我有3个下拉菜单我想显示什么,他/她已经选择了一个功能后,用户将选择或SCRPT会做的,但它必须是在脚本中显示drowdown值已经选择3选项
<?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>';
}
?>
东西,将调用write_in_div当点击我!按钮被压或者说可被用于显示3选择用户选择
输出应该是这样的,您选择1)名称2)姓和电子邮件
我应该怎么做这个PLZ给我一个简化版本,因为这对我从自己的knolgede创建我如何可以纠正这个 – Sizwe