1

我试图设置一个外键，以便我可以将照片的作者与userstable 的用户名相关联，但我无法创建表。

更新我解决了无法创建表的错误，但数据仍然没有进入表中。

CREATE TABLE IF NOT EXISTS `users` (
`username` varchar(30) NOT NULL ,
`password` varchar(40) default NULL,
`usersalt` varchar(8) NOT NULL,
`userid` varchar(32) default NULL,
`userlevel` tinyint(1) unsigned NOT NULL,
`email` varchar(50) default NULL,
`timestamp` int(11) unsigned NOT NULL,
`actkey` varchar(35) NOT NULL,
`ip` varchar(15) NOT NULL,
regdate` int(11) unsigned NOT NULL,
PRIMARY KEY  (`username`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;



CREATE TABLE IF NOT EXISTS photos (
ref int(10) unsigned NOT NULL auto_increment,
    photo varchar(75),
Firstname varchar(35),
    Lastname varchar(35),
    Age INT(3),
    author varchar(30) NOT NULL,
PRIMARY KEY (ref)

) ENGINE=InnoDB DEFAULT CHARSET=latin1;

这是我在创建表后尝试添加 FK 的内容。

ALTER TABLE photos
ADD CONSTRAINT FK_photos
FOREIGN KEY (author) REFERENCES users(username)
ON UPDATE CASCADE
ON DELETE CASCADE;

这是将照片和一些信息连同它一起插入数据库的代码...

<?php
include("/include/session.php");
if(!$session->logged_in){ header("Location: ../main.php"); } else {
}
?>



<?php
$sub=0;

ini_set( "display_errors", 0);
if(isset($_REQUEST['submited'])) {
// your save code goes here

$allowedExts = array("jpg", "jpeg", "gif", "png");
$extension = end(explode(".", $_FILES["file"]["name"]));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2097152)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "";
if (file_exists("pictures/" . $_FILES["file"]["name"]))
{
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload already exists.</b></font>";
  }

else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"pictures/" . $_FILES["file"]["name"]);
$sub= 1;
$mysqli = new mysqli("localhost", "root", "", "secure_login");

// TODO - Check that connection was successful.

$photo= "pictures/" . $_FILES["file"]["name"];
$fname = $_POST["fname"];
$lname = $_POST["lname"];
$age   =$_POST["age"];
$stmt = $mysqli->prepare("INSERT INTO photos (photo, Firstname, Lastname, Age) VALUES (?, ?, ?, ?)");

// TODO check that $stmt creation succeeded

// "s" means the database expects a string
$stmt->bind_param("ssss", $photo, $fname, $lname, $age);

$stmt->execute();

$stmt->close();

$mysqli->close();



echo "<font size='7' color='white'><b> Success! Your item has been listed.</b></font>";
echo '<meta http-equiv="refresh" content="2;url=home.php">';
}

}
}
else
{
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload is not an image or it exceeds 2MB in size.</b></font><br><font color='blue'><i>Only images under size of 2MB are allowed</i></font>.";
}
}


?>

<form action="" method="post" enctype="multipart/form-data">
<input type="hidden" name="submited" value="true" />


<?php
ini_set( "display_errors", 0);
if($sub==0)
{
?> 
<label  for="file"><font  size="5"><b>Choose Photo:</b></font></label>
<input id="shiny" type="file" name="file" onchange="file_selected = true;" required><br>
Last Name:<input  type="text" name="fname" value="<?php echo (isset($_POST['fname']) ? htmlspecialchars($_POST['fname']) : ''); ?>"required><br> 
Last Name:<input  type="text" name="lname" required><br> 
Age:<input type="text" name="age" required><br>
<input id="shiny" type="submit" value="Submit" name="submit">
<?php
}
?>


</form>
</div>
4

3 回答 3

1

帮自己一个大忙，在所有表上使用整数作为 ID 字段，并将其用作主键字段。然后让你所有的外键约束引用它。在此处查看 MySQL 文档中的第一个示例：http: //dev.mysql.com/doc/refman/5.5/en/example-foreign-keys.html

另外，发布您的错误消息，我会更新我的答案。

于 2013-01-21T23:06:33.610 回答
0

如果不存在则创建表users（ usernamevarchar(30) NOT NULL， passwordvarchar(40) 默认 NULL， usersaltvarchar(8) NOT NULL， useridvarchar(32) 默认 NULL， userleveltinyint(1) unsigned NOT NULL， emailvarchar(50) 默认 NULL， timestampint (11) unsigned NOT NULL, actkeyvarchar(35) NOT NULL, ipvarchar(15) NOT NULL, regdateint(11) unsigned NOT NULL, PRIMARY KEY ( username) ENGINE=InnoDB DEFAULT CHARSET=latin1;

如果不存在照片，则创建表 ( ref int(10) unsigned NOT NULL auto_increment, photo varchar(75), Firstname varchar(35), Lastname varchar(35), Age INT(3), author varchar(30) NOT NULL, PRIMARY KEY (ref) ) ENGINE=InnoDB 默认字符集=latin1;

一些变化，但对我有用。但是尝试在 id 列上设置外键，而不是 varchar。

于 2013-01-21T23:09:00.423 回答
0

看起来您的表定义后面需要一个逗号photo varchar(75)

CREATE TABLE IF NOT EXISTS photos (
    ref int(10) unsigned NOT NULL auto_increment,
    photo varchar(75),
    Firstname varchar(35),
    Lastname varchar(35),
    Age INT(3),
    author varchar(30) NOT NULL,
PRIMARY KEY (ref)

) ENGINE=InnoDB DEFAULT CHARSET=latin1;
于 2013-01-21T23:14:28.487 回答