PHP - Display <div> Based On Php Variable

Hello my problem here is to have a drop down menu which gathers usernames from the database. Then with a click of a button the information for that specific user selected is shown. I'm close code wise but right now it's just showing me all users. I'm displaying the info in text box's to allow an admin to change the info. 

<form action="edit.php" method="post">

	 <select name="username" id="username">
	 	  <?php 
	  
 // Connects to your Database 
  $con=mysqli_connect('localhost', 'root', '');

/* check connection */

if (mysqli_connect_errno($con)) {
    trigger_error('Database connection failed: '  . mysqli_connect_error(), E_USER_ERROR);
}

$query = "SELECT `username` FROM `bencobricks` . `users`";

$result = mysqli_query($con, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($con), E_USER_ERROR);

if($result) {
	while($row = mysqli_fetch_assoc($result)) {
		//printf ("%s\n %s\n ",
		echo	"<label for='username'>Username: </label> <option value='$row[username]'>$row[username] </option><br/>";
			
		echo "<br/><br/> ";
		
	}
} 

mysqli_close($con);

?>
	 </select>
<br />
  <input name="send" id="send" type="submit" value="Edit User" />
</form>

Then in the edit.php file i have this code: 

<?php 

// Connects to your Database
$con=mysqli_connect('localhost', 'root', '');

/* check connection */

if (mysqli_connect_errno($con)) {
	trigger_error('Database connection failed: '  . mysqli_connect_error(), E_USER_ERROR);
}

if (isset($_POST['send'])) {
 $username = $_POST['username'];
 
 $query = "SELECT * FROM `bencobricks` . `users` ";
 $result = mysqli_query($con, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($con), E_USER_ERROR); 
// Print the result

 if($result) {
while ($row = mysqli_fetch_assoc($result)) {
//	$query = "SELECT * FROM `bencobricks` . `users` WHERE `username` = username = $row[username]";
 printf ("%s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n",
			"<label for='username'>Username: </label> <input type='text' name='username' value='$row[username]'>" . "</input><br/>",
			"<label for='email'>Email: </label> <input type='text' name='email' value='$row[email]'>" . "</input><br/>",
			"<label for='membership'>Membership: </label> <input type='text' name='membership' value='$row[membership]'>" . "</input><br/>",
			"<label for='firstName'>First name: </label> <input type='text' name='firstName' value='$row[firstName]'>" . "</input><br/>",
			"<label for='lastName'>Last name: </label> <input type='text' name='lastName' value='$row[lastName]'>" . "</input><br/>",
			"<label for='gender'>Gender: </label> <input type='text' name='gender' value='$row[gender]'>" . "</input><br/>",
			"<label for='dateOfBirth'>Birthdate: </label> <input type='text' name='dateOfBirth' value='$row[dateOfBirth]'>" . "</input><br/>",
			"<label for='date'>Date: </label> <input type='text' name='date' value='$row[date]'>" . "</input><br/>",
			"<label for='sets'>Sets: </label> <input type='text' name='sets' value='$row[sets]'>" . "</input><br/>",
			"<label for='checkbox'>Checkbox: </label> <input type='text' name='checkbox' value='$row[checkbox]'>" . "</input><br/>",
			"<label for='admin'>Admin? </label> <input type='text' name='admin' value='$row[adminFlag]'>" . "</input><BR>",
			
		"<br/><br/> ");
}
}
} 
mysqli_close($con);
?>

Any help is greatly appreciated. 

Good day and Merry Christmas to all,

I just spent a good time of my christmas eve trying to figure out this problem. I hope one of you santas would be so kind as to help me with it.

First off I have two tables; employee and employee_works both connected via employee_id key.

Basically I have a parent window we'll call parent.php.
inside the parent page is a search button that once clicked will open a child window we'll call child.php

inside the child page is a list, lets say employees with name, employee_id, etc.

My main concern is this:
How do I populate parent.php based off the employee selection I made in the child window.

Example:
-Access parent.php
-Click on search
-Click on [ID: 004] [NAME: JOHN SMITH] [PHONE: 1233456] [DATE HIRED: JULY 16, 1992] <---format of a row in child.php
-child.php automatically closes and parent.php now shows all data from employee_works with the employee_id = 004

Is this even possible? I know this is vary vague and would be willing to explain more if needed. My website is built mostly on javascript and php.

I would like to know the best practice to achieve the following:

I have a list of dates related to live events for performing artists.  When someone views the web page that contains a section to display the dates, I would only want to show the dates from today into the future and not show any dates from the past.

What is the best way to accomplish this?

Thanks in advance...

So I have these session variables being stored on login. Each page stores these session variables in local PHP variables. For some reason, all pages display that variable in the URL except one page. I have done some 'echo' work just to ensure variables have values stored and that everything was being read appropriately, no issues there. The variables are apparent everywhere on the page besides the URL. Below you can find all the code I have to make this thing work....maybe someone can point me in the right direction because I am STUMPED!!! LOL

So my first page calls a popup window in javascript defined below. The link involves the following PHP

  Code: [Select]
   $customer_id = $_SESSION['cid'];
    $customer_email = $_SESSION['cust_email'];
    $customer_fname = $_SESSION['cust_first_name'];
    $customer_contact_id = $_SESSION['customer_contact_id'];
    $customer_company_id = $_SESSION['customer_company_id'];

    $c_project_id = $_GET['pid'];
    $c_project_id  = mysql_real_escape_string($c_project_id );
    $c_project_id = eregi_replace("`", "", $c_project_id);

Then comes in the Javascript...

Code: [Select]
<script type="text/javascript">
<!--
$(document).ready(function(){

$(".approval").click(function(){

v = $(this).attr("id");
url = 'deny_approval.php?cid=<?php echo "$customer_id"; ?>&company_id=<?php echo "$customer_company_id"; ?>&customer_contact_id=<?php echo "$customer_contact_id"; ?>&pid=<?php echo "$customer_project_id"; ?>&section=' + v;

window.open(url, "myWindow",
"status = 1, toolbar = no, scrollbars = yes, location = no, resizable = no, height = 600, width = 600, resizable = 0" );

});

});
//-->
</script>
And finally, the HTML....

Code: [Select]
<body>
      <a href="#" id="deny_quote_approval" class="approval">Deny Quote Approval</a><br /><br />
</body>

Any ideas?? As always, any help would be greatly appreciated.

Bl4ck Maj1k

I want to retrieve an image id from a db and show the images.
I cant get the syntax right for the image tag.Any help appreciated.
Code: [Select]
function display_covers()
{
    global $wpdb;
 $query = "select * from wp_cover";
   $result = mysql_query($query)or trigger_error("Query: $query\n<br />MySQL Error: " . mysql_error());
echo mysql_error();
  if (!$result)
  return false;
echo'<div class="wrap"><p>choose from one of the covers below</p></div>';

echo'<div id="main">';
echo'<table class="main" cellpadding="2">';
//echo"<caption>Please choose a book cover</caption>";
?>
<thead><tr><td colspan="5" ><h6 class="main">Book Covers</h6></td></tr>
</thead>
<?php
  $i=0;
  $size=3;
  echo "<tbody>";
  echo "<tr>";
  while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
  {
    /* display picture  */
     ?>
    <td class="main">
    <?php
    
    
    echo"<img src="/Applications/MAMP/htdocs/wordpress_3/wp-content/plugins/Authors2/jackets/"{.$row['pix'].}""/>";

echo"</td>";
    $i++;

    if($i==$size)
{
  echo "</tr><tr>";

  $i=0;
}

 }
}

Hi, i cant seem to get something working, should be simple but its not working for me.

I just need to only display a table if a variable in my table = a certain value.

The column in the table is called 'option1_available' and if its value is set to 'Y' i want it to display a table.

Appreciate any help, Thanks

 This is my code

Code: [Select]

$file = $file . $line;
                fclose($fh);
                echo "<script language= 'JavaScript'>alert(' . $file . ');</script>";

the alert box is not coming up. 

Please suggest a way to print the contents of the file in a alert box.

