PHP - Reports In .xls Then Email To Admin

Hello 
I am using SQL Server 2008 and Xampp server for PHP. I am trying to call SQL Server reports from PHP and I am using the code from https://ssrsphp.code...ussions/577518#   I can call the URL which connects to the Reports server from PHP but when I try to click the dropdown and run the reports I get an error. 

Fatal error: Class 'PReportException' not found in E:\PHP_SRS\ssrsphp\Factory\SSRSTypeFactory.php on line 75 


Can someone please tell me how to fix this error ? 

Thanks

Hi Guys

My site has a shout box, now and again people post inappropriate content and if a moderator is not online to mute them it doesnt look good for other users.

I have written a snippet of code to report an offender. My goal is that if 3 different people report an offender the site auto mutes them. (manual mute function already exists for moderators)

I setup a table that records the reporters name the offenders name the post id the message content and the time stamp.

I have never been good at counting num rows etc and need some help in what direction to go in.

My current code that inserts the reported content.

Code: [Select]
mysql_query("INSERT INTO `chicka_pets`.`records_sbreports` (`id`, `offender`, `reporter`, `post_id`, `post_content`, `timestamp`)
VALUES ('', '$offender', '$username', '$id', '$sbmessage', '$timestamp')");
I am guessing that somewhere after that I need some sort of count numrows on the post id where reporter is NOT the same and then an if numrows >2 set offender to mute

my problem is how to do the num rows bit where the reporters are unique?

Any help is much appreciated

Hi guys,

I want to generate reports and i want to use some fields of one table and other fields of another table. How can i use that?

Ex.   customer(customer_id, full_name,name_with_initials, address, contact_number,gender)

       transaction_table(tran_ID, account_number,account_type,transaction_type, amount, Date)


If i want to use only customer_id, tran_ID, full_name and amount how can i use them to generate reports.

I want them to be taken as fields of one table and use it..

Thanks,
Heshan

hello. I need your help please. I'm building logistics website with user panel and admin panel. I've done all login and register forms. now I want to : admin can add package with: tracking number , weight , cost , and declaration form. user can fill declaration form after admin add package to user panel. then admin can see the declared form. is it possible in php? thank you in advance

i wanting users to be able to update there email address and check to see if the new email already exists. if the email is the same as current email ignore the check.
i have no errors showing up but if I enter a email already in the db it still accepts the new email instead of bringing the back the error message.
Code: [Select]
// email enterd from form //
$email=$_POST['email'];


$queryuser=mysql_query("SELECT * FROM members WHERE inv='$ivn' ") or die (mysql_error());
 while($info = mysql_fetch_array( $queryuser )) {
  $check=$info['email'];
 // gets current email //
}


if($check!=$email){
// if check not equal to $email check the new email address already exists//
$queryuser=mysql_query("SELECT * FROM members WHERE email='$email' ");
//$result=mysql_query($sql);
$checkuser=mysql_num_rows($queryuser);
if($checkuser != 0)
{
$error= "0";
header('LOCATION:../pages/myprofile.php?id='.$error.'');


}
}

cheers

Any help would be greatly appreciated!


<?php
$host="localhost"; // Host name
$username="user"; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name


mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

 
$barcodeID=$_POST['barcode'];

echo $barcodeID;


$barcodeID = stripslashes($barcodeID);
$barcodeID = mysql_real_escape_string($barcodeID);


$sql="SELECT * FROM $tbl_name WHERE BarcodeID='$barcodeID'";




$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
$count=mysql_num_rows($result);


if($count==1){

$_SESSION['barcode'] = $barcodeSession;
$_SESSION['userlevel'] = $row['Priority'];   
   
if($row['userlevel'] == "Admin") {
header("location:AdminSection.php");
}else{
header("location:index.php");   
}   



header("location:LoggedIn.php");
}
else {   
header("location:index.php");   
}
?>




when the script has been run, I want it to redirect to either the user page or admin page depending on their priority level.

if Priority field == "Admin" then go to admin page.

Can you see anything missing?


Thank You

So i got my login down and the cookies, kinda set up
my problem is how do i make the admin panle save the true/false in the string in settings.php
id like do do it with a drop down menu to enable/disable it.
any help?

Code download

I get this error:


Code: [Select]
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\user\user.php on line 5


code:
user.php:
Code: [Select]
<?php

$get = (isset($_GET['id'])); //this means that user.php?id=1 would mean $get = 1. Note: This is not SQL Inject protected.
$users = mysql_query("SELECT * FROM users WHERE id='".$get."'");
while ($row = mysql_fetch_array($users)) 
{
echo '
Id = '.$row['id'].'
Name = '.$row['name'].'
Username = '.$row['username'].'
Password = '.$row['password'].'
Reg. on = '.$row['date'].'
';
}
 

?>

<html>
<body>
<form action='user.php' method='GET'>
Username: <input type='text' value=''>
<input type='submit' value='submit'>
</form>
<?php

//what goes here?

?>
</body>
</html>

Hey, in a nutshell the only thing in admin.php is the ability to moderate unapproved images, however, once approved, the "Approve    Delete" links are still on screen.
How it works is a user uploads an image, the filename is added to mysql and the image is added to uploads/  once I Approve an image, the image is then moved to img/ to display on the index.php (to prevent porn and anything that doesn't belong to the general public).

