PHP - Executing The Query Once

I am facing problem to execute query by assigning NULL value to a variable and then executing query.In MySQL DB four fields Mobile,landline, pincode,dob are set as integer and date(for dob) respectively.The default is set as  NULL and NULL option is selected as yes.All these fields are not mandatory.The problem is that when I edit the form my keeping the value as empty in DB these are saved as 0, 0 , 0 & 0000-00-00 inspite of Null. I have tried everything but still the defect persist. Please help me to come out of the problem 

The code, I have used:

<?php
//require_once 'includes/config.php';

      $dbusername = $_POST['email'];
        $dbfirstname = $_POST['first_name'];
        $dblastname = $_POST['last_name'];
        //$dbmobile_number = $_POST['mobile'];
      if (isset($_POST['mobile']))
      
      {
         $dbmobile_number = $_POST['mobile'];
         
      }
       
      else
      
      {
         $dbmobile_number = "NULL";
         
      }
            
      $dblandline_number = $_POST['landline'];
        $dbdob = $_POST['dob'];
      
      if(isset($_POST['is_email']))
      {
         $dbSubscribe_Email_Alert = '1';
      }
      else
      {
         $dbSubscribe_Email_Alert = '0';
      }
      
      if(isset($_POST['is_sms']))
      {
         $dbSubscribe_SMS = 0;
      }
      else
      {
         $dbSubscribe_SMS = 0;
      }
       
      
        $dbAddress_firstname = $_POST['shipping_first_name'];
        $dbAddress_lastname = $_POST['shipping_last_name'];
        $dbAddress = $_POST['shipping_address'];
        $dbcity = $_POST['shipping_city'];
        $dbpincode = $_POST['shipping_pincode'];
        $dbstate = $_POST['shipping_state'];
        $dbcountry = $_POST['shipping_country'];
      
      echo "Welcome".$dbusername;

//if($_POST['btnSave'])

      //if ($_POST['btnSave'])

//{
   
   //echo "Inside query loop";
$connect = mysql_connect("localhost","root","") or die("Couldn't connect!");
mysql_select_db("salebees") or die ("Couldn't find DB");

//$query = mysql_query("SELECT * FROM users WHERE username='$username'");

$query = mysql_query("update users set firstname = '$dbfirstname', lastname = '$dblastname', mobile_number = '$dbmobile_number', landline_number = '$dblandline_number', dob = '$dbdob', Subscribe_Email_Alert = '$dbSubscribe_Email_Alert', Subscribe_SMS = '$dbSubscribe_SMS',  Address_firstname = '$dbAddress_firstname', Address_lastname = '$dbAddress_lastname', Address = '$dbAddress', city = '$dbcity', pincode = '$dbpincode', state = '$dbstate', country = '$dbcountry'      where username = '$dbusername' ");

header("location:my_account.php");


//}

//else

//{
//die();

//}


?>

I am trying to use some php code inside of an html document and it does not work. I have a script.php file that executes perfectly when I try and use it. Here is the code I am currently using:

Code: [Select]
<?php
 echo "<title>";
 echo "the Homepage!";
 echo "</title>";
?>
Any suggestions would be great.

I'm tired and want to sleep so my mistakes are getting lots now. Anybody awake that can explain my problem with the code below. 

<?

function directoryToArray($directory, $recursive) {
        $array_items = array();
        $i = "0";
        if ($handle = opendir($directory)) {
                while (false !== ($file = readdir($handle)) && $i < "3") {
                        if ($file != "." && $file != "..") {
                                if (is_dir($directory. "/" . $file)) {
                                        if($recursive) {
                                                $array_items = array_merge($array_items, directoryToArray($directory. "/" . $file, $recursive));
                                        }
                        } else {
                                        $file = $directory . "/" . $file;
                                        $array_items[] = preg_replace("/\/\//si", "/", $file);
                                        $i++;
                                }
                        }
                }
                closedir($handle);
        }
        natsort($array_items);
        return array_reverse($array_items);
}
$data = directoryToArray('images/screenshot/', TRUE);
print_r($data);
?>


What I want is for the while loop to stop after 4 hits and exit and give me the return results. But now it continue until readdir end.

Hey there, I recently started working on a PHP calculator, I tryed testing it out and the script is not executing... *NOTE* I found the error, I forgot to assign values to the options, all is good now

Code: [Select]
<?php
/*
Script Created by: Far Cry
Started: August 20th, 2010 10:51 PM PST
Completed
*/

$title = "Calculator";

$form = "<form action='calculator.php' method='POST'>
<table> 
    <tr>
    <td><input name='firstnumber' type='text'></td>
</tr>

<tr>
   <td><input name='secondnumber' type='text'></td>
</tr>

<tr>
   <td><select name='operation'>
       <option name='multiply'>*</option>
   </select></td>
</tr>

<tr>
<td><input name='submitbtn' type='submit' value='Calculate!'></td>
</tr>
</table>
</form>";


if ($_POST['submitbtn']){

$first = ($_POST['firstnumber']);
$last = ($_POST['secondnumber']);
$operation = ($_POST['operation']); 
}
if ($operation == 'multiply'){
echo"The answer is " . $first * $last . $form;
}
echo $form;
?>
Thanks in advance.

Hey guys!
My current employer contracted someone to create a script for him. (The script itself is to create a serial number for a piece of software) It's just a PHP script that executes a file and retrieves its output.
Now I've never worked with exec() in PHP, and I don't have a reliable environment to test it in. I compiled apache2 and php 5.0.4 and set them to run on a Macintosh (10.3.9) in the office, but I'm sure I've missed some crucial setup stage (mainly, I'm missing php.ini. I did a php -i | grep php.ini to find it, and it says it is in /etc/, but when I look in my PHP/etc folder, only pear.conf is there!). PHP and apache both appear to be working fine, though.

