CODEWITHSUNDEEP
phpmysqli
MrCaptain Alex
MrCaptain Alex

Reputation: 169

PHP POST with MYSQLI

I'm trying a simple upload of datas from a table in a database with mysqli functions and sql. I don't know what i'm doing of wrong.

This is the HTML code:

<form name="formCompany" method="post" action="AddCompany.php" >
              <div class="modal-body">
                <div class="container w-75">
                    <input type="text" id="nameCompany" name="nameCompany" 
 class="form-control" placeholder="Name" aria-label="Name" aria- 
describedby="basic-addon1">
                    <br/>
                    <input type="url" id="webCompany" name="webCompany" 
 min="1900" class="form-control" placeholder="WebSite" aria-label="WebSite" 
 aria-describedby="basic-addon1">
                    <br/>
                    <input type="text" id="placeCompany" name="placeCompany" 
 class="form-control" placeholder="Place" aria-label="Place" aria- 
describedby="basic-addon1">
                    <br/>
                    <input type="email" id="emailCompany" name="emailCompany" class="form-control" placeholder="Email" aria-label="Email" aria-describedby="basic-addon1">
                    <br/>
                    <input type="text" id="noteCompany" name="noteCompany" 
 class="form-control" placeholder="Note" aria-label="Note" aria- 
 describedby="basic-addon1">
                </div>
              </div>
              <div class="modal-footer">
                <button type="submit" id="submitCompany" class="btn btn- 
primary">Save</button>
              </div>
            </form>

and this the page "AddCompany.php" :

 <?php
 include 'DB.php';



 $name = $_POST["nameCompany"];
 $place = $_POST["placeCompany"];
 $web = $_POST["webCompany"];
 $email = $_POST["emailCompany"];
 $note = $_POST["noteCompany"];

// Inserisce una nuova compagnia
$sql2="SELECT company_id FROM company ORDER BY company_id DESC LIMIT 1;";
$result=mysqli_query($connessione,$sql2);
$row=mysqli_fetch_array($result,MYSQLI_NUM);
    $last_id=$row[0];
  $id = $last_id+1;

 $sql ="insert into company (company_id, name, place, web,email,note) values 
 ('$id','$name','$place','$web','$email','$note');";



 $res1=mysqli_query($connessione, $sql);
 if($res1 !=FALSE)   {

 header("Location: Principale.php");
  }
 else
{
echo "Impossibile aggiungere Compagnia<br>";
 echo "<a href='Principale.php'>Clicca Qui</a> per tornare alla pagina 
principale.";
}
 ?>

It always print me the else because it finds $res1=FALSE

Why?

Upvotes: 1

Views: 654

Answers (2)

Shan Biswas
Shan Biswas

Reputation: 87

Try this one, you will see what you are doing wrong

DB.php

define ("DB_USER", "Database username");
define ("DB_PASSWORD", "Database password");
define ("DB_DATABASE", "Database name");
define ("DB_HOST", "Hostname");

//@ $con = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
@ $con = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
if($con->connect_error)
    die("Unable to connect to database : ".$con->connect_error);

Change the query executions

$res1=$con->query($sql);
if($res1){
   header("Location: Principale.php");
}
else
{
  echo "Error -> ".mysqli_error($con);
  echo "Impossibile aggiungere Compagnia<br>";
  echo "<a href='Principale.php'>Clicca Qui</a> per tornare alla pagina principale.";
}

Upvotes: 1

pedram shabani
pedram shabani

Reputation: 1680

name is keyword in Mysql you have to us `

your query must to be something like this 

$sql ="insert into company (company_id, `name`, place, web,email,note) values 
 ('$id','$name','$place','$web','$email','$note');";

Upvotes: 0

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