Question

如何在我的页面中的“我的帐户”链接下显示仅登录用户的信息？我尝试在两个表之间设置一个外键，因为我有一个用于登录的表，另一个用于插入照片，信息内容以及将照片的作者与我在表{{1}中的用户名相关联}。我的目的是只显示登录用户的用户信息和照片。

问题是，当我尝试设置外键时，它设置了外键但数据没有进入表中。在我上传图片并插入一些信息的页面上，它向我显示上传成功，但当我去我的数据库中看到什么都没有。任何帮助？我已经坚持了很长时间。

以下是我创建表格的方法。

users

这是我在创建表格后尝试添加FK的内容..

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;

最后这是我的Upload.php

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

这是main.php页面，其中显示用户登录时的用户名

<?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 Photo has been Uploaded.</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($sub==0)
{
?> 
<label  for="file"><font  size="5"><b>Choose Photo:</b></font></label>
<input id="shiny" type="file" name="file" onchange="file_selected = true;" required><br>
First Name:<input  type="text" name="fname" value="<?php echo (isset($_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>

Answer 1

您必须实际将用户ID添加到查询中：

$photo= "pictures/" . $_FILES["file"]["name"];
$fname = $_POST["fname"];
$lname = $_POST["lname"];
$age   =$_POST["age"];
$username = "?";//Add code here to get the username you are interested in
$stmt = $mysqli->prepare("INSERT INTO photos (photo, Firstname, Lastname, Age, author) VALUES (?, ?, ?, ?,?)");

// TODO check that $stmt creation succeeded

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

您不会显示抓取用户信息的代码，但假设您当前的查询类似于：

SELECT * FROM users WHERE username = 'someUser';

您可以更改此选项以包含照片（分组以防用户可以拥有多张照片：

SELECT * FROM users u
LEFT JOIN photos p ON (p.author = u.username)
WHERE u.username = 'someUser'
GROUP BY u.username;

Answer 2

您在author列上放置了一个约束，并使其不为空。这意味着photos表中的每条记录都需要填充author，此列中的值必须出现在users列的username表格中。

在INSERT查询中，您没有为author设置任何值，因此由于约束和表定义（NOT NULL）而失败。

要使其正常运行，您必须确保始终填充author列，或使其可以为空以允许匿名上传。

Answer 3

您需要在插入中设置author字段。由于您的外键约束，任何没有此值的插入尝试都将失败。

您还应该检查并记录任何MySQL错误。如果您已经这样做了，那么您已经知道插入因外键约束而失败。

设置外键只显示用户信息

3 个答案: