如何将表单输入中的用户 ID 值与数据库中的记录进行匹配 - How can I match a user id value from a form input to records in my database?

嗨，我正在尝试将表单输入中的用户 ID 值与数据库中的记录进行匹配。我正在尝试在下拉选择菜单中显示与表单输入中的用户 ID 值匹配的数据列（itemname）。我不知道我是否正确定义了我的变量或我做错了什么，但看不到我错过了什么。任何帮助将不胜感激。谢谢。

<?php
// Create the connection to the database
$con=mysqli_connect("xxx","xxx","xxx","xxx");

// Check if the connection failed
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  die();
}

  if (isset($_POST['userid']))
{
    $userid= $_POST['userid'];
   $query = "SELECT itemname
        FROM seguin_orders
  WHERE username = '".($userid)."'";
    $result = mysqli_query($con,$query);
}
?>

带下拉列表的表单

<form action="xxx" method="post" name="form1">

<select name="xxx"><option value="">-- Select One --</option>

 <?php 
        while ($row = mysqli_fetch_assoc($result)) 
        {
         echo '<option value =" ' . $row['itemname'] . ' ">' . $row['itemname'] . '</option>';
        }
    ?>
</select>

<input id="input1" name="input1" type="text" />
</br>
<input id="userid" name="userid" type="text" value="demo@gmail.com" readonly="readonly"/>

<button type="submit" value="Submit">Submit</button>

</form>

==AJAX FORM的解决方案==

订单.php

<?php
// Create the connection to the database
$con=mysqli_connect("xxx","xxx","xxx","xxx");

// Check if the connection failed
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  die();
}

  if (isset($_GET['userid']))
{
    $userid= $_GET['userid'];
   $query = "SELECT itemname
        FROM seguin_orders
  WHERE username = '".($userid)."'";
    $result = mysqli_query($con,$query);
    while ($row = mysqli_fetch_assoc($result)) 
        {
         echo '<option value =" ' . $row['itemname'] . ' ">' . $row['itemname'] . '</option>';
        }
}
?>

带有表单的原始页面

这只是基本和简单的，供您理解。您可以更改它并使其更安全。请阅读评论以了解

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Document</title>
</head>
<body>
<form action="#">
    <div class="pre-dropdown">  <!-- This class is here to hide this mini form after submit with jquery -->
        <input type="text" name="userid" id="userid">
        <button id="submitId">Submit Id</button>
    </div>
    <div class="dropdown">  <!-- This is hidden because there is no data, but when userid is submited, this will be visible -->
        <select name="xxx" id="dropdown-select">
            <option value="">-- Select One --</option>
        </select>
    </div>
</form>
<script src="http://code.jquery.com/jquery.min.js"></script>
<script>
     $(function(){
        $('.dropdown').hide();  // Hide dropdown div when page is loaded, mbetter way will be with css, but it's enough for now

        $('#submitId').on('click', function(e){  // Things to do when 'Submit Id' button is clicked
            var userid = $('#userid').val(); // Grab user id from text field
            e.preventDefault(); // Prevent form from submit, we are submiting form down with ajax.
             $.ajax({
              url: "orders.php",
              data: {
                userid: userid
              }
            }).done(function(data) {
                $('.pre-dropdown').hide();  // Hide mini form for user id
               $('.dropdown').show(); // show dropdown
              $('#dropdown-select').append(data); // append results from orders.php to select
            });
        });
     });
</script>
</body>
</html>

根据需要更改表单。我隐藏pre-dropdown因为如果用户再次提交userid，我们将附加结果两次。

你犯了两个错误

您错过了 if 条件结束 } 的结尾，并删除了分号;从 while 循环也缺少字符串结尾。

简而言之，你需要一个好的IDE，它可以告诉你代码中的基本错误

。

尝试将代码的回显更改为类似这样的内容 ->

echo "<option value ='"". $row['itemname'] ."'">". $row['itemname'] . "</option>";

或者尝试使用 mysqli_fetch_array（）而不是 mysqli_fetch_assoc（）

以及更多将 sql 查询更改为类似这样的东西

$query = "SELECT itemname FROM seguin_orders WHERE username = '$userid'";