Executing query with php variables

Question

I have to query a DB where i stored credentials for login (for homework). This is my php code but it's not working.

<?php
$user = "cardatabase";
$password = "";
$host = "";
$database = "my_cardatabase";
$connession = mysql_connect($host, $user, $password);
mysql_select_db($database, $connession);


$id = $_REQUEST['id'];
$password = $_REQUEST['password'];

$query = "select Password
          from Client
          where IDClient = '".$id."'";
$result = mysql_query($query, $connession);

echo $result;

            if ($result == $password) {
                echo ok;
            } else {
                echo error;}
mysql_close();
?>

I tried to print the value coming from my html page and it's correct, so there should be a problem with this code but i can't figure it out.

I assume that the IDClient is of numeric type so you have extra quote around the $id. — Hamlet Hakobyan
– Hamlet Hakobyan, Commented Feb 16, 2014 at 16:55

azhar · Accepted Answer · 2014-02-16 16:56:11Z

1

replace the code

mysql_select_db($database, $connessione);

to

mysql_select_db($database, $connession);

answered Feb 16, 2014 at 16:56

azhar

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Comments

sabbir · Accepted Answer · 2014-02-16 17:12:21Z

change the following code ::

<?php     
  $connession = mysql_connect($host, $user, $password) or die( mysql_error() . " <br/> could not connect to server");
  mysql_select_db($database, $connessione) or die( mysql_error() . " <br/> could not connect datbase");


  $id = $_REQUEST['id'];
  $password = $_REQUEST['password'];

   $query = "select Password
      from Client
      where IDClient='{$id}'";

   $result = mysql_query($query, $connession) or die(mysql_error());

   if( mysql_num_rows($result) > 0 )  {
       $row = mysql_fetch_array($result) or die(mysql_error());

       $dbPassword = $row['Password'];
       if( strcmp($password, $dbPassword) == 0 )  {
           echo "ok";
       }
       else {
           echo "Error";
       }
   }
   else {
       echo "error";
   }

   mysql_close();
?>

user1710796 · Accepted Answer · 2014-02-16 16:57:45Z

0

You have an extra e on the end of your second variable name on the following line:

mysql_select_db($database, $connessione);

answered Feb 16, 2014 at 16:57

user1710796

Comments

dave · Accepted Answer · 2014-02-16 16:57:57Z

0

First problem:

$connession = mysql_connect($host, $user, $password);
mysql_select_db($database, $connessione);

You have an extra e, so that should be:

mysql_select_db($database, $connession)

Second problem: mysql_query returns a resource - you need to do something like:

$result = mysql_query($query, $connession);
$rows = mysql_fetch_assoc($result);
echo $rows[0]['password'];

answered Feb 16, 2014 at 16:57

dave

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2 Comments

r4id4 Over a year ago

it prints nothing, meaning query was unsuccessful? anyway the connession/e was my misprint is this question on the .php it was correct

dave Over a year ago

echo $rows[0]['password']; should actually be echo $rows[0]['Password'];

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