Question

起初这听起来很简单，但实际上我找不到实现这个的方法。

我有Spring网络应用程序，有休息控制器和数据库。我的网络应用中有用户，我想打开特定页面，其中将显示该用户的详细信息。那么如何将有关点击内容的信息（链接到该特定用户的详细信息）发送到另一个显示用户详细信息的页面？

这是我尝试过的，当我加载个人资料页面时，它上面没有任何内容：的index.html

from lxml import html
import requests

class page_crawler(object):

    main_link = "https://www.yellowpages.com/search?search_terms=pizza&geo_location_terms=San%20Francisco%2C%20CA"
    base_link = "https://www.yellowpages.com"

    def __init__(self):

        self.links = [self.main_link]


    def crawler(self):
        for link in self.links:
            self.get_link(link)

    def get_link(self, link):

        print("Running page "+ link)
        page = requests.get(link)
        tree = html.fromstring(page.text)
        item_links = tree.xpath('//h2[@class="n"]/a[@class="business-name"][not(@itemprop="name")]/@href')
        for item_link in item_links:
            return self.base_link + item_link

        links = tree.xpath('//div[@class="pagination"]//li/a/@href')
        for url in links:
            if not self.base_link + url in self.links:
                self.links += [self.base_link + url]



class Info_grabber(page_crawler):

    def __init__(self, plinks):
        page_crawler.__init__(self)
        self.plinks = [plinks]

    def passing_links(self):
        for nlink in self.plinks:
            print(nlink)
            self.crawling_deep(nlink)

    def crawling_deep(self, uurl):

        page = requests.get(uurl)
        tree = html.fromstring(page.text)

        name = tree.findtext('.//div[@class="sales-info"]/h1')
        phone = tree.findtext('.//p[@class="phone"]')
        try:
            email = tree.xpath('//div[@class="business-card-footer"]/a[@class="email-business"]/@href')[0]
        except IndexError:
            email=""

        print(name, phone, email)


if __name__ == '__main__':

    crawl = Info_grabber(page_crawler)
    crawl.crawler()
    crawl.passing_links()

profile.html

这里我使用相同的<html ng-app="app" ng-controller="appController"> <head > <script src="js/angular.min.js"></script> <script src="js/App.js"></script> <p ng-click = "setProfileDetails(john)"><a href="profile.html">profile</a></p>文件，因此我可以读取相同的控制器。

App.js

App.js

<body ng-controller = "appController">
<div ng-model = "profileDetails">
{{profileDetails.username}}
</div>
</body>

Answer 1

您应该公开一个返回用户详细信息的端点，并提供一些用户标识符，例如userId。因此，当有人点击用户时，将userId发送到API并检索响应并显示它。您只需要进行简单的API调用即可。缺少有关前端实施的详细信息。

Answer 2

在jsp页面上：

<form action="user-details/${userid}">
//list of users or whatever you want
</form>

控制器上的

@RequestMapping("user-details/{userid}")
public String showUserDetails(@PathVariable int userid, Model model) {
    model.addAttribute("userDetails",userDAO.getUserDetailsById(userid);
    return "/ac_user/user-details";
}

如何根据单击的内容打开特定的用户配置文件

2 个答案: