php/mysql dynamic form fields are not inserted in my table

Question

I have a form that has a variable number of input fields, and i now try to get these values in my database. I got this code from another question here and all the replies where implying that they got it working..so i think i'm doing something wrong here.

I get no error, it just enters one empty entry/row in my database every time i submit the form. The $_POST array is filled with all the data i need, it shows when i print_r it.

<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {


  if (
     !empty($_POST['homeTeam']) && !empty($_POST['awayTeam']) && !empty($_POST['homeWin']) && !empty($_POST['awayWin']) && 
     is_array($_POST['homeTeam']) && is_array($_POST['awayTeam']) && is_array($_POST['homeWin']) && is_array($_POST['awayWin']) && 
     count($_POST['homeWin']) === count($_POST['awayWin'])
  ) {
      $homeTeam_array = $_POST['homeTeam'];
      $awayTeam_array = $_POST['awayTeam'];
      $homeWin_array = $_POST['homeWin'];
      $awayWin_array = $_POST['awayWin'];

      for ($i = 0; $i < count($homeTeam_array); $i++) {

          $homeTeam = mysql_real_escape_string($homeTeam_array[$i]);
          $awayTeam = mysql_real_escape_string($awayTeam_array[$i]);
          $homeWin = mysql_real_escape_string($homeWin_array[$i]);
          $awayWin = mysql_real_escape_string($awayWin_array[$i]);

          $sql = "INSERT IGNORE INTO CalcOdds (homeTeam, awayTeam, homeWin, awayWin) VALUES ('$homeTeam', '$awayTeam', '$homeWin', '$awayWin')"; 
          $conn->query($sql);
          $conn->close();

      }
  }
  echo "<pre>";
  print_r($_POST);
  echo "</pre>";
  echo 'Done!';
}

?>

WARNING: When using mysqli you should be using parameterized queries and bind_param to add user data to your query. DO NOT use manual escaping and string interpolation or concatenation to accomplish this because you will create severe SQL injection bugs. Accidentally unescaped data is a serious risk. Using bound parameters is less verbose and easier to review to check you’re doing it properly. — tadman
– tadman, Commented Oct 9, 2018 at 19:59
Note: The object-oriented interface to mysqli is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete mysql_query interface. Before you get too invested in the procedural style it’s worth switching over. Example: $db = new mysqli(…) and $db->prepare("…") The procedural interface is an artifact from the PHP 4 era when mysqli API was introduced and should not be used in new code. — tadman
– tadman, Commented Oct 9, 2018 at 19:59
Why are you mixing mysqli_ and mysql_ functions and where are you actually connecting to your database? — Cyclonecode
– Cyclonecode, Commented Oct 9, 2018 at 20:03
Thanks @tadman for the warning, i'll look into this.I only just picked up coding again after i while and really need to do some studying :) — Awesom-o
– Awesom-o, Commented Oct 9, 2018 at 20:28

Kholy · Accepted Answer · 2018-10-09 20:28:08Z

3

I think the problem is because you have $conn->close(); inside the for loop try to add it after the loop like this:

     for ($i = 0; $i < count($homeTeam_array); $i++) {

          $homeTeam = mysql_real_escape_string($homeTeam_array[$i]);
          $awayTeam = mysql_real_escape_string($awayTeam_array[$i]);
          $homeWin = mysql_real_escape_string($homeWin_array[$i]);
          $awayWin = mysql_real_escape_string($awayWin_array[$i]);

          $sql = "INSERT IGNORE INTO CalcOdds (homeTeam, awayTeam, homeWin, awayWin) VALUES ('$homeTeam', '$awayTeam', '$homeWin', '$awayWin')"; 
          $conn->query($sql);            
      }
      $conn->close();

answered Oct 9, 2018 at 20:28

Kholy

4413 silver badges10 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

user9407009 · Accepted Answer · 2018-10-09 20:08:25Z

0

Instead of doing !empty() I'd do isset()
By the looks of things you haven't actually established a connection to your database.
Make sure that the data actually gets through the if() statement by using a echo() for example.

answered Oct 9, 2018 at 20:08

user9407009

Collectives™ on Stack Overflow

php/mysql dynamic form fields are not inserted in my table

2 Answers 2

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related