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Editorial Team
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Asked: 2026-06-15T08:25:19+00:00

I have this simple code where I am trying to have 3 fields, first

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I have this simple code where I am trying to have 3 fields, first name, last name, and email, tag to mysql database from a website form. When I test the form and hit “register”
the error comes up that “An Error Has Occured. The item was not added.” How can I debug this, as I’m not sure what point the error is coming in?

<html>
<head>
<title> Nomad - New User Registration Results</title>
</head>

<body>
   <h1>Nomad - New User Registration Results</h1>
 <?php
 // create short variable names
 $First_Name=$_POST['First_Name'];
 $Last_Name=$_POST['Last_Name'];
 $Email=$_POST['Email'];
 if(!$First_Name || !$Last_Name || !$Email) {
 echo "You have not entered all the required details.<br />"
 ."Please go back and try again.";
 exit;
 }
 @ $db=new mysqli('localhost','nomad_steve','steven','nomad_prod');
 if (mysqli_connect_errno()) {
 echo "Error: Could not connect to database. Get it together Steve!";
 exit;
 }
 $query = "insert into nomad_prod values
      (' " .$First_Name.",' " .$Last_Name."',' ".$Email."')";
  $result=$db->query($query);
 if ($result) {
 echo $db->affected_rows." book inserted into database.";
  } else {
 echo "An Error Has Occured. The item was not added.";
  }
 $db->close();
 ?>
 </body>
   </html>
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1 Answer

  1. Editorial Team

    Try this may solve your problem

    $query = "insert into Player(first_name,last_name,email) values('" .$First_Name."','".$Last_Name."','".$Email."')";
$result=$db->query($query);

    I have checked and got it you have to specify your column name where you have to enter your data.

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