Question

My code doesn’t insert any records to mysql. What is wrong? I am really confused.
I have designed a form and I want to read data from text box and send to the database.

<?php
if(isset($_post["tfname"]))
    {
        $id=$_post["tfid"];
        $name=$_post["tfname"];
        $family=$_post["tffamily"];
        $mark=$_post["tfmark"];
        $tel=$_post["tftell"];

$link=mysql_connect("localhost","root","");
if (!$link)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("university",$link);
$insert="insert into student (sid,sname,sfamily,smark,stel) values ($id,'$name','$family',$mark,$tel)";

mysql_query($insert,$link);
    }
mysql_close($link);
?>

Answer 1

You’d better to put quotation mark for id, mark and tel after values in your query. Also as @Another Code said, you must use $_POST instead of $_post in your code. Try this and tell me the result:

<?php
if(isset($_POST["tfname"])) {
   $id=$_POST["tfid"];
   $name=$_POST["tfname"];
   $family=$_POST["tffamily"];
   $mark=$_POST["tfmark"];
   $tel=$_POST["tftell"];

   $link=mysql_connect("localhost","root","");
   if (!$link) {
      die('Could not connect: ' . mysql_error());
   } else {
      mysql_select_db("university",$link);
      $insert="insert into student 
               (sid,sname,sfamily,smark,stel) values 
               ('$id','$name','$family','$mark','$tel')";
      mysql_query($insert,$link) or die (mysql_error());
      mysql_close($link);
   }
} else {
   die('tfname did not send');
}
?>

Use mysql_query($insert,$link) or die (mysql_error()); to fetch the error message.