【发布时间】:2015-12-21 13:47:31
【问题描述】:
我是编程新手。我想从数据库中获取记录的用户 ID 并存储在 SESSION 中,然后将该会话 ID 插入另一个表中。我想我成功了,但不幸的是有一个错误让我发疯。任何帮助将不胜感激!
数据库btrs 有表booking 有库名
booking_id,customer_id,route_id
dbConfig.php
<?php
$dbhost = "localhost";
$dbname = "btrs";
$dbuser = "root";
$dbpass = "";
$conn =mysql_connect($dbhost, $dbuser, $dbpass) or die("MySQL Error: " . mysql_error());
mysql_select_db($dbname) or die("MySQL Error: " . mysql_error());
?>
Logincheck.php
<?php
session_start();
include('dbConfig.php');
$error = ''; // Variable To Store Error Message
if (isset($_POST['submit'])) {
if (empty($_POST['email']) || empty($_POST['password'])) {
$error = "email or Password is invalid";
}
else {
$email = $_POST['email'];
$password = $_POST['password'];
$email = stripslashes($email);
$password = stripslashes($password);
$email = mysql_real_escape_string($email);
$password = mysql_real_escape_string($password);
$query = mysql_query("select * from member where password='$password' AND email='$email'");
$count = mysql_num_rows($query);
if ($count > 0 ) {
$row = mysql_fetch_array(mysql_query("select * from member where password='$password' AND email='$email'"));
$id = $row['id'];
// echo $id;
$_SESSION['id']=$id;
header("location: profile.php");
} else {
$error = "Username or Password is invalid";
}
}
}
?>
booking.php
<?php
include"dbConfig.php";
include "logincheck.php";
if (isset($_POST['submit'])) {
$from1 = $_POST['from'];
$destination1 = $_POST['destination'];
$query = mysql_query("SELECT route_id FROM route WHERE pick_from='$from1' AND destination='$destination1'");
$row = mysql_fetch_array($query);
$route_id = $row['route_id'];
// echo $route_id;
if ($row!=NULL) {
$query1 = "INSERT INTO booking(customer_id,route_id) VALUES ('".$_SESSION['id']."','$route_id')";
if ($query1==1) {
echo"very goodddd";
}
}
else {
echo "good";
}
}
?>
【问题讨论】:
-
这是一个错字:
Logincheck.php(大写“L”)然后在你的脚本中:include "logincheck.php";? -
既然你是新手,这里给你一个小提示。不要费心使用 mysql_ 函数。 MySQL 扩展自 PHP 5.5 起正式弃用,并在 PHP 7 中完全删除。它没有积极的开发,使用它会使您的代码不那么面向未来。相反,应该使用 MySQLi 或 PDO_MySQL 扩展。