PHP Dropdown and database

Question

I'm new to php. I have a dropdown menu in my form and the dropdown options are coming from a database and I'm trying to insert the selected options in the dropdown menu to a separate table in my database. The query seems to be getting executed but the team name values are not being inserted into the database. This is the code for the form. Any help is much appreciated!

<form class="form-register" method="POST" enctype="multipart/form-data">
Match Type
  <select class="form-control" name="MatchType" value="Match Type">
<option value="Select one">Select One</option>
<option value="T20">Twenty20 Match</option>
<option value="OneDay">One-Day Match</option>
<option value="Test">Test Match</option> </select>
Home Team
<?php  
mysql_select_db('cricket_system');
$sql = "SELECT TeamName FROM teams";
  echo "<select class='form-control' name='Team1' value='Team1'>";
    while ($row = mysql_fetch_array($result)) {
      echo "<option value='". $row['TeamName']."'>". $row['TeamName']."</option>";
    }
 echo "</select> "; 
 ?> 
  Away Team
  <?php  
mysql_select_db('cricket_system');
    $sql1 = "SELECT TeamName FROM teams";
    $result1 = mysql_query($sql1);
   echo "<select class='form-control' name='Team2' value='Team2'>";
    while ($row = mysql_fetch_array($result1)) {
      echo "<option value='". $row['TeamName']."'>". $row['TeamName']."</option>";
    }
 echo "</select> "; 
 ?>   
   Date (yyyy/mm/dd)
<input type="text" id="Date" name="Date" class="form-control" placeholder="Date (yyyy/mm/dd)" required>
<br><button class="signupbutton" type="submit" name="submit" >Add Match</button> <br> <br>
 </form>
<?php
include('includes/database.php');
mysql_select_db('cricket_system');
if(isset($_POST['submit'])){
 $Team1 = $_POST['Team1'];
 $Team2 = $_POST ['Team2'];
 $MatchType = $_POST['MatchType'];
  $insert = "INSERT INTO matches (Team1, Team2, Date, MatchType) values 
  ('$Team1', '$Team2', '$Date', '$MatchType')";
  $add = mysql_query($insert);
  if ($add) {
      echo "<script>alert('Match has been successfully added.')</script>";
  }
  else {
      echo mysql_error();
  }
}
mysql_close();
?>

`$sql` is not being executed, so that explains why your Home Teams wont appear. — Ash, Apr 01 '16 at 16:26
[Your script is at risk for SQL Injection Attacks.](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) — Jay Blanchard, Apr 01 '16 at 16:27
Please [stop using `mysql_*` functions](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php). [These extensions](http://php.net/manual/en/migration70.removed-exts-sapis.php) have been removed in PHP 7. Learn about [prepared](http://en.wikipedia.org/wiki/Prepared_statement) statements for [PDO](http://php.net/manual/en/pdo.prepared-statements.php) and [MySQLi](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) and consider using PDO, [it's really pretty easy](http://jayblanchard.net/demystifying_php_pdo.html). — Jay Blanchard, Apr 01 '16 at 16:27
@JayBlanchard Are you a bot or a human? Just becasue you post those same two comments on every question with `mysql_*` ;) — Ash, Apr 01 '16 at 16:29
I am most certainly not a bot! ¯\\_(ツ)_/¯ I post those comments (sometimes others) because they are important @ash — Jay Blanchard, Apr 01 '16 at 16:30

score 2 · Answer 1 · answered Apr 01 '16 at 16:25

You are missing the $Date variable that i added in the code

<form class="form-register" method="POST" enctype="multipart/form-data">
Match Type
  <select class="form-control" name="MatchType" value="Match Type">
<option value="Select one">Select One</option>
<option value="T20">Twenty20 Match</option>
<option value="OneDay">One-Day Match</option>
<option value="Test">Test Match</option> </select>
Home Team
<?php  
mysql_select_db('cricket_system');
$sql = "SELECT TeamName FROM teams";
  echo "<select class='form-control' name='Team1' value='Team1'>";
    while ($row = mysql_fetch_array($result)) {
      echo "<option value='". $row['TeamName']."'>". $row['TeamName']."</option>";
    }
 echo "</select> "; 
 ?> 
  Away Team
  <?php  
mysql_select_db('cricket_system');
    $sql1 = "SELECT TeamName FROM teams";
    $result1 = mysql_query($sql1);
   echo "<select class='form-control' name='Team2' value='Team2'>";
    while ($row = mysql_fetch_array($result1)) {
      echo "<option value='". $row['TeamName']."'>". $row['TeamName']."</option>";
    }
 echo "</select> "; 
 ?>   
   Date (yyyy/mm/dd)
<input type="text" id="Date" name="Date" class="form-control" placeholder="Date (yyyy/mm/dd)" required>
<br><button class="signupbutton" type="submit" name="submit" >Add Match</button> <br> <br>
 </form>
<?php
include('includes/database.php');
mysql_select_db('cricket_system');
if(isset($_POST['submit'])){
 $Team1 = $_POST['Team1'];
 $Team2 = $_POST ['Team2'];
$Date= $_POST ['Date'];
 $MatchType = $_POST['MatchType'];
  $insert = "INSERT INTO matches (Team1, Team2, Date, MatchType) values 
  ('$Team1', '$Team2', '$Date', '$MatchType')";
  $add = mysql_query($insert);
  if ($add) {
      echo "<script>alert('Match has been successfully added.')</script>";
  }
  else {
      echo mysql_error();
  }
}
mysql_close();
?>

it would be better you just highlight the few lines of code you modified instead of the whole thing :) — Ash, Apr 01 '16 at 16:27
I have that in my original code just accidently missed it out here. It still doesn't work. — M. Aleem, Apr 01 '16 at 16:30
@UmairKhan no error, the query supposedly is getting executed as the alert appears and other fields are being added to the database just the two teams are being left blank in the database. — M. Aleem, Apr 01 '16 at 16:41

score 0 · Answer 2 · answered Apr 01 '16 at 16:53

0

The SQL for the first dropdown never gets populated as the $result variable is never set therefore the $_POST["Team1"] is null or empty/unset. The first dropdown would have never worked. Try add the following:

$result = mysql_query($sql);

What I would suggest is to add the code and then submit it again once all the values are in both Team dropdowns.

As a simple check do the following when assigning the team names

$Team1 = isset($_POST['Team1']) ? $_POST['Team1'] : '';

If this inserts an empty value for Team1 in the database table then it proofs the above dropdown was the problem

answered Apr 01 '16 at 16:53

Conrad Lotz

8,200
3
23
27

If you look at my code, I already have $result = mysql_query($sql); – M. Aleem Apr 02 '16 at 13:43
Still don't see it. there is one `$result1 = mysql_query($sql1);` for the `Away Team` but not the `Home Team` – Conrad Lotz Apr 02 '16 at 18:11

PHP Dropdown and database

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