我的 PHP 表单不会发布到数据库中。我对 php 和 mysql 的连接理解得相当好,但我对这个很困惑。当我在表单上点击提交时,它不会回显我(用户)输入的值。日期显示为 1969-12-31,而不是用户提交的日期。如果有人可以提供帮助,那就太好了。我的代码如下
表单代码为:
<form method="POST" action="add_event.php" id="create_event">
<label for="event_name">Event Name:</label>
<input type="text" id="event_name"><br />
<label for="date">Date:</label>
<input class="datepicker" type="date" id="date"><br />
<label for="zip_code">Zip Code:</label>
<input type="text" id="zip_code" maxlength="5"><br />
<label for="description">Description</label>
<textarea id="description" rows="5" columns="10"></textarea>
<br>
<input type="submit" name="submit">
</form>
add_event.php 插入代码是:
<?php
require_once '../app_config.php';
require_once '../database_connection.php';
require_once '../authorize.php';
session_start();
// Authorize any user, as long as they're logged in
authorize_user();
//Get the user ID of the user to show
$user_id = $_SESSION['user_id'];
$select_query = "SELECT first_name, last_name FROM users WHERE user_id = " . $user_id;
// Run the query
$result = mysql_query($select_query);
if ($result) {
$row = mysql_fetch_array($result);
$first_name = $row['first_name'];
$last_name = $row['last_name'];
}
$name = $first_name . ' ' . $last_name;
$event_name = trim($_POST['event_name']);
$date = trim($_POST['date']);
$zip_code = trim($_POST['zip_code']);
$description = trim($_POST['description']);
// $date = "2012-08-22";
$newdate = date("Y-m-d", strtotime($date));
// $event_name = "test";
// $zip_code = "22153";
// $description = "test";
$insert_sql = sprintf("INSERT INTO events " .
"(name, user_profile_id, event_name, date, zip_code, description) " .
"VALUES ('%s', %d, '%s', '%s', '%s', '%s');",
mysql_real_escape_string($name),
mysql_real_escape_string($user_id),
mysql_real_escape_string($event_name),
mysql_real_escape_string($newdate),
mysql_real_escape_string($zip_code),
mysql_real_escape_string($description));
//insert the user into the database
mysql_query($insert_sql);
echo $insert_sql;
?>
非常感谢提前。