Php 使用数据库中的选定条目在我的网页中创建下拉列表

我正在创建一个网页，其中有人输入一个名称，并在下拉列表中显示以该名称注册的市场。 这是我的部分代码

            <form name="login" action="index_submit" method="get" accept-charset="utf-8">
        <ul>
            <li>
                <label for="seller">Seller Name/ID</label>
                <input type="text" id="username" placeholder="Seller Username/ID" required>
            </li>
             <?php

             ?>
            <li>
                <label for="marketplace">Choose a marketplace</label>
                <select name="url" onchange="menu_goto(this.form)">
                <?php
                    include ('config.php');
                    $username = $_POST['username'];
                    $query = "SELECT Marketplace FROM Details WHERE Seller_name='$username' OR Seller_ID='$username'";
                    $result = mysql_query($dbcon, $query) or die('error getting data');
                    while ($row = mysql_fetch_array($result)){
                    echo "<option value="marketplace1">" . $row['Marketplace'] . "</option>";
                    }
                ?>
                </select>
            </li>

卖方名称/身份证


选择一个市场
在您的文档中，您应该得到如下内容：

Parse error: syntax error, unexpected 'marketplace1' (T_STRING), expecting ',' or ';' in C:\xampp\htdocs\testing\index.php on line 2

假设您已使用@yaso建议的HTML格式

这样更改代码时，应将错误设置为显示在浏览器窗口上

我还假设你是用一个类似

example.com/myscript.php？username=OneThatExists

<?php
    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    include ('config.php');

    $username = $_POST['username'];
    $query = "SELECT Marketplace FROM Details WHERE Seller_name='$username' OR Seller_ID='$username'";
    $result = mysql_query($query,$dbcon) or die('error getting data');

    while ($row = mysql_fetch_array($result)){
        echo "<option value="marketplace1">" . $row['Marketplace'] . "</option>";
    }
?>

首先，你把mysql\u query（）的参数搞错了，请执行
$result=mysql\u query（$query，$dbcon）
，或者你使用的是MYSQLI\u扩展，它只是一个输入错误，应该是
MYSQLI\u query（$dbcon，$query）
弃用的mysql*它们不再维护，正式弃用。使用mysqli或PDOsearch有关错误报告的信息查看此信息，以了解t'internet@RiggsFolly，我正在使用mysql和Internet更改$dbcon和$query，但没有任何帮助。结果仍然是一样的，我不确定错误报告（E_ALL）做了什么，尽管在我的php脚本中添加它并没有什么不同。我是否需要为我的php代码添加submit以读取查看器输入的用户名的市场？如果您在服务器设置中测试此问题，请查看
php错误日志，因为实时错误可能不会显示在浏览器上，但可能会记录到错误日志中
                <?php
                include ('config.php');
                $username = $_POST['username'];
                $query = "SELECT Marketplace FROM Details WHERE Seller_name='$username' OR Seller_ID='$username'";
                $result = mysql_query($dbcon, $query) or die('error getting data');
                $res='';
                while ($row = mysql_fetch_array($result)){
                $res .='<option value="marketplace1">' . $row['Marketplace'] . '</option>';
                }
             echo $res;
             ?>

echo "<option value="marketplace1">" . $row['Marketplace'] . "</option>";
                    ^            ^ ---------> Not escaped

echo "<option value=\"marketplace1\">" . $row['Marketplace'] . "</option>";

echo '<option value="marketplace1">' . $row['Marketplace'] . '</option>';

error_reporting(E_ALL);

Parse error: syntax error, unexpected 'marketplace1' (T_STRING), expecting ',' or ';' in C:\xampp\htdocs\testing\index.php on line 2

<?php
    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    include ('config.php');

    $username = $_POST['username'];
    $query = "SELECT Marketplace FROM Details WHERE Seller_name='$username' OR Seller_ID='$username'";
    $result = mysql_query($query,$dbcon) or die('error getting data');

    while ($row = mysql_fetch_array($result)){
        echo "<option value="marketplace1">" . $row['Marketplace'] . "</option>";
    }
?>

<?php
    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    include ('config.php');

    if ( isset($_POST['username']) ) {

        $_POST['username'] = sanityCheck($_POST['username']);

        $query = "SELECT Marketplace 
                  FROM Details 
                  WHERE Seller_name='{$_POST['username']' 
                     OR Seller_ID='{$_POST['username']}'";

        $result = mysql_query($query, $dbcon) or die('error getting data');

        while ($row = mysql_fetch_array($result)){
            echo "<option value="marketplace1">" . $row['Marketplace'] . "</option>";
        }
    } else {
        // do whatever you want to do if no parameter was entered
    }
?>