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🎨 优化菜单数据返回结果，只返回顶级菜单的rank值

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This commit is contained in:
李寻欢 2024-02-02 11:45:19 +08:00
parent 1d1ca21815
commit 63c8c3a292
2 changed files with 9 additions and 6 deletions
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2
model/vo/menu/menu.go
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@ -15,7 +15,7 @@ type Item struct {
type ItemMeta struct { type ItemMeta struct {
Title string `json:"title"` // 标题 Title string `json:"title"` // 标题
Icon string `json:"icon"` // 图标 Icon string `json:"icon"` // 图标
Rank int `json:"rank"` // 排序 Rank int `json:"rank,omitempty"` // 排序
Roles []string `json:"roles"` // 当前菜单所属的角色代码 Roles []string `json:"roles"` // 当前菜单所属的角色代码
Auths []string `json:"auths"` // 当前菜单包含的按钮，如果传入了用户Id，返回的将会是权限内所有的按钮 Auths []string `json:"auths"` // 当前菜单包含的按钮，如果传入了用户Id，返回的将会是权限内所有的按钮
} }

3
service/menu/utils.go
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@ -30,7 +30,10 @@ func toTree(records []menuRecordItem, pid string) (tree []menu.Item) {
var meta menu.ItemMeta var meta menu.ItemMeta
meta.Title = record.Title meta.Title = record.Title
meta.Icon = record.Icon meta.Icon = record.Icon
// rank只有顶级菜单才返回rank值
if record.ParentId == nil || *record.ParentId == "" {
meta.Rank = record.Sort meta.Rank = record.Sort
}
if record.RoleCode != "" { if record.RoleCode != "" {
meta.Roles = strings.Split(record.RoleCode, ",") meta.Roles = strings.Split(record.RoleCode, ",")
} }