显示MySQL中特定行的数据

I'm building a simple bug tracking tool. When you create a new project, all the info you fill in in the form, gets stored in the database.

When you create the new project you get redirected to a unique project page. On top of the page it shows the name of the project, but it's not the name of the project I just created, it always shows the name of the first project in the MySQL table.

How can I show the name of the project I just created?

With this query I retrieve the data from the database.

$query = "SELECT CONCAT(name) 
AS name FROM projects";
$result = @mysql_query ($query)

With this I show the project name, but it always shows the name of the first record in the table.

    <?php 
    if ($row = mysql_fetch_array ($result))
        echo '<h5>' . $row['name'] . '</h5>'; 
    ?>

It isn't yet SQL Injection prove and is far from complete... But I'm really struggling with this problem.

Try using ORDER BY.

$query = "SELECT CONCAT(name) 
AS name FROM projects ORDER BY id DESC";

This would show the most recent project (assuming you have an ID column).

However, a much better way is to have an ID variable on the page.

$query = "SELECT CONCAT(name) 
AS name FROM projects WHERE id=?";

You need an AUTO_INCREMENT field on your table for a unique identifier (at least, you really should). Then you can do something like this:

<?php
$sql = new MySQLi('localhost', 'root', '', 'database');

$sql->query('INSERT INTO `projects` (`name`) VALUES ("Test Project");');
$projectID = $sql->insert_id; // Returns the auto_increment field value of the last insert query performed
// So this assumes you have a field in your table called "id" in this example

$res = $sql->query('SELECT CONCAT(`name`) AS `name` FROM `projects` WHERE `id` = '.$projectID.';');

if ($row = $res->fetch_assoc()) {
    echo '<h5>'.$row['name'].'</h5>';
}
?>

Since you were calling for a redirect to the unique project page, you should have something like this: header("Location: project.php?id=$projectID");

Then, on project.php, you can attempt to fetch the project with the query above, only your query's WHERE clause should be something like:

'`id` = '.intval($_GET['id']).';'

Technically, you could pass all the project info along to the next page as a request or a session cookie and save yourself a query altogether. Just make sure you keep the id handy so it's easy to update the record.