当前位置: 首页>>代码示例>>PHP>>正文


PHP Sql::get_user_from_name方法代码示例

本文整理汇总了PHP中Sql::get_user_from_name方法的典型用法代码示例。如果您正苦于以下问题:PHP Sql::get_user_from_name方法的具体用法?PHP Sql::get_user_from_name怎么用?PHP Sql::get_user_from_name使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在Sql的用法示例。


在下文中一共展示了Sql::get_user_from_name方法的2个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的PHP代码示例。

示例1: set_include_path

 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
// FIXME: Use a host that better supports PHP include paths...
set_include_path(get_include_path() . PATH_SEPARATOR . '/customers/tasktaste.com/tasktaste.com/httpd.www/phpincludes');
require_once 'TaskTaste/tasktaste.php';
$project_urlname = Utils::get_name_from_get(PROJECT_NAME);
$project_ownername = Utils::get_name_from_get(USERNAME);
$project = NULL;
$project_owner = Sql::get_user_from_name($project_ownername);
if ($project_owner) {
    $project = Sql::get_project_from_url($project_owner->get_id(), $project_urlname);
}
$tasks = NULL;
$project_name = "<Unknown project>";
$project_owner_name = "<Unknown owner>";
$project_description = "This project does not exist yet.";
$project_id = 0;
$worked_per_week = -1;
$owner_logged_in = FALSE;
if ($project) {
    $project_name = $project->get_name();
    $project_description = $project->get_description();
    $project_id = $project->get_id();
    $tasks = Sql::get_tasks_from_project_id($project_id);
开发者ID:rafaelbeckel,项目名称:task-taste,代码行数:31,代码来源:manage-project.php

示例2: authenticate_user

 /**
  * Authenticate a user using a password. Returns the user id of
  * the user if authentication was successful or -1 otherwise.
  */
 public static function authenticate_user($username, $password)
 {
     $userid = -1;
     $user = Sql::get_user_from_name($username);
     if ($user && $password && User::authenticate($user, $password)) {
         $userid = $user->get_id();
     }
     return $userid;
 }
开发者ID:rafaelbeckel,项目名称:task-taste,代码行数:13,代码来源:sql.php


注:本文中的Sql::get_user_from_name方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。