Question

我是php和AJAX的新手，我最近被指派开发假期管理系统。我遇到了一个小问题，我从mysql中获取数据并以表格格式显示，因为一个字段被设计为下拉列表，所以对于我收到的数据并完美显示，现在我的问题是，我想从中选择值下拉列表，当我点击按钮时，它应该在数据库中更新它。我的代码是

display_record.php

function save_row(leave_ID)
{
          
         
	var hods=document.getElementById("hods_val"+leave_ID).value;

	
		
	
 $.ajax
 ({
  type:'post',
  url:'modify_records.php',
  data:{
	edit_row:'save_row',
	row_id:leave_ID,  
	hods_val:hods
  },
  success:function(response) {
   if(response=="success")
   {
   
  document.getElementById("hods_val"+leave_ID).innerHTML=hods;
  
    document.getElementById("edit_button"+leave_ID).style.display="block";
    document.getElementById("save_button"+leave_ID).style.display="none";
   }
  }
 });
}

$host="localhost";
$username="root";
$password="";
$databasename="hrdb";
$con=mysqli_connect($host,$username,$password);
$db=mysqli_select_db($con,$databasename);

$select =mysqli_query($con,"SELECT * FROM leavedb");


?>

<table align="center" cellpadding="10" border="1" id="user_table" class="data-table">
<caption class="title">Leave Requested by staff Members</caption>

<tr>
<th>Emp ID</th>
<th>Emp name</th>
<th>Leave Token</th>
<th>Leave Type</th>
<th>Applied Date</th>
<th>Leave From</th>
<th>Leave To</th>
<th>Req.Days</th>
<th>Reason</th>
<th>Available credit</th>
<th>HOD Status</th>
<th>Status</th>
</tr>
<?php
while ($row=mysqli_fetch_array($select)) 
{
 ?>
 <tr id="row<?php echo $row['leave_ID'];?>">
  <td id="eid_val<?php echo $row['leave_ID'];?>"><?php echo $row['emp_ID'];?></td>
  <td id="empn_val<?php echo $row['leave_ID'];?>"><?php echo $row['emp_ID'];?></td>
  <td id="lid_val<?php echo $row['leave_ID'];?>"><?php echo $row['leave_ID'];?></td>
  <td id="lt_val<?php echo $row['leave_ID'];?>"><?php echo $row['leave_type'];?></td>
  <td id="dor_val<?php echo $row['leave_ID'];?>"><?php echo $row['date_of_request'];?></td>
  <td id="df_val<?php echo $row['leave_ID'];?>"><?php echo $row['date_from'];?></td>
  <td id="dt_val<?php echo $row['leave_ID'];?>"><?php echo $row['date_to'];?></td>
  <td id="rd_val<?php echo $row['leave_ID'];?>"><?php echo $row['req_days'];?></td>
  <td id="rs_val<?php echo $row['leave_ID'];?>"><?php echo $row['reason'];?></td>
  <td id="ad_val<?php echo $row['leave_ID'];?>"><?php echo $row['available_days'];?></td>
 <td id="hods_val<?php echo $row['leave_ID'];?>"> 
  
 <select id="hods_val" name="hods_val"class="aaa">
  <option value="Approved">Approved</option>
  <option value="NotApproved">not Approved</option>
  <option value="MeetMe">Meet Me</option>
  <option value="hods1" selected><?php echo $row['hod_status'];?></option>
</select>
  </td>
     <td>
<input type='button' class="edit_button" id="edit_button<?php echo $row['leave_ID'];?>" value="edit" onclick="edit_row('<?php echo $row['leave_ID'];?>');">
 <input type='button' name="submitt"class="save_button" id="save_button<?php echo $row['leave_ID'];?>" value="Execute" onclick="save_row('<?php echo $row['leave_ID'];?>');">

 
 
 
  </td>
 </tr>
 <?php
 
 

 



}


 



?>


</table>



modify_records.php


<?php
$host="localhost";
$username="root";
$password="";
$databasename="hrdb";

$connect=mysqli_connect($host,$username,$password);
$db=mysqli_select_db($connect,$databasename);



if(isset($_POST['edit_row']))
	

 $row=$_POST['row_id'];
 
$hods111=$_POST['hods_val'];
 

if( mysqli_query($connect,"update leavedb set hod_status='$hods111' where leave_ID='$row'");)
 
 	{
			
			echo '<script language="javascript">';
			echo 'alert("Successfully approved")';
			echo '</script>';	
			
	}
 
 
 echo "success";
 exit();
}

?>

Answer 1

你不能在dom中使用超过1个相同的id。

所以不要这样：

<select id="hods_val" name="hods_val"class="aaa">

为您的选择提供唯一ID：

<select id="select_hods_val<?php echo $row['leave_ID'];?>" name="hods_val"class="aaa">

之后，您可以按如下方式选择下拉列表：

document.getElementById("select_hods_val"+leave_ID)...

php ajax下拉列表在编辑后没有更新

1 个答案: