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
– Ash, Commented Apr 1, 2016 at 16:26
Please stop using mysql_* functions. These extensions have been removed in PHP 7. Learn about prepared statements for PDO and MySQLi and consider using PDO, it's really pretty easy. — Jay Blanchard
– Jay Blanchard, Commented Apr 1, 2016 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
– Ash, Commented Apr 1, 2016 at 16:29

Umair Khan · Accepted Answer · 2016-04-01 16:25:56Z

2

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();
?>

answered Apr 1, 2016 at 16:25

Umair Khan

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6 Comments

Ash Over a year ago

it would be better you just highlight the few lines of code you modified instead of the whole thing :)

M. Aleem Over a year ago

I have that in my original code just accidently missed it out here. It still doesn't work.

M. Aleem Over a year ago

Why isn't the $sql being executed?

M. Aleem Over a year ago

@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.

Umair Khan Over a year ago

you should remove value from the select tag

|

Conrad Lotz · Accepted Answer · 2016-04-01 16:53:41Z

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 1, 2016 at 16:53

Conrad Lotz

8,9533 gold badges25 silver badges28 bronze badges

2 Comments

M. Aleem Over a year ago

If you look at my code, I already have $result = mysql_query($sql);

Conrad Lotz Over a year ago

Still don't see it. there is one $result1 = mysql_query($sql1); for the Away Team but not the Home Team

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