Updating table rows from different table data not working

I have two tables in my database. One called wp_tickets and another one called wp_tickets_regenerated.

I have two tables in my database. One called wp_tickets and another one called wp_tickets_regenerated.

I'm trying to update some of the data in the wp_tickets_regenerated table from the wp_tickets table if the lottery_id matches so i end up with a new table with different ticket_number generated for each order placed for that product.

The code I have so far is throwing up an error:

 WordPress database error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM wp_tickets
        WHERE lottery_id = wp_tickets.lottery_id' at line 6 for query UPDATE wp_tickets_regenerated

Can anyone help please?

wp_tickets table

+---------+-----------+-----------+---------------+----------+------------+
| user_id | full_name | answer_id | ticket_number | order_id | lottery_id |
+---------+-----------+-----------+---------------+----------+------------+
| 0       | test      | 1         | 3             | 791      | 790        |
+---------+-----------+-----------+---------------+----------+------------+
| 0       | test      | 1         | 5             | 791      | 790        |
+---------+-----------+-----------+---------------+----------+------------+
| 0       | test      | 1         | 10            | 791      | 790        |
+---------+-----------+-----------+---------------+----------+------------+

wp_tickets_regenerated

   +---------+-----------+-----------+---------------+----------+------------+
    | user_id | full_name | answer_id | ticket_number | order_id | lottery_id |
    +---------+-----------+-----------+---------------+----------+------------+
    |         |           |           | 1             |          | 790        |
    +---------+-----------+-----------+---------------+----------+------------+
    |         |           |           | 2             |          | 790        |
    +---------+-----------+-----------+---------------+----------+------------+
    |         |           |           | 3             |          | 790        |
    +---------+-----------+-----------+---------------+----------+------------+

What i have so far is:

 $wpdb->query(
       'UPDATE wp_tickets_regenerated
        SET user_id = wp_tickets.user_id,
            full_name = wp_tickets.full_name,
            answer_id = wp_tickets.answer_id,
            order_id = wp_tickets.order_id
        FROM wp_tickets
        WHERE lottery_id = wp_tickets.lottery_id'
  );


PHP / MySQL - Count how many table in database contain Specific name

I'm building an admin to control an interactive game, and I'm using PHP/MySql to store all the data for all the rounds. Each of the tables that I want to load into the game via PHP are titled "Puzzle*" with the * being an incremental index number. There are a few other tables in my database for global game data and bonus rounds.

I'm building an admin to control an interactive game, and I'm using PHP/MySql to store all the data for all the rounds. Each of the tables that I want to load into the game via PHP are titled "Puzzle*" with the * being an incremental index number. There are a few other tables in my database for global game data and bonus rounds.

What I want to be able to do is count the number of tables in my database that contain the name "Puzzle", and then use that total number as a randomizer to select a puzzle randomly. I plan to have roughly 50 puzzles, so an automated way of selecting a random puzzle, and then if I want to select another puzzle - to remove that previous puzzle from the list so that there is no chance of it being repeated until I start the whole game over again.

MySQL Database

Database Name: puzzleGame
Table 1: puzzle1
Table 2: puzzle2
Table 3: puzzle3
Table 4: puzzle4
Table 5: globals
Table 6: bonus


Inserting session array into mysql database

I am still learning bits of PHP and have been stuck on this issue regarding the POST data from radio buttons within a form being inserted into a mysql database from a session array. Currently, I have 2 questions within my survey and they consist of radio button options for a user to select from. The relevant answers can be selected and they are stored in a session array:

I am still learning bits of PHP and have been stuck on this issue regarding the POST data from radio buttons within a form being inserted into a mysql database from a session array. Currently, I have 2 questions within my survey and they consist of radio button options for a user to select from. The relevant answers can be selected and they are stored in a session array:

array(1) { ["choice"]=> array(2) { [0]=> string(11) "56-65 years" [1]=> string(17) "Prefer not to say" } }

However I was not too sure if the array should be as such. The two answers chosen from the radio options in the above example are "56-65 years" and "Prefer not to say".

Following on from the session array, I am trying to insert each of those answers in to a mysql database. I have a column created to insert each answer (i.e. there are two columns (question 1 and question 2) to store the two answers from the example above).

I am aware that my code is open to SQL injection and plan on fixing that after overcoming the inserting of data.

