php - 当从表中更新列行时，然后在该列的所有行中提取

Question

我做了一个表格，属性代码很少

<table class="table table-hover table-striped">
  <thead>
          <tr>
           <th>Id</th>
           <th>File-name</th>
           <th>Purpose</th>
           <th>Recieved-By </th>
           <th>Processed-By</th>
           <th>Adress</th>
          <th>Contact-No</th>
          <th>Date</th>
           <th>Update</th>
           </tr>`

这里是从数据库表中除更新之外的所有列中获取数据

                  <tbody>
                       <?php
                            if ( $search )
                            {
                                $p_query = "select * from files where recieved_by like '%$search%' or processed_by like  '%$search%' or   purpose like  '%$search%' or file_name like '%$search%' order by id desc limit $page_start_from, $total_num_page";
                            } 
                            else 
                            {

                                $p_query = "select * from files order by id desc limit $page_start_from, $total_num_page";
                            }
                          $p_run=mysqli_query($con,$p_query);
                            if(mysqli_num_rows($p_run)){
                             while($row=mysqli_fetch_array($p_run))
                             {
                                $c_id=$row['id'];
                                $file=$row['file_name'];
                                $purpose=$row['purpose'];
                                $recieve=$row['recieved_by'];
                                $processed=$row['processed_by'];
                                $address=$row['address'];
                                $contact=$row['contact_no'];
                                $date=$row['date'];
                                $show_status=$row['show_status'];


                            ?>
                                            <tr>

                        <td><a  href="post.php?post_id=<?php echo $c_id?>"><?php echo $c_id;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $file;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $purpose;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $recieve;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $processed;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $address;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $contact;?></a></td>
                        <td><a href="post.php?post_id=<?php echo $c_id?>"><?php echo $date;?></a></td>

现在在更新列中，当文件表数据库中的更新列中有值时，更新列中就会发生变化，那么它将在更新列中显示，否则它将打开一个<强> Action 是一个按钮，有一个表单代码在这里

                                                <td>
                                                    <?php if( !empty($show_status)){
                                                           echo $show_status;
                                                                                }
                                                           else
                                                             {
                                                    ?>
                                                    <div class="btn-group">
                                                      <button type="button" class="btn btn-default dropdown-toggle" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false" style=" padding-top: 3px; padding-bottom: 3px;">
                                                        Action <span class="caret"></span>
                                                      </button>
                                                      <ul class="dropdown-menu" style="margin-left:-50px;">
                  <?php 
                      if(isset($_POST['submit-form']))
                      {
                          $update_date=$_POST['reason-date'];
                          $status=$_POST['reason'];
                          $p_id=$_POST['idvalue'];
                          $d_query="INSERT INTO update_table (id, reason, update_date) VALUES ('$p_id', '$status', '$update_date'); update files set show_status ='  $status' where id=' $p_id'";
                    if(mysqli_multi_query($con,$d_query))
                    {
                        $msg="file have been submitted ";
                        header('location:index.php');
                    }
                    else
                    {
                        $error_msg="Already updated this file";
                    }


}?>
                         <script type="text/javascript">
                                      function checkvalue()
                                        {
                                              var ureason=document.forms["sform"]["ureason"].value;
                                              var udate=document.forms["sform"]["udate"].value;
                                            if (ureason=="")
                                            {
                                                alert("Status Field is blank ");
                                                return false;
                                            }
                                             if (udate=="")
                                            {
                                                alert("Date Field is blank ");
                                                return false;
                                            }


                                        }
                            </script>
                            <form role="banner" class="actionform" action="index.php" method="POST" name="sform"  onsubmit=" return checkvalue()">
                                <input type="hidden" name="idvalue" value="<?php echo $c_id; ?>">
                                 <label  style="font-weight: normal;">Enter Reason*</label><br>
                                <input  type="text" name="reason" placeholder="call or letter or filed " id="ureason"><br>
                                <label style="font-weight: normal;">Update Date:*</label><br>
                                <input  type="date" name="reason-date" placeholder="enter date here" id="udate"> 
                                <br><br>
                                <input type="submit" name="submit-form" id="actionid" value="Submit" style="background: #3596e0; ">
                       <?php
                   if(isset($error_msg))
                   {
                   echo "<span style='color:red; margin-bottom:10px;'  class='pull-right'>$error_msg</span>"; 
                       exit();
                   }
                   else if(isset($msg))
                   {
                       echo "<span style='color:green;' class='pull-right'>$msg</span>";
                       exit();
                   }?>
                            </form>
                                                      </ul>
                                                    </div>
                                                    <?php 
                                                        } ?>

                                                </td>
                                            </tr>
                                            <?php 
                                      }
                                    }
                                            else {
                                                 echo "<h3> NO Related Table is Found Here </h3>";
                                                }
                                   ?>
                                        </tbody>
                                    </table>

现在的问题是，当我第一次填写更新操作表单时，它会正常，但是当我要进行第二个或第三个操作时，上面的行更新功能未填充，它不断显示检查表单列是否为空的错误，如何停止这，当我填写一行更新列时，它应该只影响该特定行，请有人帮助我，我在这部分停留了两天。

Answer 1

由于您使用的是INSERT方法，因此请使用UPDATE方法进行更新。

第一次，由于不存在数据，您的 INSERT 方法可以按您想要的方式工作，但之后您必须使用 UPDATE 方法进行更新。

有关更新，请参阅 https://www.w3schools.com/sql/sql_update.asp

对于插入，请参阅 https://www.w3schools.com/sql/sql_insert.asp