Php 在用户选择3选项后显示drowdown值

我有3个下拉列表，我想显示用户在选择功能或scrpt后将选择的内容，但必须在脚本中

<?php
$resource_names = mysql_query("SELECT DISTINCT NAME FROM selections ORDER BY id ASC");
$names = array();
while($row = mysql_fetch_row($resource_names)){
    $names[] = $row[0]    
} 
$resource_surnames = mysql_query("SELECT DISTINCT SURNAME FROM selections ORDER BY id ASC"); 
$surnames = array();
while($row = mysql_fetch_row($resource_surnames)){
    $surnames[] = $row[0];
} 
$resource_emails   = mysql_query("SELECT DISTINCT EMAIL FROM selections ORDER BY id ASC");    
$emails = array();
while($row = mysql_fetch_row($resource_emails)){
    $emails[] = $row[0];
}
if(count($emails) <= 0 || count($surnames) <= 0 || count($emails) <= 0){
    echo 'No results have been found.';
} else {

    // Display form
    echo '<form name="form" method="post" action="test.php">';

    //Names dropdown:
    echo '<select name="id" id="names">';
    foreach($names as $name) echo "<option id='$name'>$name</option>";
    echo '</select>';

    //Surnames dropdown
    echo '<select name="id" id="surnames">';
    foreach($surnames as $surname) echo "<option id='$surname'>$surname</option>";
    echo '</select>';

    //Emails dropdown
    echo '<select name="id" id="emails">';
    foreach($emails as $email) echo "<option id='$email'>$email</option>";
    echo '</select>';

    echo "<button id='write_in_div'>Click me!</button>";

    echo '</form>';

}
?>

单击“我”时将调用write_in_div的内容！按钮是按下或任何其他方法，可用于显示用户选择的3个选择

---

输出应该类似于您选择的名称1）姓名2）姓氏和电子邮件

您的html选择中有错误每个选择都有相同的名称“id”，它们都需要唯一，以便您可以检测

您需要检测用户是否已提交表单

if(isset($_POST["select_name"])) {
     echo $_POST["select_name"];
}

---

你的表格上有个大错误。 在表单中，每个选择和输入必须具有唯一的名称。您需要使用此名称在php脚本中检索提交的值

我想你有这个：

<?php
$resource_names = mysql_query("SELECT DISTINCT NAME FROM selections ORDER BY id ASC");
$names = array();
while($row = mysql_fetch_row($resource_names)){
    $names[] = $row[0]    
} 
$resource_surnames = mysql_query("SELECT DISTINCT SURNAME FROM selections ORDER BY id ASC"); 
$surnames = array();
while($row = mysql_fetch_row($resource_surnames)){
    $surnames[] = $row[0];
} 
$resource_emails   = mysql_query("SELECT DISTINCT EMAIL FROM selections ORDER BY id     ASC");    
$emails = array();
while($row = mysql_fetch_row($resource_emails)){
    $emails[] = $row[0];
}
if(count($emails) <= 0 || count($surnames) <= 0 || count($emails) <= 0){
    echo 'No results have been found.';
} else {

// Display form
echo '<form method="post" action="test.php">';

//Names dropdown:
echo '<select name="names">';
foreach($names as $name) echo "<option id='$name'>$name</option>";
echo '</select>';

//Surnames dropdown
echo '<select name="surnames">';
foreach($surnames as $surname) echo "<option id='$surname'>$surname</option>";
echo '</select>';

//Emails dropdown
echo '<select name="emails">';
foreach($emails as $email) echo "<option id='$email'>$email</option>";
echo '</select>';

echo '<button id="write_in_div">Click me!</button>';

echo '</form>';
}

在文件末尾无效

注2：在编写html文件时，请注意“和”。html属性的值是用“而不是”来编写的。
我应该怎么做请给我一个简化的版本，因为这是我从自己的knolgede创建的关于如何更正此问题的知识查看select声明，它们都有一个唯一的名称。第一个名为“names”，因此在php脚本中，您可以使用$\u REQUEST['names']获取此select的选定值