PHP从表中选择数据，不起作用

Question

好吧，我不太了解 PHP，当有人提交表单时，它需要检查该日期时间戳是否已经在 mysql 表中，然后执行某些操作。

localhost.php:

<?php 
$hostname_localhost = "localhost";
$database_localhost = "bookings";
$username_localhost = "username";
$password_localhost = "password";
$localhost = mysql_pconnect($hostname_localhost, $username_localhost, $password_localhost) or trigger_error(mysql_error(),E_USER_ERROR);
?>

主文件:

<?php require_once('../Connections/localhost.php'); ?>
<?php
$date = $_POST['date'];

$result = mysql_query("SELECT * FROM `bookingtable` WHERE `Date of Session` = '$date'");
if(mysql_num_rows($result) > 0) {
    echo 'the date was found, please try another';
} else {
    echo 'the date wasnt found';
};
?>

它曾经有效，但现在它总是说它不在表中，而实际上它在表中。

Answer 1

在 localhost.php 末尾写入以下行

mysql_select_db($database_localhost);