Question.php

     <form method="POST" action="process.php">
      <ul class="choices">
  &lt;?php while($row = $answers-&gt;fetch_assoc()):?&gt;
    &lt;li&gt;&lt;input type="radio" name="choice" value="&lt;?php echo $row['answerText']; ?&gt;"&gt;

    &lt;?php echo $row['answerText'] ?&gt;&lt;/li&gt;
    &lt;?php endWhile; ?&gt;

  &lt;/ul&gt; 
  &lt;input type="submit" name="submit" value="Next Question"&gt;
  &lt;input type="hidden" name="number" value="&lt;?php echo $number; ?&gt;"&gt;

</form>

Process.php

<?php

    if(isset($_POST['submit'])){

$number = $_POST['number'];
$next = $number+1;

$selected_choice = $_POST['choice'];
$_SESSION['choice'][] = $selected_choice;

foreach ($_SESSION['choice'] as $key) {

$user_responses = $key;

}

$insertdata = "INSERT INTO userAnswersTable('question1', 'question2') VALUES ($_SESSION['choice'])";
$resulting = $mysqli->query($insertdata) or die($mysqli->error.LINE);

}
?>

I would expect each of the user responses to be in an array from which each individual selected choice within the session array can be stored in each corresponding column within the mySql table. Any suggestions would be very helpful.

Flexible PHP 7- MySQLi database class (+ download)

Flexible PHP 7- MySQLi database class (+ download)

We all know it: Write annoying code over and over again! That’s over now. With this PHP MySQLi database class you can easily connect to your MySQL database.

This class is very simple and flexible for any project. It also ensures that no SQL-Injectionattacks can be carried out. You may use the class below freely for your projects (also commercially) and of course extend it as you like. And so you use it:

1. Include class in your project structure

You copy the complete class and create a new file in your project structure, e.g. Database.php and paste the complete code there.

<?php
/**
* Simple Database class for PHP7+
* The class contains main functions for your database. For a detailed documentation, see: https://webdeasy.com/
* created 08.11.2017
* 
* @author LH
*/
class Database {
  private $host, $database, $username, $password, $connection;
  private $port = 3306;

/**
*

  • Sets the connection credentials to connection to your database
  • @param string $host - the host of your database
  • @param string $username - the username of your database
  • @param string $password - the password of your database
  • @param string $database - your database name
  • @param integer $port - the port of your database
  • @param boolean $autoconnect - to auto connect to the database after settings connection credentials
    /
    function __construct($host, $username, $password, $database, $port = 3306, $autoconnect = true) {
    $this->host = $host;
    $this->database = $database;
    $this->username = $username;
    $this->password = $password;
    $this->port = $port;
    if($autoconnect) {
    $this->open();
    }
    }
    /
    *
  • Open the connection to your database
    /
    function open() {
    $this->connection = new mysqli($this->host, $this->username, $this->password, $this->database, $this->port);
    }
    /
    *
  • Close the connection to your database
    /
    function close() {
    $this->connection->close();
    }
    /
    *
  • Execute your query
  • @param string $query - your sql query
  • @return the result of the executed query
    /
    function query($query) {
    return $this->connection->query($query);
    }
    /
    *
  • Escape your parameter
  • @param string $string - your parameter to escape
  • @return the escaped string
    */
    function escape($string) {
    return $this->connection->escape_string($query);
    }
    }
    ?>
2. Include database class

Your program code runs in a different PHP file. There you add the following code to include the class:

require_once("path/to/your/file/Database.php");

What’s the different between require and includeThis!

3. Create instance

To connect to the database, we need to create an instance of the class. For this we need the access data to the database. You can find the optional parameters of the construct in the class. A call could look like this:

$database = new Database("localhost", "testuser", "verySafePassword", "ourDatabase");

An attempt is made to establish a connection to the database with the access data transferred. This happens automatically if the $autoconnect parameter is set to true (see line 26).

If you don’t get any errors, the connection to the database is established successfully. Now you can formulate your SQL queries.

4. Escape parameters!

That’s probably the most important step. To prevent SQL injections, every parameter you use in the query must be escaped. This can prevent harmful SQL statements – whether intentional or unintentional – from getting into your query and thus into your server system. The following lines must be inserted before each query:

$parameter = $_POST["id"];
$parameter = $db->escape($parameter);

In this example, the id parameter of the POST request is passed to PHP and escaped by the second line.

5. Your queries

You can pass the queries to the query($query) function. The query is sent to the database and the function returns the result.

Example of an Insert Query

$query = "INSERT INTO users (id, name) VALUES (1, 'Peter Parker');";
$db->query($query);

Example of a select query with output of the result

$query = "SELECT name FROM users WHERE id = 1";
$result = $db->query($query);

while($row = mysqli_fetch_assoc($result)) {
echo "Name: " . $row["name"];
}
</pre>

At the end of your program you should call the close() function to close open database connections.

$db->close();

Ultimately, these are the functions of the database class. I kept them as simple as possible, but still tried to include all important and security relevant functions. And yes: I am aware that PHP and the MySQLi class have much more functions to offer. But this class is just for simple, fast and flexible applications that only require a simple database connection.

Safety instructions or suggestions for improvement are welcome in the comments, so that I can add them! 

Originally published at webdeasy.de on 26. May 2019

====================================================================

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