PHP结构 - 在一个页面或多个页面上有多个IF GET语句？

Question

我已经用PHP开发了几年，但我不断质疑的一件事是我是否应该像这样构建我的页面：

            if(getValue('action') == "jobsFilter" && getValue('jobType') == "open")
            {
                $job->FetchJobs($jobStatus = 1);
                foreach($job->result AS $pulledJob)
                {
                ?>
                <div class = "openJob panelJob col-xs-12">
                    <h2><?php echo $pulledJob['jobTitle'] ?></h2>
                    <a href = "?action=viewJob&jobId=<?php echo $pulledJob['Id'] ?>" class = "viewJobBtn ">View Job</a>
                </div>
                <?php
                }       
            }else if(getValue('action') == "jobsFilter" && getValue('jobType') == "active")
            {
                $job->fetchAllJobsAppliedToUser($jobStatus = 1);
                foreach($job->result AS $pulledJob)
                {
                ?>
                <div class = "openJob panelJob col-xs-12">
                    <h2><?php echo $pulledJob['jobTitle'] ?></h2>
                    <a href = "?action=viewJob&jobId=<?php echo $pulledJob['Id'] ?>&jobStatus=1" class = "viewJobBtn ">View Details</a>
                </div>
                <?php
                } 
            }else if(getValue('action') == "viewJob" && isset($_GET['jobId']))
            {
                $job->FetchJobs($jobStatus = 1, $jobId = $_GET['jobId']);
                foreach($job->result AS $pulledJob)
                {
                ?>
                <div class = "viewJob panelJob col-xs-12">

                    <div class = "pulledJobInfo">
                        <h2><?php echo $pulledJob['jobTitle'] ?></h2>
                        <p><?php echo $pulledJob['JobDescription'] ?></p>
                    </div>

                    <form method = "post" action = "?action=acceptJob">
                        <input type = "submit" name = "acceptJobSubmitBtn" class = "acceptJobBtn fullWidthButton ctaButton" value = "Let me help" />
                        <input type = "hidden" name = "jobId" value = "<?php echo $pulledJob['Id'] ?>" />
                    </form>
                </div>
                <?php
                }                    
            }

或者我是否应该有一个单独的页面，对于打开的作业称为open-jobs.php，对于已关闭的作业称为closed-jobs.php等。

如果有人能够对最佳解决方案有所了解，那就太棒了！

Answer 1

选项A：

       if(getValue('action') == "jobsFilter" && getValue('jobType') == "open")
       {
           $job->FetchJobs($jobStatus = 1);
           $file="view1"
       }else if(getValue('action') == "jobsFilter" && getValue('jobType') == "active")
       {
           $job->fetchAllJobsAppliedToUser($jobStatus = 1);
           $file="view2"

       }else if(getValue('action') == "viewJob" && isset($_GET['jobId']))
       {
           $job->FetchJobs($jobStatus = 1, $jobId = $_GET['jobId']);
           $file="view3"                                
       }

       if($job != null){
         foreach($job->result AS $pulledJob)
         {
          include($file)
         }   
       } 

选项B：

       // Logic what to display
       switch(getValue('action')){

        case "jobsFilter":

          switch (getValue('jobType')){
            case "open":
              $job->FetchJobs($jobStatus = 1);
              $file="view1"
              break;
            case "active":
              $job->fetchAllJobsAppliedToUser($jobStatus = 1);
              $file="view2" 
              break;             
          }

        case "viewJob":
          $job->FetchJobs($jobStatus = 1, $jobId = $_GET['jobId']);
          $file="view3"  
          break;
       }

       // Execute the display
       if($job != null){
         foreach($job->result AS $pulledJob)
         {
          include($file)
         }   
       }