PHP - Check Box

This is suppose to check a table to see if it has a value in it.  I get this error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /homepages/6/d365637117/htdocs/EE Live/Active/Inc/session.php on line 33


This is the code
if($checkProfile)
{
    $result = mysql_query("SELECT prof_pic FROM members_profile WHERE id = '".$_SESSION['EEID']."'");
   $mem_num_rows = mysql_num_rows($result);
   if($mem_num_rows =1)
         {
         
            
         }

What is the best way to check for a Date in a Form?

I see there is a CheckDate() function, but how does that know what format the Date is in?

Should I use a Regular Expression?

My Form Field is just a Text Box...

Thanks,


Debbie

Gud pm!!
So i made this script to check if a username and password exist but it kinda fails me.
It gives me a database connection error from my confirm_query() which means it returned false?
so here's my code:

Code: [Select]
  function check_if_exist($username, $password)
{
global $connection;
$query = "SELECT username, hashed_password";
$query .= "FROM users ";
$query .= "WHERE username = '$username' ";
    $query .= "AND hashed_password = '$password' ";
$query .= " LIMIT 1";
$result_set = mysql_query($query, $connection);
confirm_query($result_set);
if(mysql_num_rows($result_set) == 1)
{
return true;
}
else
{
return false;
}
}

my confirm_query():

Code: [Select]
function confirm_query($result)
{
if(!$result)
{
die("Database connection failure ". mysql_error());
}
}

I have a another script to check if the email exist and it works perfectly, i dunno why this won't work properly.. 
maybe another simple mistake..
Any ideas or Violent reaction? 
Thanks!

hey gys im trying to build a permissions system based on the users group.

Groups can be 1, 2, or 3.

Here is the code im using:


<?php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("chat") or die(mysql_error());
$result = mysql_query("SELECT * FROM `members` WHERE username='".$_SESSION['username']."'") or die(mysql_error());

if($result["group"]==1){
echo 'yes';
}
else {
echo 'no';
}
?>

And here is how my db looks like:



The thing is, when Im logged in with the user test

and run that script, its saying no as the group im in isnt 1.

Why is this happening?  Can someone help me fix my code.

Hello..

I have Two HTML form and one is a Check boxes form that enable users to select their category. Then I need to display second form according to the categories they selected in first form. I use this code in first form to validate form submission..

      if ( isset( $_POST['category']) && sizeof( $_POST['category']) <= 3) {
      
         $category = $_POST['category'];         
         
      } else {
      
         $errors[] = 'Please select atleast 1, not more than 3 categories';
      }

If errors array is empty I did this..

      if ( empty( $errors )) {  // If everything's OK      
               
         $_SESSION = $category;
         
         $url = 'http://localhost/lanka_institute/tutorsignup/select_subjects.php? // Define the URL.
         ob_end_clean(); // Delete the buffer.
         header("Location: $url");
         exit(); // Quit the script.   
         
      }   


Can I know and is this correct? I display category list from mysql category table, it has category name and category id

this is my html part from first page

      echo '<td width="50%"><input type="checkbox" name="category[]"  vlaue="' . $info['category_id'] . '" />&nbsp;&nbsp;' . $info['category_name'] . '</td>';


any help appreciated.

Thanks in advance..

Hi,

I am trying to see if the domain name in the string is in an array.

eg;

Code: [Select]
mystring='http://google.com/somethingelse';

myarray[]='google.com';
myarray[]='anotherdomain.com';
in this case return TRUE

Code: [Select]
mystring='http://phpfreaks.com/adir';

myarray[]='google.com';
myarray[]='anotherdomain.com';

in this case return FALSE

How would I do this?

Thanks,

-mme

I have the variables $cap and $url
If the $cap contains any of the strings ".jpg,.JPG,.gif,GIF"
I have to concatenate $url.$cap

Hi guys,

I am trying to get a admin panel, which when the user is logged in, it will check if there user access is(say for this post) 9...

If there access is 9 in the database then direct to admin panel if not return them home.

Thanks guys

I am new to PHP.

I developed this class, I wonder if there's anything wrong or that I can improve. I could not test it because I'm in school.

Thanks in advance.

