how to insert multiple variable in databse table through sql?

Question

I want insert multiple variable as VALUES in my database table. here is my code

include 'database_connect.php';

$namef=  $_post["fnam"];
$lnamel= $_post["lnam"];
$addres= $_post["addres"];
$mail= $_post["email"];
$countr= $_post["countr"];
$db= $_post["do"];
$m= $_post["m"];
$pasword= $_post["pas"];

$insertSql= "INSERT INTO abcd(Id, First_name, Last_name, Address, Email, Country, Date_of_birth, Mobile_number, Password, Status) 
VALUES(NULL,'$namef', '$namel', '$addres', '$mail', '$countr', '$db', '$m', '$pasword', '0')";

if(mysql_query($insertSql))
    header("location: login_success.php");
else
    header("location: wrong_inf.php");

here only in Id and status field are inserted and all other fields are going empty even though i am giving some input in those inputs.

I don't know how to solve this.

Note:here Id is integer type and status is tinyint type and all other field is varchar type.

It seems odd that you are setting your ID to NULL. Is that really what you want to do? — TecBrat, Feb 25 '14 at 11:04
Please make sure to carefully read and understand this as well: http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?rq=1 It will help prevent your DB from being hacked. — TecBrat, Feb 25 '14 at 11:04

krishna · Answer 1 · 2014-02-25T11:10:41.223

For using post variable you need use

$_POST['fieldname']

and not as

$_post["fieldname"]

Because the $_POST array is case sensitive in php

Also always check whether query is executed correctly or not using mysql_error() and the proceed

$query = mysql_query($insertSql) or die(mysql_error());

if($query)
  {
   // updated
  }
else
   {
    // not updated
   }

And at last i advice you not use mysql_* functions and start use mysqli_* function or PDO

score 1 · Accepted Answer · answered Feb 25 '14 at 11:05

1

PHP Variables are case-sensitives

try

$namef=$_POST["fnam"];
$lnamel=$_POST["lnam"];

ETC....

answered Feb 25 '14 at 11:05

user1844933

3,296
2
25
42

score 1 · Answer 3 · answered Feb 25 '14 at 11:11

write $_REQUEST OR $_POST instead of $_post

also use exit after calling header(".."); and escape before sql execution to prevent injections

your modified code is

include 'database_connect.php';

$namef=  $_REQUEST["fnam"];
$lnamel= $_REQUEST["lnam"];
$addres= $_REQUEST["addres"];
$mail= $_REQUEST["email"];
$countr= $_REQUEST["countr"];
$db= $_REQUEST["do"];
$m= $_REQUEST["m"];
$pasword= $_REQUEST["pas"];

$namef = mysql_real_escape_string($namef);
$lnamel = mysql_real_escape_string($lnamel);
$addres = mysql_real_escape_string($addres);
$mail= mysql_real_escape_string($mail);
$countr = mysql_real_escape_string($countr);
$db = mysql_real_escape_string($db);
$m = mysql_real_escape_string($m);
$pasword = mysql_real_escape_string($pasword);

$insertSql= "INSERT INTO abcd(Id, First_name, Last_name, Address, Email, Country, Date_of_birth, Mobile_number, Password, Status) 
VALUES(NULL,'$namef', '$namel', '$addres', '$mail', '$countr', '$db', '$m', '$pasword', '0')";

if(mysql_query($insertSql))
{
    header("location: login_success.php");
    exit;
}
else
{
    header("location: wrong_inf.php");
    exit;
}

NOTE: mysql_* is deprecated now user mysqli_* OR PDO

score 0 · Answer 4 · answered Feb 25 '14 at 11:04

You could start by writing out your SQL string after you build it to see if the values are being added properly:

echo $insertSql;

If there are blank items in there, check your $_POST data are being passed through properly:

var_dump($_POST);

(And as krishna pointed out, you're using lowercase $_post when you should be using uppercase $_POST.)

Also, though you didn't ask about it, you shouldn't write data directly from the $_POST array to your database without sanitizing it first. Best bet is to use PDO, but if you're going to use the mysql_* functions, then use mysql_real_escape_string on each item you are writing to the database, like this:

$sql = 'INSERT INTO `table` (`column`) VALUES ("' . mysql_real_escape_string($_POST['item']) . '")';

score 0 · Answer 5 · answered Feb 25 '14 at 11:06

0

one of the error is undefine variable is $namel in sql

replaece it with $lnamel

answered Feb 25 '14 at 11:06

wild

340
1
3
14

how to insert multiple variable in databse table through sql?

5 Answers5