PHP - 复选框不会崩溃并在数据库中得到更新

那么，我的目标是获得所选复选框的值，然后爆炸数组，以便它是一个字符串格式，然后我可以已经插入数据库。PHP - 复选框不会崩溃并在数据库中得到更新

echo " 
<div class='modal fade' id='modalRejectCEO' role='dialog'> 
    <div class='modal-dialog modal-lg'> 
     <div class='modal-content'> 
     <div class='modal-header'> 
      <button type='button' class='close' data-dismiss='modal'>&times;</button> 
      <h4 class='modal-title'>Reject ".$strings["project"].": ".$projectDetail->pro_name[0]."</h4> 
     </div> 
     <div class='modal-body'> 
      <h5>Please specify the reason for rejecting the project ".$projectDetail->pro_name[0].". (eg. Reason is too vague)</h5> 
      <form> 
      <input type='checkbox' name='projectInfo[]' value='location'><b>Location: </b>".nl2br($projectDetail->pro_street_address[0]).", ".nl2br($projectDetail->pro_city_province[0])." <br><br> 
      <input type='checkbox' name='projectInfo[]' value='date'><b>Date: </b>".$projectDetail->pro_start_date[0]." - ".$projectDetail->pro_end_date[0]." <br><br> 
      <input type='checkbox' name='projectInfo[]' value='summary'><b>Project Description: </b><br><br>".nl2br($projectDetail->pro_description[0])."<br><br> 
      <input type='checkbox' name='projectInfo[]' value='objectives'><b>Objectives: </b>".nl2br($projectDetail->pro_objectives[0])."<br><br> 
      </form> 

      <div class='input-group'> 
       <span class='input-group-addon' id='basic-addon1' style='width:500px;'>Reason for Rejection:</span> 
       <input type='text' id='reasonCEO' name='reasonCEO' class='form-control' style='width:500px;' placeholder='' aria-describedby='basic-addon1'> 
      </div> 

     </div> 
     <div class='modal-footer'> 
      <input type='submit' id='rejectCEO' class='btn btn-primary' name='rejectCEO'> 
     </div> 
     </div> 
    </div> 
    </div> 

    ";

这是触发SQL

的更新方法

if(isset($_POST['rejectCEO']) && $_POST['reasonCEO'] != ""){ 
     $selectedReason = implode(', ',$_POST['projectInfo']); 

     $query3="UPDATE projects 
     SET project_status = '3', reason = '".$_POST['reasonCEO']."', reason_referrer = '".implode(',',$_POST['projectInfo'])."' 
     WHERE id =".$idProj."";  

     retrieveTable($query3); 
     header('Location: listprojectsprojectproposal.php'); 
}

来源

2017-03-06 Marco M

一件事，你关闭了你的''太早。可能是你的代码失败的原因。 –

错误报告会帮助你在这里。 –

@ Fred-ii- Hi fred！请把它作为答案发送。将接受你的答案，因为它解决了问题 –