Code: [Select]
<?



class user {



var $userID,

$schoolID,

$userName,

$userPass,

$dbHost,

$dbUser,

$dbName,

$dbPass,

$dbUserTable;

$dbSchoolTable;



function dbInfo() {

$this->dbHost = 'localhost';

$this->dbUser = '';

$this->dbName = '';

$this->dbPass = '';

$this->dbUserTable = '';

$this->dbSchoolTable = '';

}



function registerUser($userName, $userPass) {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);



$query = "INSERT INTO $this->dbUserTable VALUES (NULL, \"$userName\", \"$userPass\")";

$result = mysql_query($query);



if(!$result) {
echo "Fail.";
} else {
$this->userID = mysql_insert_id();
}



mysql_close($dbLink);



$this->userName = $userName;

$this->userPass = $userPass;

}



function registerSchool($schoolName) {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);



$query = "INSERT INTO $this->dbSchoolTable VALUES (NULL, \"$schoolName\")";

$result = mysql_query($query);



if(!$result) {
echo "Fail.";
} else {
$this->schoolID = mysql_insert_id();
}



mysql_close($dbLink);



$this->schoolName = $schoolName;

}


function userLogin() {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);



$query = "SELECT * FROM $this->dbUserTable WHERE userName = \"$this->userName\" AND userPass = \"$this->userPass\" LIMIT 1";

$result = mysql_query($query);



if(!$result) {
echo "Fail.";
} else {
while($row = mysql_fetch_array($result)) {
session_start();
$_SESSION['userID'] = $row['userID'];
session_write_close();
}
}



mysql_close($dbLink);

}


function changePass($newPass) {

$dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);

if(!$dbLink) die("Could not connect to database: " . mysql_error());



mysql_select_db($this->dbName);

$query = "SELECT * FROM $this->dbUserTable WHERE userName = \"$this->userName\" LIMIT 1";

$result = mysql_query($query);

if(!$result) {
echo "Fail.";
} else {

$query = "UPDATE $this->dbUserTable SET userPass = \"$newPass\" WHERE userName = \"$this->userName\"";

$result = mysql_query($query);



if(!$result) {
echo "Fail";
} else {
$this->userPass = $newPass;
}
}



mysql_close($dbLink);

}



}

?>

does anyone have code for checking to make sure an email address is typed in correct

Hi,

I'd like to know what the most efficient code is for checking the first 2 digits of a string.

eg a strings first 2 digits must  equal '07'

thanks,

Code: [Select]
        public function getBonusChildren($userID)
        {
if($userID != NULL)
{
$sql = 'SELECT COUNT(*) AS count, userID FROM jos_backoffice_users WHERE parentID= ' . $userID . ';';
$stmt = conn::getInstance()->prepare($sql);
$stmt->execute();
                $obj = $stmt->fetchAll(PDO::FETCH_ASSOC);
if(is_object($obj))
{
$obj->count += $this->getBonusChildren($obj->userID);
}
}
                return $obj->count;
}
I get no error.  It just always return null.   How can I get this done?

Hi, I am trying to perform a login check and redirect the form to index.php if the username and password are correct. Also the login page check if the fields are entered or not.

The issue is that the login check is performing but not redirecting to index.php. Also I'm not getting any error after performing the login check.  Let me know if I'm making any errors in below code.  the include file only contains the database details.

Code: [Select]

<?php include("connections.php"); ?>
<?php
if(isset($_POST['submit']))
{
$errors=array();
$required_fields=array('username', 'pass');
foreach($required_fields as $fields)
{
    if(!isset($_POST[$fields]) || empty($_POST[$fields]))
    {
        $errors[]=$fields;
    }
}

if(empty($errors))
{
   $username=trim($_POST['username']);
   $password =trim($_POST['pass']);
    $query= "SELECT * FROM users WHERE username='{$username}' AND hashed_password='{$password}' ";
    $result=mysql_query($query);
    $result_set=mysql_fetch_array($result);
        if(mysql_num_rows($result_set) == 1)
    {
   header('location: index.php');
   exit;
    }
    else
    {
    echo 'Invalid password';
    }
}
   
} 
 ?>

Hi ,

I have a login page that i acces via www.mydomain.com/logib.php?last_visited_page.php
however the passed in variable last_visited_page.com is not always set .

once the login script has run , when the variable it is not set i would want jump to myaccount.php
otherwise jump to last_visited_page.php

How do i check for no variable beinf present and therefore jump to myaccount.php

Many Thanks
Fraser