentPassword,
      };
      await apiAccountSettings.validatePassword(tenantName, body);
    },
    onSuccess: () => {
      return true;
    },
    onError: (error) => {
      toast.error(t('Current password is incorrect'));
      throw error; // Throw the error to propagate it
    },
    mutationKey: ['ValidatePassword', { tenant: tenantName }],
  });
};

const { mutate: validatePassword } = useValidatePassword();

const Somefn = async () => {
  try {
    await new Promise((resolve, reject) => {
      validatePassword(validatePayload, {
        onSuccess: resolve,
        onError: reject,
      });
    });
  } catch (error) {
    toast.error(t("Something went wrong"));
    return;
  }
};