fix: 修改未登录状态下的返回值,提供更明确的错误信息
This commit is contained in:
@@ -79,10 +79,10 @@ const showMe = async (show = true) => {
|
||||
const localToken = storage.getItem('token');
|
||||
if (!token && !localToken) {
|
||||
console.log('请先登录');
|
||||
return;
|
||||
return { code: 40400, message: '请先登录' };
|
||||
}
|
||||
let me = await queryLogin.getMe(token);
|
||||
if (me.code === 401) {
|
||||
if (me?.code === 401) {
|
||||
me = await queryLogin.getMe();
|
||||
}
|
||||
if (show) {
|
||||
@@ -113,6 +113,9 @@ const command = new Command('me')
|
||||
res = await showMe(false);
|
||||
isRefresh = true;
|
||||
}
|
||||
if (res.code === 40400) {
|
||||
return
|
||||
}
|
||||
if (res.code === 200) {
|
||||
if (isRefresh) {
|
||||
console.log(chalk.green('refresh token success'), '\n');
|
||||
|
||||
Reference in New Issue
Block a user