1

大家好，我正在使用 codeigniter 构建项目管理系统。

我的模型中有一个函数来显示一个项目以及与之关联的任务。$projectId = 1 仅用于测试目的。

function getAllProjects($projectId = 1)
 {

  $this->db->select('*');    
  $this->db->from('projects');
  $this->db->where('projects.projectId', $projectId);
  $this->db->join('projectTasks', 'projects.projectId = projectTasks.projectId');
  $this->db->join('tasks', 'projectTasks.taskId = tasks.taskId');
  $projects = $this->db->get();
  return $projects;
}

然后在我看来我让它显示结果

<?php if($projects->num_rows() > 0): ?>
  <table width="100%">
    <?php foreach($projects->result() as $p): ?>
      <tr>

        <td><?php echo $p->projectId; ?></td>
        <td><?php echo $p->projectName; ?></td>
        <?php foreach($projects->result() as $g): ?>
             <td><?php echo $g->taskName; ?></td>
             <td><?php echo $g->taskHours; ?></td>
             <td><?php echo $g->taskCost; ?></td>
        <?php endforeach; ?>


        </td>
      </tr>
    <?php endforeach; ?>
  </table>
<?php else: ?>
    <p>No projects at this time.</p>
<?php endif; ?>

哪个返回这个

  ID - Project Name - Task Name - Task Hours - Task Cost (**to show you what field is what**)
  1 The Range 702   Contact Form      10         100            Custom Logo Desgin  10  100 Custom Login From   5   75
  1 The Range 702   Contact Form      10         100            Custom Logo Desgin  10  100 Custom Login From   5   75
  1 The Range 702   Contact Form      10         100            Custom Logo Desgin  10  100 Custom Login From   5   75

有人可以告诉我如何制作它，因此它只显示一次项目信息，然后显示与项目关联的每个任务。现在，对于与项目关联的每个任务，它都会再次列出项目和所有任务。

表

 -----------------------
 |       projects      |
 -----------------------
 |  projectId          | (Primary)
 |  projectName        |
 |  projectHours       |
 |  projectDeadline    |
 |  projectStartDate   |
 |  projectTasks       |
 |  projectUsers       |
 |  projectNotes       |
 |                     |
 -----------------------

 -----------------------
 |     projectTasks    |
 -----------------------
 |  projectTasksId     | (Primary)
 |  projectId          | (FK project->projectId)
 |  taskId             | (FK tasks->taskId)
 |                     |
 -----------------------

 --------------------
 |      tasks       |
 --------------------
 |  taskId          | (Primary)
 |  taskName        |
 |  taskHours       |
 |  taskCost        |
 |                  |
 --------------------
4

2 回答 2

0

创建一个变量来存储每次迭代的项目 ID，然后与实际项目 ID 进行比较，如果不同，则显示项目信息。像这样的东西。

<?php
 $beforeProject = '';
 foreach($projects->result() as $p):
?>
    <td><?php if ($beforeProject != $p->projectId) { echo $p->projectId; } ?></td>

     //rest of code

   <?php $beforeProject = $p->projectId; ?>
<?php endforeach; ?>
于 2013-10-24T19:29:42.717 回答
0

While I think Jorge Campos solution was close it won't prevent the tasks from being listed out multiple times.

   <?php $ProjectId = ''; ?>
      <?php if($projects->num_rows() > 0): ?>
        <table width="100%">
          <?php foreach($projects->result() as $p): ?>
            <tr>
              <?php if ($ProjectId != $p->projectId) { ?>
              <td><?php echo $p->projectId; ?></td>
              <td><?php echo $p->projectName; ?></td>

                <?php foreach($projects->result() as $g): ?>
                    <td><?php echo $g->taskName; ?></td>
                    <td><?php echo $g->taskHours; ?></td>
                    <td><?php echo $g->taskCost; ?></td>
               <?php endforeach; ?>

             <?php } ?>
       <?php $ProjectId = $p->projectId; ?>

            </tr>
          <?php endforeach; ?>
        </table>
      <?php else: ?>
        <p>No projects at this time.</p>
      <?php endif; ?>

This solution will make it so the information is only displayed once for each projectId.

于 2013-10-24T20:00:21.963 回答