我试图让PHP中的下拉框得到验证，这样如果值为Null，它将输出错误。但我似乎无法让这个工作。
我已尝试过其他网站上的解决方案以及此网站上的其他答案，但它们似乎没有起作用。
<?php
include("config.php");
$sql = "SELECT ID, Name, Surname, Address, TelNo, Email, Paid, Method,
Form, GMS, CanVolunteer FROM Tbl_Social";
if(isset($_POST['form'])) {
if($_POST['form'] == 'NULL') {
echo 'Please select an option from the dropdown menu.';
}
elseif ($_POST['form'] == 'ID') {
echo 'You have selected ID.';
}
}
?>
<!DOCTYPE html>
<html>
<div class="dropdown">
How do you want to find a player?<br>
<select name="form">
<option value="">Select a choice </option>
<option value="ID">ID</option>
<option value="Name">Name</option>
<option value="Surname">Surname</option>
<option value="Address">Address</option>
<option value="TelNo">Telephone Number</option>
<option value="Email">Email</option>
</select>
<input type="text" name="varQuery">
<input type="submit">
</div>
</html>
如果表单为空，我希望网页显示“请在下拉框中输入值”的行。
只需给出一个值选择一个像
<option value="select">Select a choice </option>
并检查是否选择了选择。
if($_POST['form'] == 'select') {
echo 'Please select an option from the dropdown menu.';
}
elseif ($_POST['form'] == "ID") {
echo 'You have selected ID.';
}
顺便说一句，如果你想发送POST
<form action ="POST">... <form>
更有用。
所有代码：
<?php
if(isset($_POST['clickedButton'])) {
if($_POST['form'] == 'select') {
echo 'Please select an option from the dropdown menu.';
}
elseif ($_POST['form'] == "ID") {
echo 'You have selected ID.';
}}
?>
<!DOCTYPE html>
<html>
<div class="dropdown">
How do you want to find a player?<br>
<form method ="POST" >
<select name="form">
<option value="select">Select a choice </option>
<option value="ID">ID</option>
<option value="Name">Name</option>
<option value="Surname">Surname</option>
<option value="Address">Address</option>
<option value="TelNo">Telephone Number</option>
<option value="Email">Email</option>
<input type="text" name="varQuery">
<button type="submit" name ="clickedButton" >ENTERS </button>
</select>
</form>
isset()已经检查变量是否存在且不为空。您可以将if语句合并为一个：
if(!isset($_POST['form'])) {
echo 'Please select an option from the dropdown menu.';
}
elseif ($_POST['form'] == 'ID') {
echo 'You have selected ID.';
}