Using Javascript, how do I do an alert and show the value of a variable?

thanks

I am building a website that uses session variables extensively.  The site is getting a little complicated, and I am starting to lose track of which session variables are assigned, and what their values are.  Is there some way where I can create a piece of code that will display all of the existing session variables and their values?

Trying to figure out how to make it so name is a link to the profile when its echo
anyone know how to do this?

im at a huge stand still


Code: [Select]
<?php 

                                                $sql = "SELECT name FROM users WHERE DATE_SUB(NOW(),INTERVAL 5 MINUTE) <= lastactive ORDER BY id ASC"; 
$query = mysql_query($sql) or die(mysql_error());
$count = mysql_num_rows($query);
$i = 1;
while($row = mysql_fetch_object($query)) {
 $online_name = htmlspecialchars($row->name);


   echo '<a href="Inbox.php">"'[$goauld]'</a>"'; ?>

I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here.  It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere.

<?PHP require('connect.php'); ?>

<form action ='' method='post'> <select name="id">

<?php
$extract = mysql_query("SELECT * FROM cars");
   
   while($row=mysql_fetch_assoc($extract)){
   $id = $row['id'];
   $make= $row['make'];
   $model= $row['model'];
   $year= $row['year'];
   $color= $row['color'];

echo "<option value=$id>$color $year $make $model</option>
";}?>

</select>

Which attribute would you like to change?<br />
<input type="radio" name="getchanged" value="make"/>Make<br />
<input type="radio" name="getchanged" value="model"/>Model<br />
<input type="radio" name="getchanged" value="year" />Year<br />
<input type="radio" name="getchanged" value="color"   />Color<br /><br />

   <br /><input type='text' value='' name='tochange'>
   <input type='submit' value='Change' name='submit'>

</form>

//This is where I need help...

<?PHP
if(isset($_POST['submit'])&&($_POST['tochange'])){
mysql_query("
   UPDATE cars
   SET '$_POST[getchanged]'='$_POST[tochange]'
   where id = '$_POST[id]'
         ");}?>

Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment.

My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted."
The program should not accept any numbers greater than 100 or any characters.

Once I do this, I must take a second number and do a similar thing.

Finally, I must have a statement show up at the bottom stating which number is greater.

Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters.

Any help you can provide will be greatly appreciated!

I have just re-installed Xampp and suddenly my sites are now displaying lots of:

Notice: Use of undefined constant name - assumed 'name' in ...
Notice: Use of undefined constant price - assumed 'price' in ...

this is an example of the line its refering too:

$defineProducts[1001] = array(name=>'This is a product', price=>123);

My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user  and $priv are  undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working.

The code I'm using:

<?php
session_start();
function verify() {
	//verify that the user is logged in via the login page. Session_start has already been called. 
	if (!isset($_SESSION['loggedin'])) {
	header('Location: /index.html'); 
	exit;
	}
	//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges   // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. 
        $user === $_SESSION['name'];
    //Connect to Databse
	$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
	if (!$link) {
  	  echo "Error: Unable to connect to MySQL." . PHP_EOL;
    	echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    	echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    	exit;
        }
        //SQL Statement to lookup privlege information. 
       if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
         //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
            while ($row = $result->fetch_assoc()) {
           $priv === $row["su"];
            }
        }
        // close SQL connection.
        mysqli_close($link);

 // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special 
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. 

        if ($priv !== 1) {
        echo $_SESSION['name'];
        echo "you have privlege level of $priv";
        echo "<br>";
        echo 'Your account does not have the privleges necessary to view this page';
        exit;
        }
        
}
verify();
?>

 

Hello everyone,

I can get Test 2 to successfully operate the if statement using a variable variable.

But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed?

Code: [Select]

<?php
# TEST 1
$_SESSION[test_variable] = "abcd";
$session_variable_name = "_SESSION[test_variable]";
if ($$session_variable_name == "abcd")
{
echo "<br>line 373, abcd<br>";
}

# TEST 2
$test_variable = "efgh";
$test_variable_name = "test_variable";
if ($$test_variable_name == "efgh")
{
echo "<br>line 379, efgh<br>";
}

?>
Many thanks,

Stu