Question

我有一个Rest-Application，它有一个方法可以从LDAP中获取所有用户。 spring方法看起来像：

@ResponseStatus(HttpStatus.OK)
@RequestMapping(value = "users", method = RequestMethod.GET, consumes = "application/json", produces = "application/json")
public List<User> getUsers(@RequestParam(
        @RequestParam(required = false, value = "start") Long start,
        @RequestParam(required = false, value = "sort") String sort,
        @RequestParam(required = false, value = "asc") boolean asc)
        throws Exception {

    LdapContext ctx= .....
    List<User> Users;

    try {
        Users = searchUsers(ctx, "ou=users", start, sort, asc);
    } catch (Exception e) {
        ....
    } finally {
        ctx.close();
    }
    return Users;
}

搜索所有用户的方法是：

private List<User> searchUsers(LdapContext ctx, String ou, Long start, 
            String sort, boolean asc) throws Exception {

        NamingEnumeration<SearchResult> enumResult = null;
        List<User> LdapUserList = null;
        int totalSize = 0;

        try {
            SearchControls searchCtrls = new SearchControls();
            searchCtrls.setSearchScope(SearchControls.SUBTREE_SCOPE);
            searchCtrls.setReturningAttributes(attributes);

            // Create a sort control that sorts based on sort-field
            String sortKey = sort;
            ctx.setRequestControls(new Control[] { new SortControl(sortKey, false) });

            String filter = "(&(objectClass=inetOrgPerson))";

            enumResult = ctx.search(ou, filter, searchCtrls);


            // Iterate over sorted search results
            while (enumResult != null && enumResult.hasMore()) {

                totalSize++;                
                SearchResult result = (SearchResult) enumResult.next();
                System.out.println("Result: "+result.getName());

                User ldapUser = new User(); 

                ldapUser.add(......)
                // some code to fullfil the ldapUser with attributes

                UserList.addElement(ldapUser);
            }

        } catch (Exception e) {
            ....
        }

        System.out.println(totalSize);
        return UserList;

    }

我试图通过链接“/ users？sort = uid”来调用我的spring应用程序。

问题是，我从我的ou获得所有用户，但他们没有排序。如果我打印出变量“sort”，我得到正确的“uid”。所以@RequestParam也在工作。如何在服务器端对ldap响应的结果进行排序？错误在哪里？

最后，我想在代码“asc”和“start”中添加更多param，用于排序顺序和start-index。

感谢您的帮助！

Spring @RequestParams结合LDAP查询

0 个答案: