PHP - Can't Insert Session Username Into Database

Hi,

my goal is to be able to create a web based interface for someone who has no programming skills or interest to be able to maintain a list of usernames and passwords for protecting a page of links on a website.

so, I've created a page that can write to a database, and I can see that it is working, I can read entries, etc.

now, I need to know how to make a script that will check against that database for valid UN/PW combinations.

what is the best method for this?

James

Hi,

I'm am trying to use a search engine that search's for a username in the database and displays the information back. I have searched for a script but none of them has helped me. I have tried to use this without any luck:

mysql_connect("localhost", "", "") or die(mysql_error()); 
 mysql_select_db("dbsystem") or die(mysql_error()); 
$todo=$_POST['todo'];
if(isset($todo) and $todo=="search"){
$search_text=$_POST['search_text'];
$type=$_POST['type'];

$search_text=ltrim($search_text);
$search_text=rtrim($search_text);

if($type<>"any"){
$query="select * from users where name = '$search_text'";
}else{
$kt=split(" ",$search_text);//Breaking the string to array of words
// Now let us generate the sql
while(list($key,$val)=each($kt)){
if($val<>" " and strlen($val) > 0){$q .= " name like '%$val%' or ";}

}// end of while
$q=substr($q,0,(strlen($q)-3));
// this will remove the last or from the string.
$query="select * from users where $q ";
} // end of if else based on type value
echo $query;
echo "<br><br>";
$nt=mysql_query($query);
echo mysql_error();
while($row=mysql_fetch_array($nt)){
echo "$row[name]<br>";
}
// End if form submitted


}else{
echo "<form method='post' action=''><input type='hidden' name='todo' value='search' />
<input type='text' name='search_text' /><input type='submit' value='Search' /><br>
<input type='radio' name='type' value='any' checked />Match any where <input type='radio' name='type' value='exact' />Exact Match

</form>
";
}


I will be grateful if you can help with this.

Im trying to create a website where users login, and then when they add a new entry to the database there name is put as the author. This is how my tables are set up. One table is named job and has the columns id, jobtext, jobdate, and authorid. Another table is called author. This table contains the columns id, username, password, and name. Authorid from the job table matches with id from the author table. When a user logins in this code is used to register the name...session_start();
$_SESSION['myusername'] = $_POST['myusername'];
$_SESSION['mypassword'] = $_POST['mypassword'];
header("location: index.php");
}
else {
echo "Wrong Username or Password";
}

This is the form users use to add a new entry...
if (isset($_GET['add']))
{
$pagetitle = 'New Job';
$action = 'addform';
$text = '';
$authorid = '';
$id = '';
$button = 'Add job';

include $_SERVER['DOCUMENT_ROOT'] . '/jobs/includes/db.inc.php';

// Build the list of authors
$sql = "SELECT id, name FROM author";
$result = mysqli_query($link, $sql);
if (!$result)
{
$error = 'Error fetching list of authors.';
include 'error.html.php';
exit();
}

while ($row = mysqli_fetch_array($result))
{
$authors[] = array('id' => $row['id'], 'name' => $row['name']);
}

// Build the list of categories
$sql = "SELECT id, name FROM category";
$result = mysqli_query($link, $sql);
if (!$result)
{
$error = 'Error fetching list of categories.';
include 'error.html.php';
exit();
}

while ($row = mysqli_fetch_array($result))
{
$categories[] = array(
'id' => $row['id'],
'name' => $row['name'],
'selected' => FALSE);
}

include 'form.html.php';
exit();
}

if (isset($_GET['addform']))
{
include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php';

$text = mysqli_real_escape_string($link, $_POST['text']);
$author = mysqli_real_escape_string($link, $_POST['author']);

if ($author == '')
{
$error = 'You must choose an author for this job.
Click &lsquo;back&rsquo; and try again.';
include 'error.html.php';
exit();
}

$sql = "INSERT INTO job SET
jobtext='$text',
jobdate=CURDATE(),
authorid='$author'";
if (!mysqli_query($link, $sql))
{
$error = 'Error adding submitted job.';
include 'error.html.php';
exit();
}

$jobid = mysqli_insert_id($link);

if (isset($_POST['categories']))
{
foreach ($_POST['categories'] as $category)
{
$categoryid = mysqli_real_escape_string($link, $category);
$sql = "INSERT INTO jobcategory SET
jobid='$jobid',
categoryid='$categoryid'";
if (!mysqli_query($link, $sql))
{
$error = 'Error inserting job into selected category.';
include 'error.html.php';
exit();
}
}
}

header('Location: .');
exit();
}


Form.html.php =
<?php include_once $_SERVER['DOCUMENT_ROOT'] .
'/includes/helpers.inc.php'; ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title><?php htmlout($pagetitle); ?></title>
<meta http-equiv="content-type"
content="text/html; charset=utf-8"/>
<style type="text/css">
textarea {
display: block;
width: 100%;
}
</style>
</head>
<body>
<?php session_start(); ?>

<h1><?php htmlout($pagetitle); ?></h1>
<form action="?<?php htmlout($action); ?>" method="post">
<div>
<label for="text">Type your job he </label>
<textarea id="text" name="text" rows="3" cols="40"><?php
htmlout($text); ?></textarea>
</div>
<div>
<label for="author">Author:</label>
<select name="author" id="author">
<option value="">Select one</option>
<?php foreach ($authors as $author):?>
<option value="<?php htmlout($author['id']); ?>"<?php
if ($author['id'] == $authorid)
echo ' selected="selected"';
?>><?php htmlout($author['name']); ?></option>
<?php endforeach; ?>
</select>
</div>
<fieldset>
<legend>Categories:</legend>
<?php foreach ($categories as $category): ?>
<div><label for="category<?php htmlout($category['id']);
?>"><input type="checkbox" name="categories[]"
id="category<?php htmlout($category['id']); ?>"
value="<?php htmlout($category['id']); ?>"<?php
if ($category['selected'])
{
echo ' checked="checked"';
}
?>/><?php htmlout($category['name']); ?></label></div>
<?php endforeach; ?>
</fieldset>
<div>
<input type="hidden" name="id" value="<?php
htmlout($id); ?>"/>
<input type="submit" value="<?php htmlout($button); ?>"/>
</div>
</form>
</body>
</html>


Right now, under authors, it displays all the authors in the database. I want it to just show/submit the authorid of the logged in user.

Alright, I've been assigned a project at work. I did not develop the application and the individual who did used CodeIgnited framework and mysql as the db.   Here's the problem, I'm not given much OT to do this and in our meeting the best way to proceed was to replicate the database for different parts of the organization. Basically we are a subsidiary and have been using an application that other groups within the organization want to use. Usually I would reconfigure the db schema and add org ids and in the user table add the appropriate organization to go to. However, they are not giving me enough time to do that.   So what I'm thinking is to just create a copy of the database we use (just the structure) and create a new database.   What I want to know is how to use mysql to check to see if a user exists in one database and if they don't then to go on to the next database. I understand this is a very sloppy way to do it, but it's the way we are moving forward.   I found the code to connect to the db in CodeIgnitor... how can I connect to a database, check to see if the user exists, then close that db connection and try the next database?  

    /**
     * Select the database
     *
     * @access    private called by the base class
     * @return    resource
     */
    function db_select()
    {
        return @mysql_select_db($this->database, $this->conn_id);
    }

  Thanks in advance.

Just want to know if it is possible to do something like this? If so, how?

Code: [Select]
$qry = "SELECT * FROM users WHERE password='$password' AND username='$_SESSION['SESS_USER_NAME']'";

Hi, I need to insert some code into my current form code which will check to see if a username exist and if so will display an echo message. If it does not exist will post the form (assuming everything else is filled in correctly). I have tried some code in a few places but it doesn't work correctly as I get the username message exist no matter what. I think I am inserting the code into the wrong area, so need assistance as to how to incorporate the username check code.

$sql="select * from Profile where username = '$username';
$result = mysql_query( $sql, $conn )
      or die( "ERR: SQL 1" );
if(mysql_num_rows($result)!=0)
{
process form
}
else
{
echo "That username already exist!";
}


the current code of the form


<?PHP
//session_start();
require_once "formvalidator.php";
$show_form=true;

if (!isset($_POST['Submit'])) {



$human_number1 = rand(1, 12);



$human_number2 = rand(1, 38);



$human_answer = $human_number1 + $human_number2;



$_SESSION['check_answer'] = $human_answer;
}

if(isset($_POST['Submit']))
{





if (!isset($_SESSION['check_answer'])) {
echo "<p>Error: Answer session not set</p>";
}


if($_POST['math'] != $_SESSION['check_answer']) {
echo "<p>You did not pass the human check.</p>";
exit();
}


   $validator = new FormValidator();
    $validator->addValidation("FirstName","req","Please fill in FirstName");







$validator->addValidation("LastName","req","Please fill in LastName");
$validator->addValidation("UserName","req","Please fill in UserName");
$validator->addValidation("Password","req","Please fill in a Password");
$validator->addValidation("Password2","req","Please re-enter your password");
$validator->addValidation("Password2","eqelmnt=Password","Your passwords do not match!");
$validator->addValidation("email","email","The input for Email should be a valid email value");
$validator->addValidation("email","req","Please fill in Email");
$validator->addValidation("Zip","req","Please fill in your Zip Code");
$validator->addValidation("Security","req","Please fill in your Security Question");
$validator->addValidation("Security2","req","Please fill in your Security Answer");
    if($validator->ValidateForm())
    {
        $con = mysql_connect("localhost","uname","pw") or die('Could not connect: ' . mysql_error());
        mysql_select_db("beatthis_beatthis") or die(mysql_error());
$FirstName=mysql_real_escape_string($_POST['FirstName']); //This value has to be the same as in the HTML form file
$LastName=mysql_real_escape_string($_POST['LastName']); //This value has to be the same as in the HTML form file
$UserName=mysql_real_escape_string($_POST['UserName']); //This value has to be the same as in the HTML form file
$Password= md5($_POST['Password']); //This value has to be the same as in the HTML form file
$Password2= md5($_POST['Password2']); //This value has to be the same as in the HTML form file
$email=mysql_real_escape_string($_POST['email']); //This value has to be the same as in the HTML form file
$Zip=mysql_real_escape_string($_POST['Zip']); //This value has to be the same as in the HTML form file
$Birthday=mysql_real_escape_string($_POST['Birthday']); //This value has to be the same as in the HTML form file
$Security=mysql_real_escape_string($_POST['Security']); //This value has to be the same as in the HTML form file
$Security2=mysql_real_escape_string($_POST['Security2']); //This value has to be the same as in the HTML form file



$sql="INSERT INTO Profile (`FirstName`,`LastName`,`Username`,`Password`,`Password2`,`email`,`Zip`,`Birthday`,`Security`,`Security2`) VALUES ('$FirstName','$LastName','$UserName','$Password','$Password2','$email','$Zip','$Birthday','$Security','$Security2')"; 
//echo $sql;
if (!mysql_query($sql,$con)) {

 die('Error: ' . mysql_error());

} else{



mail('email@gmail.com','A profile has been submitted!',$FirstName.' has submitted their profile',$body);

echo "<h3>Your profile information has been submitted successfully.</h3>";
}

mysql_close($con);
        $show_form=false;
    }
    else
    {
        echo "<h3 class='ErrorTitle'>Validation Errors:</h3>";

        $error_hash = $validator->GetErrors();
        foreach($error_hash as $inpname => $inp_err)
        {
            echo "<p class='errors'>$inpname : $inp_err</p>\n";
        }        
    }
}

if(true == $show_form)
{
?>

Im trying to insert a name into my database but I need it to go into the column that matches the sessions user id


Code: [Select]
"INSERT INTO users SET name = '".$_POST['name']."'";

how do I tell it to put name into name column with id matching Session id.

really appreicate if someone could please help me with this.

$result = mysqli_query($con,"SELECT * FROM $_SESSION['datakey']";

hello
I want query from one table and insert in another table on another domain .
each database on one domain name.
for example http://www.site.com $con1
and http://www.site1.com $con.

can anyone help me?
my code is :
<?php



$dbuser1 = "insert in this database";
$dbpass1 = "insert in this database";
$dbhost1 = "localhost";
$dbname1 = "insert in this database";

// Connecting, selecting database

$con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1)
   or die('Could not connect: ' . mysql_error());



mysql_select_db($dbname1) or die('Could not select database');


$dbuser = "query from this database";
$dbpass = "query from this database";
$dbhost = "localhost";
$dbname = "query from this database";

// Connecting, selecting database

$con = mysql_connect($dbhost, $dbuser, $dbpass)
   or die('Could not connect: ' . mysql_error());

mysql_select_db($dbname) or die('Could not select database');

//query from database
$query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1");
    while($row=mysql_fetch_array($query)){
$result=$row[0];
$text=$row[1]."</br>Size:(".$row[4].")";
$alias=$row[2];
$link = '<a target="_blank" href='.$row[3].'>Download</a>';
echo $result;
}  
//insert into  database
mysql_query("SET NAMES 'utf8'", $con1); 
$query3= " INSERT INTO `jos_content`
 (`id`,
  `title`,
   `alias`,
    `) VALUES 
                        (NULL, 
                        '".$result."', 
                        '".$alias."', 
                       '')";
if (!mysql_query($query3,$con1))
  {
  die('Error: text add' . mysql_error());
  }


            mysql_close($con);
                        mysql_close($con1);
             


?>

ISSUE.
A User enters information into a form.
If the 'username' is already taken, a 'message' in Red and with larger font-size will be returned, for example,
"The username $username already exists."
If the username is 'mattd' then the message should say, "The username mattd already exists."

Within my php application, I have included 'inline html'.
Here is part of the code:   ....
if (mysql_num_rows($query_run)==1) {
// it will never = more than one because only
//one user will or will not exist
?>

<html>
</body>
<h1><font color="#FF0066">The username <?php echo $username; ?>already exists.</h1>
</body>
</html>

<?php
}else{
//start the registration process
$query = "INSERT INTO `Names` VALUES
....   1. At one point I did get this: "The username mattd already exists."
2. But now I only get "The username already exists."
I am not retrieving the $username variable.
  This screenshot is found he
http://imgur.com/lIwLZ1G


thanks.

While we're on the subject, is there a way to ensure that the first letter of a name is captalized, and the rest lowercase?  Or is this best handled later on, when the name is being used and called from the DB.

PS: some of us comment are code as to WHAT we are doing because we're just not that good yet, and we need to explain it to ourselves.

Hi

A part of my site allows users to send messages to other users. When a member is logged on, they see a panel on the left with a link to the messages page. If there is a message they have not seen, it looks like messages(1). As this panel is on every page, the message(1) is displayed on every page.

My question is a general one which i've always wondered about - I determine whether all messages have been read or not from the database. Should I go once to the database when user logs on, and save this value to a session, or should i go to the database each time the member goes to a new page....

The reason I ask is because I am saving a lot of data in the session already so where do I draw the line between saving stuff to a session and just repeatedly going to the database..

I've begun to play around with object oriented php and I was wondering how I would go about writing a method that inserts values into a table.    I have a class, called queries, and I made this very simple function:

class queries {

    public function insert($table, $field, $value){

    global $dbh;
 
    $stmt = $dbh->prepare("INSERT INTO $table ($field) VALUES (:value)");
    $stmt->bindParam(":value", $value);
    $stmt->execute();
    }

}

$queries = new queries();

how do I make the $image piece so that it inserts "images/$image.php" into the database?

   VALUES('$name', '$id', '$image','$event')";

Hi guys,
I have a table that i want all the entries to be inserted all at once.
Mine is just inserting only one of the item, how can i get it to work? I also attached a sample table.

Thanks

	if(isset($_POST['send'])){
    
    $category = $_POST['category'];
    $item_type = $_POST['item_type'];
    $item_size = $_POST['item_size'];
    $product_name = $_POST['product_name'];
    $item_qty = $_POST['item_qty'];
    $price = $_POST['price'];
    $total_price = $_POST['total_price'];
    $item_price = $_POST['item_price'];
    
$sql = "
INSERT INTO shopping(
trans_ref,
category,
product_name,
item_type,
item_size,
item_qty,
item_price,
price,
date
)
	VALUES(
:trans_ref,
:category,
:product_name,
:item_type,
:item_size,
:item_qty,
:item_price,
:price,
NOW()
)
";
$stmt = $pdo->prepare($sql);
$stmt->execute(array(
':trans_ref' => $_SESSION['trans_ref'],
':category' => $category,
':product_name' => $product_name,
':item_type' => $item_type,
':item_size' => $item_size,
':item_qty' => $item_qty,
':item_price' => $item_price,
':price' => $price
));
	
if ( $stmt->rowCount()){    
    echo '<div class="alert bg-success text-center">SHOPPING COMPLETED</div>';
}else{
    echo '<div class="alert bg-danger text-center">SHOPPING NOT COMPLETED. A PROBLE OCCURED</div>';
}
	
}