I know what's happening, because I've got while loops to display the image while looping through the mysql database, so once the image is moved, the links are still on screen, displaying an "Approve   Delete" for every image in the database.  Also another thing that happens is the images on index.php are blank until approved.  How can I work around this?

Here is the index.php when an image hasn't been approved:
http://www.xodiac.net/1.png

And here is the admin.php displaying Approve and Delete once an image has been approved:
http://www.xodiac.net/2.png

Hi, I am new here 🙂

I have been learning PHP and am currently working on my own OOP MVC CMS.  I am up to the stage where I would like to start working on the admin area, but I am not sure how best to organise things.

Should I create admin specific Controllers and Models?

In Views, should I create a sub directory Admin, and place all admin view templates within it?

Are there any good books on OOP/MVC you would recommend?

 

 

 

Hey, So i have an admin.php page that lists all of the users in the database and im wondering how i can add functions so the administrator can delete / ban the user from the webpage i'm not sure on how you would select the user?

Hallo I have a problem. This is my code:

<?php include 'connect.php';
?>

<html>
<head>
<title>Admin Insert page!</title>
</head>
<body>
<?php
error_reporting(-1);ini_set('display_errors',1);
if (isset($_POST['submit'])){
$name = $_POST['name'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM users WHERE user='$name' AND password='$password'");
$num = mysql_num_rows($result);
if($num == 0){
echo "Bad login, go <a href='login.php'>back</a>";
}else{
session_start();
$_SESSION['name'] = $name;
header("Location: admin.php");
}
}else{
?>
<form action='login.php' methody='post'>
Username: <input type='text' name='name'/><br />
Password: <input type='password' name='password'/><br />
<input type='submit' name='submit' value='Login' />

</body>
</html>

I try to use console to find the problem but I didn't.... I know that there is some problem with $num Can somebody help me? Thank you.
Edited by Artur, 19 October 2014 - 12:11 PM.

Hello I am trying to add an IF statement to my login script so that if the username entered is 'admin' it directs to 'adminpage.php

Here is my script:

<?php
include ("connection.php");


session_start();
// Collect data from form and save in variables
//See if any info was submitted 
$Username = $_GET['Username'];  
//Clean data - trim space 
$Username = trim ( $Username);  
//Check its ok - if not then add an error message to the error string 
if (empty($Username))  
    $errorString = $errorString."<br>Please supply Username.";
//See if any info was submitted 
$Password = $_GET['Password'];  
//Clean data - trim space 
$Password = trim ( $Password);  
//Check its ok - if not then add an error message to the error string 
if (empty($Password))  
    $errorString = $errorString."<br>Please supply Password.";        

//  Query to search the user table
 $query= "SELECT * FROM Users WHERE Username='$Username' AND  Password='$Password'";

// Run query through connection
  $result = mysql_query ($query); 

$row = mysql_fetch_assoc($result);

// if rows found set authenticated user to the user name entered 
if (mysql_num_rows($result) > 0) { 
$_SESSION["authenticatedUser"] = $Username;

$_SESSION['UserID'] = $row['UserID'];

// Relocate to the logged-in page
header("Location: loggedon.php");
} 
else
// login failed redirect back to login page with error message
{
$_SESSION["message"] = "Could not connect as $Username " ;
header("Location: login.php");
}
?>

Thank you

Hey guys, I've set up a database with a login and logout script for my site..

There is a TINYINT value called admin and it either equals 1 or 0 depending on whether the user is an admin or not..

The registration script works perfectly to create the table value and the login script works fine for the site..

The question I had was if I wanted to add a link to the bottom of every page that said:

Go to Administration Panel and make it only viewable by ADMINS I figured this little script would work..


Here would be the end of the page:
Code: [Select]
  <br />
  <center>Copyright &copy 2010 <a href="http://www.website.com">www.WEBSITE.com</a></center>

  <?php include('includes/start_admincheck.php'); ?>
    <center><a href="<?php echo $homedir .'admin.php'; ?>">Go to Administration Panel</a></center>
  <?php include('includes/end_admincheck.php'); ?>

</body>

</html>
Inside start_admincheck.php I have:

(NOTE: $cUsername refers to a setcookie and $cAdmin does as well.. They are defined on my Variable page included at the top.)
Code: [Select]
<?php include('variables/variables.php'); ?>

<?php
  mysql_connect("$mysql_hostname", "$mysql_username", "$mysql_password") or die(mysql_error());
  mysql_select_db("$mysql_database") or die(mysql_error());

  if(isset($cUsername))
  {
    $check = mysql_query("SELECT * FROM users WHERE username = '$cUsername'")or die(mysql_error());

    while($info = mysql_fetch_array( $check ))
    {
      if (($cAdmin == 1) && ($info['admin'] == 1))
      {
?>
And this is the end_admincheck.php
Code: [Select]
<?php include('variables/variables.php'); ?>

<?php
      }

      else
        die(); 
    } 
  }
  else
    die();
?>
?>
I get this Parse error thrown at the bottom of the page:

Code: [Select]
Parse error: syntax error, unexpected $end in /*******/includes/start_admincheck.php on line 15
Naturally I would checkout line 15 in start_admincheck.php, but normally when I get an $end error it is the last line of the code and leaves me lost..

Something I'm missing guys?

As always, thanks in advance 

Hello,

Do you know where I can download a nice looking PHP admin dashboard for free?

Thanks in advance for the help