| | |
| | | import { |
| | | queryDepTree, |
| | | select_Role_ByPageAndCount, |
| | | |
| | | roleResInserts, |
| | | |
| | | roleLayer_selectResByRole, |
| | | roleLayer_inserts, |
| | | roleLayer_deletes |
| | |
| | | if (std.length == 0) { |
| | | return |
| | | } |
| | | |
| | | const data = await roleLayer_inserts(std); |
| | | |
| | | if (data.code != 200) { |
| | | this.$message.error('授权失败'); |
| | | } else { |
| | |
| | | return cloneData.filter((father) => { |
| | | // 循环所有项 |
| | | let branchArr = cloneData.filter((child) => father.layerid == child.pid); // 对比ID,分别上下级菜单,并返回数据 |
| | | |
| | | branchArr.sort(function (a, b) { |
| | | return a.orderNum - b.orderNum |
| | | }) |
| | | |
| | | branchArr.length > 0 ? (father.children = branchArr) : ""; // 给父级添加一个children属性,并赋值 |
| | | // 属于同一对象问题,例如:令 a=b、c=1 ,然后再令 b.c=c , 那么 a.c=b.c=c=1 ;同理,后续令 c.d=2 ,那么 a.c.d 也是=2; |
| | | // 由此循环多次后,就能形成相应的树形数据结构 |
| | |
| | | async getuserList() { |
| | | this.Layertree = []; |
| | | this.layeroption = []; |
| | | |
| | | const data = await roleLayer_selectResByRole({ roleid: this.userlistData.roleid }); |
| | | if (data.code !== 200) { |
| | | return this.$message.error('根据ID查询资源失败'); |
| | |
| | | } |
| | | } |
| | | var val = this.setTreeData(data.result); |
| | | |
| | | |
| | | val.sort(function (a, b) { |
| | | return a.orderNum - b.orderNum |
| | | }) |
| | | this.Layertree = val |
| | | this.$refs.layerFrom.setCheckedKeys(std); |
| | | }, |