乙太的小屋 - 课程设计
type: channel
link: https://www.oisec.cn/index.php/category/image9/
description: 毕业设计，课程设计等
items: 服装生产销售管理系统
items: 道路交通规划课程设计

服装生产销售管理系统
type: item
date: 2025-01-20T11:28:59+08:00
description: 前言麻烦给个三连吧~~~课程·任务书和源码都在：通过网盘分享的文件：服装生产销售管理系统.zip链接: https://pan.baidu.com/s/12Bz7XWsq5EJdztW_F4NaNQ?pwd=yf5g 提取码: yf5g文件中的 综合课程设计.pdf 是java课程和软件工程的综合设计题目，我的选题是服装生产销售管理系统，当时不会写可，最后一晚上搞定的，从网站找了一个类似的项目，源码文档数据库都在原始版本的文件夹里面，最后我修改的那本的在oisec修改版文件及中，只有最后一个导出报表的还没有实现，由于课程结束就不再想去动他了，希望有大佬能够帮忙实现吧，传递为爱发电。鄙人由于才疏学浅，课设时间紧迫，亟需展示效果，所以改的可能不是很好，期待指导。修改的部分登录界面划分四个权限用户login.vue <el-form class="login-form" label-position="left" :label-width="2 == 3 ? '56px' : '0px'"> 服装生产销售管理系统 <el-form-item :label="2 == 3 ? '用户名' : ''" :class="'style'+2"> <svg-icon icon-class="user" /> <el-input placeholder="请输入用户名" name="username" type="text" v-model="rulesForm.username" /> </el-form-item> <el-form-item :label="2 == 3 ? '密码' : ''" :class="'style'+2"> <svg-icon icon-class="password" /> <el-input placeholder="请输入密码" name="password" type="password" v-model="rulesForm.password" /> </el-form-item> <el-form-item v-if="0 == '1'" class="code" :label="2 == 3 ? '验证码' : ''" :class="'style'+2"> <svg-icon icon-class="code" /> <el-input placeholder="请输入验证码" name="code" type="text" v-model="rulesForm.code" /> {{ item.num }} </el-form-item> <el-form-item label="角色" prop="loginInRole" class="role"> <el-radio v-for="item in menus" v-if="item.hasBackLogin=='是'" v-bind:key="item.roleName" v-model="rulesForm.role" :label="item.roleName" >{{item.roleName}}</el-radio> </el-form-item> <el-button type="primary" @click="login()" class="loginInBt" style="padding:0;font-size:16px;border-radius:4px;height:44px;line-height:44px;width:100%;backgroundColor:rgba(13, 102, 171, 1); borderColor:rgba(13, 102, 171, 1); color:rgba(255, 255, 255, 1)">{{'1' == '1' ? '登录' : 'login'}}</el-button> <el-form-item class="setting"> <!-- 修改密码 --> </el-form-item> </el-form> 登录的后端接口需要添加用户的权限角色对比认证UserController.java /** * 登录 */ @IgnoreAuth @PostMapping(value = "/login") public R login(String username, String password, String role, String captcha, HttpServletRequest request) { UserEntity user = userService.selectOne(new EntityWrapper<UserEntity>().eq("u…

道路交通规划课程设计
type: item
date: 2025-01-15T16:23:41+08:00
description: 前言麻烦给个三连吧~~~课程·任务书和源码都在：通过网盘分享的文件：交通规划课程设计.zip链接: https://pan.baidu.com/s/1HULrttoIKGPb85Vr6xXGQQ?pwd=esqw 提取码: esqw没有残缺，很完整的一个项目，如果觉得对你有用，求打赏，最后还有个优化改进后的代码，都用python的面向对象，输出结果都是一样的，避免查重可用。——交通需求预测中“四阶段法”的实践2.a容量限制-增量加载法最短路法.py#OD分布流量矩阵输入 from numpy import * flowm = mat([[0,0,0,0,0,0,0,0], [0,0,0,0,350,400,0,500], [0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0], [0,300,0,0,0,100,0,350], [0,400,0,0,200,0,0,700], [0,0,0,0,0,0,0,0], [0,500,0,0,350,700,0,0]]) flow1m = 0.6*flowm flow2m = 0.3*flowm flow3m = 0.1*flowm flow1, flow2, flow3 = flow1m.tolist(), flow2m.tolist(), flow3m.tolist() Q12 = Q13 = Q24 = Q34 = Q35 = Q46 = Q56 = Q57 = Q67 = 0 Q21 = Q31 = Q42 = Q43 = Q53 = Q64 = Q65 = Q75 = Q76 = 0 #图矩阵化 def graph2adjacent_matrix(graph): vnum = len(graph) dict = {'1':0,'2':1,'3':2,'4':3,'5':4,'6':5,'7':6} adjacent_matrix = [[0 if row==col else float('inf') for col in range(vnum)] for row in range(vnum)] vertices = graph.keys() for vertex in vertices: for edge in graph[vertex]: w,u,v = edge adjacent_matrix[dict.get(u)][dict.get(v)]=w return adjacent_matrix #floyd算法 def floyd(adjacent_matrix): vnum = len(adjacent_matrix) a = [[adjacent_matrix[row][col] for col in range(vnum)] for row in range(vnum)] nvertex = [[-1 if adjacent_matrix[row][col]==float('inf') else col for col in range(vnum)] for row in range(vnum)] for k in range(vnum): for i in range(vnum): for j in range(vnum): if a[i][j]>a[i][k]+a[k][j]: a[i][j]=a[i][k]+a[k][j] nvertex[i][j] = nvertex[i][k] return nvertex, a #判断最短路径 def route(allroute,nvertex): for i in allroute.keys(): sp = int(i[5:6]) dp = int(i[:-2:-1]) allroute[i].append(sp) while True: if nvertex[allroute[i][-1]-1][dp-1] != dp: allroute[i].append(nvertex[allroute[i][-1]-1][dp-1]) else: allroute[i].append(dp) break return allroute #创建路段流量矩阵 allflow = [[], [0,0,Q12,Q13], [0,Q21,0,0,Q24], [0,Q31,0,0,Q34,Q35], [0,0,Q42,Q43,0,0,Q46], [0,0,0,Q53,0,0,Q56,Q57], [0,0,0,0,Q64,Q65,0,Q67], [0,0,0,0,0,Q75,Q76]] #3次循环分配 time=1 flows=[flow1,flow2,flow3] for flow in flows: #出行时间输入…