php inside php with $_session echo

Question

I am having an issue, I am trying to create a form registration but this page will be accessible only by staff people. So I have: <?php if($_SESSION['id']) to verify the users if they are logged in and display the echo to this users.

I am having trouble to insert the Birthday registration as I use PHP to calculate the years, months and days. Here is the part of the code

if($_POST['submit']=='Rregjistro')
                {
// If the Register form has been submitted

$err = array();

if(strlen($_POST['emri'])<4 || strlen($_POST['emri'])>32)
{
    $err[]='Emri duhet te permbaje nga 3 deri 32 karaktere!';
}

if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['emri']))
{
    $err[]='Emri juaj permban karakter te palejueshme!';
}
if(strlen($_POST['atesia'])<4 || strlen($_POST['atesia'])>32)
{
    $err[]='Atesia duhet te permbaje nga 3 deri 32 karaktere!';
}

if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['atesia']))
{
    $err[]='Atesia juaj permban karakter te palejueshme!';
}

if(strlen($_POST['mbiemri'])<4 || strlen($_POST['mbiemri'])>32)
{
    $err[]='Mbiemri duhet te permbaje nga 3 deri 32 karaktere!';
}

if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['mbiemri']))
{
    $err[]='Mbiemri juaj permban karakter te palejueshme!';
}

if(strlen($_POST['vendlindja'])<4 || strlen($_POST['vendlindja'])>32)
{
    $err[]='Vendlindja duhet te permbaje nga 3 deri 32 karaktere!';
}

if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['vendlindja']))
{
    $err[]='Vendlindja juaj permban karakter te palejueshme!';
}
if(strlen($_POST['vendbanimi'])<4 || strlen($_POST['vendbanimi'])>32)
{
    $err[]='Vendbanimi duhet te permbaje nga 3 deri 32 karaktere!';
}

if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['vendbanimi']))
{
    $err[]='Vendbanimi juaj permban karakter te palejueshme!';
}
if(strlen($_POST['telefon'])<4 || strlen($_POST['telefon'])>10)
{
    $err[]='Telefoni duhet te permbaje nga 5 deri 10 numra!';
}

if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['telefon']))
{
    $err[]='Telefoni juaj permban karakter te palejueshme!';
}


if(!count($err))
{


    $_POST['emri'] = mysql_real_escape_string($_POST['emri']);
    $_POST['atesia'] = mysql_real_escape_string($_POST['atesia']);
    $_POST['mbiemri'] = mysql_real_escape_string($_POST['mbiemri']);
    $_POST['vendlindja'] = mysql_real_escape_string($_POST['vendlindja']);
    $_POST['vendbanimi'] = mysql_real_escape_string($_POST['vendbanimi']);
    $_POST['telefon'] = mysql_real_escape_string($_POST['telefon']);
    $_POST['zona'] = mysql_real_escape_string($_POST['zona']);
    // Escape the input data


    mysql_query("INSERT INTO antaret(ID,emri,atesia,mbiemri,ditelindja,vendlindja,vendbanimi,telefon,zona,regtime)
                    VALUES('".$_POST['emri']."','".$_POST['atesia']."','".$_POST['mbiemri']."','".$date."','".$_POST['vendlindja']."','".$_POST['vendbanimi']."','".$_POST['telefon']."','".$_POST['zona']."',NOW())");


}
                }

I guess I have something wrong in the code but is missing to my eyes and also I have no error response so I guess is something with the query.

Thank you

mysql is deprecated since PHP 7 you should really use alternatives like mysqli or PDO — Shogunivar
– Shogunivar, Commented Apr 4, 2017 at 10:04
if($_POST['submit']=='Rregjistro') is not opening any brackets but you're closing an extra one at the end. That would normally be a syntax error which any IDE would be very vocal about. — apokryfos
– apokryfos, Commented Apr 4, 2017 at 10:04
actually I had this old secure login script and I was trying to adopt somehow, the original script works fine I have no error in the Dreamweaver IDE and neither on server side the only thing is that I am not able to put data in DB (MariaDB).I have php 5.5 in my server — sinaps1
– sinaps1, Commented Apr 4, 2017 at 10:12
You SQL query is incorrect, you are trying to insert into 10 columns but you are providing only 9 values, you are missing 'ID' value in your query. — Echoes
– Echoes, Commented Apr 4, 2017 at 10:30

Echoes · Accepted Answer · 2017-04-04 10:31:43Z

1

You SQL query seems incorrect, the column count doesn't match, you are trying to insert into 10 columns but you are providing only 9 values, you are missing 'ID' value in your query.

answered Apr 4, 2017 at 10:31

Echoes

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Comments

symcbean · Accepted Answer · 2017-04-04 12:15:49Z

0

"I am having trouble inserting ..." is not a useful diagnostic. The mysql extension (and its replacements) provide an API to capture why something has failed:

$qry=("INSERT INTO antaret
(ID, emri,
 atesia, mbiemri,
 ditelindja, vendlindja,
 vendbanimi, telefon,
 zona, regtime) 
VALUES
('".$_POST['emri']."','".$_POST['atesia']."','"
  .$_POST['mbiemri']."','".$date."','"
  .$_POST['vendlindja']."','".$_POST['vendbanimi']."','"
  .$_POST['telefon']."','".$_POST['zona']."',
  NOW()
)");
$result = mysql_query($qry);
if (false===$result) {
   print "Query failed: " . mysql_error() . "<br /><br />" . $qry;
}

This might have given you a clue that there is a difference in the number of columns declared in the statement and the number of values. Formatting your SQL so it is readable might have given you a clue.

BTW you should read up on SQL injection before you expose this on the internet.

answered Apr 4, 2017 at 12:15

symcbean

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1 Comment

sinaps1 Over a year ago

Thank You! I just solved my issue using mysqli and not mysql

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