在两个不同的表中添加记录

Question

I created a table which stores the users details 我创建了一个存储用户详细信息的表

CREATE TABLE IF NOT EXISTS `users` (
  `userid` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(25) NOT NULL,
  `username` varchar(25) NOT NULL,
  `password` varchar(25) NOT NULL,
  PRIMARY KEY (`userid`),
  KEY `userid` (`userid`),
  KEY `userid_2` (`userid`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=18 ;

this one is to keep the first name and the last name from two textboxes 这是为了保留两个文本框中的名字和姓氏

CREATE TABLE IF NOT EXISTS `data` (
  `dataid` int(11) NOT NULL AUTO_INCREMENT,
  `firstname` varchar(25) NOT NULL,
  `lastname` varchar(25) NOT NULL,
  PRIMARY KEY (`surveyid`),
  KEY `firstname` (`firstname`),
  KEY `firstname_2` (`firstname`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

and this table has userid and dataid as foreign keys. 并且该表具有userid和dataid作为外键。

CREATE TABLE IF NOT EXISTS `jointable` (
`jointableid` int(11) NOT NULL AUTO_INCREMENT,
`dataid` int(11) NOT NULL,
`userid` int(11) NOT NULL,
PRIMARY KEY (`jointableid`),
KEY `surveyid` (`dataid`,`userid`),
KEY `userid` (`userid`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

ALTER TABLE `jointable`
  ADD CONSTRAINT `lookup_ibfk_2` FOREIGN KEY (`userid`) REFERENCES `users` (`userid`) ON DELETE CASCADE ON UPDATE NO ACTION,
  ADD CONSTRAINT `lookup_ibfk_1` FOREIGN KEY (`dataid`) REFERENCES `data` (`dataid`) ON DELETE CASCADE ON UPDATE NO ACTION;

my page which inserts the data is 我插入数据的页面是

<?php 
session_start();

if ($_SESSION['username'])
{
echo "Welcome, ".$_SESSION['username']."! <a href='logout.php'>Logout</a>";
}
else
die("You must be logged in!!");
?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<script src="question.js"></script>
<title>Questionnaire</title>
<style type="text/css"> 
    span {color: #FF00CC}
</style>
</head>
<body  background="images/good.jpg">
<h1></h1>
<form name="quiz" method="post" action="submitdata.php">

First Name: <input type="text" name="firstname" id="fname"/>
<p></p>
Last Name: <input type="text" name="lastname" id="lname"/>
<input type="submit" name="submitbutton" value="Go"></input>
<input type="reset" value="clear all"></input>
</form>
</body>
</html>

and finally submitdata.php is the page which stores the data in the database 最后，submitdata.php是将数据存储在数据库中的页面

<?php
session_start();

if ($_SESSION['username'])
{
echo "Welcome, ".$_SESSION['username']."! <a href='logout.php'>Logout</a>";
}
else
die("You must be logged in!!");
$con=mysql_connect ("localhost","****","****");
mysql_select_db("project",$con);
@$firstname=$_POST['firstname'];
@$lastname=$_POST['lastname'];

$s="INSERT INTO data(`firstname`,`lastname`) VALUES ('$firstname','$lastname')";
echo "You have successfully submit your questions";
 mysql_query ($s);
?>

Furthermore I have this login page 此外，我有此登录页面

<?php 
session_start();

@$username = $_POST['username'];
@$password = $_POST['pass'];

if(@$_POST['Submit']){
if($username&&$password)
{
$connect = mysql_connect("localhost","****","****") or die("Cannot Connect");
mysql_select_db("project") or die("Cannot find the database");

$query = mysql_query("SELECT * FROM users WHERE username='$username'");
$numrows = mysql_num_rows($query);
if($numrows!=0)
{
    while ($row = mysql_fetch_assoc($query))
    {
        $dbusername = $row['username'];
        $dbpassword = $row['password'];
    }
    if($username==$dbusername&&$password==$dbpassword)
    {
        echo "You are login!!!!! Continue now with the survey <a href='mainpage.php'>here</a>";
        $_SESSION['username']=$username;
    }
    else
    {
        echo "<b>Incorrect Password!!!!</b>";
    }
}
else
    //die("That user does not exist");
    echo "<b>That user does not exist</b>";
}
else
echo "<b>You must enter a username and a password</b>";
}
?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>   
<title>Login Page</title>
</head>
<body>

<h2>Login Page</h2>
<form name="loginform" method='POST'>

<fieldset>
<legend>Form</legend>
    <label>Username: <input type="text" name="username"/><span>*</span></label><br/>
    <label>Password: <input type="password" name="pass"/><span>*</span></label>
    <input class="placeButtons" type="reset" value='Reset'/>
    <input class="placeButtons" type="submit" name="Submit" value='Login'/>
    <a href='registration.php'>Register</a>
</fieldset>

</form>
</body>
</html>

So my aim is the user who login or register, fill the page and then add the record to the data (with the data) and jointable (with the ids of the record) at the same time. 因此，我的目标是用户登录或注册，填写页面，然后将记录同时添加到数据（带有数据）和可联合的（带有记录的ID）中。 The problem is with the registration and login pages which I have I can insert the users data and also I can insert the data in the DATA table but I do not have a record in jointable. 问题出在我拥有的注册和登录页面上，我可以插入用户数据，也可以在DATA表中插入数据，但是我没有可连接的记录。 Why is this happen? 为什么会这样呢？

Answer 1

To answer your question: "why is this happen?" 要回答您的问题：“为什么会这样？”

Because you are not inserting any data in join_table and therefore it wont do it for you. 由于您没有在join_table中插入任何数据，因此不会为您完成此操作。

Solution: 解：

You need whats called lastInsertId to be returned once you insert a record in data table. 在数据表中插入记录后，您需要返回称为lastInsertId的内容 。 Here is an example in PDO and I would recommend using PDO: 这是PDO中的示例，我建议使用PDO：

//prepare statement
$stmt_data = $this->db->prepare('
                 INSERT INTO data(`firstname`,`lastname`) 
                 VALUES (:fname, :lname)
             ');

//bind parameter/variables
$stmt_data->bindParam(':fname', $firstName, PDO::PARAM_STR);
$stmt_data->bindParam(':lname', $lastName, PDO::PARAM_STR);

//insert row
$res = $stmt_data->execute();

//get last isnerted ID
if ($res) {
    $id = $this->db->lastInsertId();

    //NOW INSERT INTO jointable with this id
    $stmt_jointable = $this->db->prepare('
                          INSERT INTO join_table(`userid`, dataid`)
                          VALUES (:uid, :dataid)
                      ');

    $stmt_jointable->bindParam(':uid', $userID, PDO::PARAM_INT);
    $stmt_jointable->bindParam(':dataid', $id, PDO::PARAM_INT);

    //insert
    $stmt->execute();
}

TIPS: 提示：

Get rid of mysql_* functions and move on using PDO or MySQLi 摆脱mysql_ *函数并继续使用PDO或MySQLi
For jointable you dont need autoincrement primary key, you can just make both user ID and data ID be the primary keys, since same user can not have multiple data records. 对于可jointable您不需要自动递增主键，只需将用户ID和数据ID都设为主键，因为同一用户不能有多个数据记录。 Isnt it 1-1 relationship? 不是1-1关系吗？

在两个不同的表中添加记录

问题描述

1 个解决方案

解决方案1
1 已采纳 2013-08-09 18:48:02

在两个不同的表中添加记录

问题描述

1 个解决方案

解决方案1 1 已采纳 2013-08-09 18:48:02

解决方案1
1 已采纳 2013-08-09 18:48:02