在表列中显示多个记录不起作用

本文介绍了在表列中显示多个记录不起作用的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我需要从表特定列获取所有数据并将其显示为html表列（选择选项）。但没有任何表现。我做错了什么？ 我尝试了什么：

I need to get all the data from a table particular column and display into a html table column as (select option). But nothing is showing. what did i do wrong? What I have tried:

<form action="Report_customerHistory.php" method="POST"> <table align="center" border="0" align="center" style="width:40%"> <tr> <th> </th> <th> </th> <th> </th> </tr> <tr> <td> <select name="VehicleNo"> <option selected="" value="Default">(Please select.)</option> <?php include_once('db.php'); $result=mysqli_query($con,"SELECT * FROM vehicle_customer"); while($rawR1 = mysqli_fetch_array($result)) { $rawsData1=$rawR1['veh_reg_no']; ?> <option value="<?php echo $rawsData1; ?>"><?php echo $rawsData1;?></option> <?php } ?> </select> </td> <td> </td> <td> </td> </tr> </table>

推荐答案

result = mysqli_query（ result=mysqli_query(

con， SELECT * FROM vehicle_customer）; while（ con,"SELECT * FROM vehicle_customer"); while(

rawR1 = mysqli_fetch_array（ rawR1 = mysqli_fetch_array(