When I try and run the PHP script that executes the .ELF file, I get a blank page. Is this because I'm running it in OS X?  should I be running it in Linux?
Also, I don't know how to Enable exec() on my server, or to see if it's enabled. Basically, I just need someone to walk me through the basics of this. The script and executable SHOULD work fine because the contractor got it working fine.
Below is the script, and attached is the executable.
Thanks! :3

<?php

/*
inUID = 456890
inSerialNumUID = 123456
inMachineID_s = &6dA-8P@K-Xsq
*/

$p1 = "456890";
$p2 = "123456";
$p3 = "\&6dA-8P@K-Xsq";

exec("./create_activation_code.elf $p1 $p2 $p3", $output);

foreach ($output as $output_str)
{
echo $output_str."<br>";
}

?>

Hello,

PowerShell script stored locally can never be executed after I click on button:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Testing PowerShell</title>
</head>
<body>
<?php
 
// If there was no submit variable passed to the script (i.e. user has visited the page without clicking submit), display the form:
if(!isset($_POST["submit"]))
{
    ?>
    <form name="testForm" id="testForm" action="get-process.php" method="post" />
        Your name: <input type="text" name="username" id="username" maxlength="20" /><br />
        <input type="submit" name="submit" id="submit" value="Do stuff NowNew" />
    </form>
    <?php    
}
// Else if submit was pressed, check if all of the required variables have a value:
elseif((isset($_POST["submit"])) && (!empty($_POST["username"])))
{
   // Display the alert box 
   echo '<script>alert("Welcome to Geeks for Geeks")</script>';
   
   // Get the variables submitted by POST in order to pass them to the PowerShell script:
    $username = $_POST["username"];
    // Best practice tip: We run out POST data through a custom regex function to clean any unwanted characters, e.g.:
    // $username = cleanData($_POST["username"]);
         

$psPath = "C:\\Windows\\SysWOW64\WindowsPowerShell\\v1.0\\powershell.exe";
$psDIR = "C:\\TestNew\\";
$psScript = "pscripta.ps1";
$runScript = $psDIR. $psScript;
$runCMD = $psPath." ".$runScript." 2>&1"; 

echo "\$psPath  $psPath <br>";
echo "\$psDIR  $psDIR <br>";
echo "\$psScript  $psScript <br>";
echo "\$runScript  $runScript <br>";
echo "\$runCMD   $runCMD  <br>";

exec( $runCMD,$out,$ret);

echo "<pre>";
print_r($out);
print_r($ret);
echo "</pre>";

}
// Else the user hit submit without all required fields being filled out:
else
{
    echo "Sorry, you did not complete all required fields. Please go back and try again.";
}
?>
</body>
</html>

 

Thank you for your help!

OK, so I have the following file that has database functions:

<?php

function dbDelete($param){
$conDelete = mysql_connect($url,'username','password',true);
mysql_select_db('my_db',$conDelete);
mysql_query($param,$conDelete);
mysql_close();
   if(isset($conDelete)){
   mysql_close($conDelete);
   }
}
?>

I include the above file in a session handling file.  But for some reason the new file can't call the above function as follows:
<?php
include_once 'the above stated file';
function timeOut(){
        dbDelete("DELETE FROM acctussessi WHERE acctussessi_usid = '$sessius[0]'");
   $_SESSION = array();
   setcookie(session_name(),'',time()-4200);
   session_destroy();
}
//CHECK FOR TIMEOUT REQUEST
switch($_GET['xyz']){
   case 'timeout':
   timeOut();
   header("Location: /?xyz=timedout");
   break;
   case 'logout':
   timeOut();
   header("Location: /?xyz=loggedout");   
   break;
   case 'refresh':
   dbUpdate("UPDATE acctussessi SET acctussessi_time = ".time()." WHERE acctussessi_unid ='".$token."'");   
   break;
}
?>

dbDelete isn't being called inside of the TimeOut() function.  The only way it works is by doing the following:

<?php
function timeOut(){
   $_SESSION = array();
   setcookie(session_name(),'',time()-4200);
   session_destroy();
}
//CHECK FOR TIMEOUT REQUEST
switch($_GET['xyz']){
   case 'timeout':
      dbDelete("DELETE FROM acctussessi WHERE acctussessi_usid = '$sessius[0]'");
   timeOut();
   header("Location: /?xyz=timedout");
   break;
   case 'logout':
      dbDelete("DELETE FROM acctussessi WHERE acctussessi_usid = '$sessius[0]'");
   timeOut();
   header("Location: /?xyz=loggedout");   
   break;
   case 'refresh':
      dbUpdate("UPDATE acctussessi SET acctussessi_time = ".time()." WHERE acctussessi_unid ='".$token."'");   
   break;
}
?>
But this is annoying and repetitive.  What is going on?

Thanks in advance.

I am playing around with php but the script doesnt execute to the last point and with no error... 

 

What i am trying to do is checking the input of the user and comparing it what is in db and then applying conditions. So i have 2 files. update.php and home.php. 

home.php is my html form page and i included 'update.php' in it

In my update.php, i have these codes

 

<?php

error_reporting(E_ALL);


include_once('database.php');

if(isset($_POST['submit']))

{

$q = $_GET['q'];

$next = $q + 1;

if($q == 26)
{
echo '<h2>You have Completed the Exam.Congrats!!</h2>';


}
elseif($q <= 25){
    
$sql = "SELECT * FROM Students WHERE Email=?";
$stmt = $conn->prepare($sql); 
$stmt->bind_param('s', $_SESSION['login']);
$stmt->execute();
$result = $stmt->get_result(); $row = $result->fetch_assoc();

$Mark = $row['Mark'];


$Correct = $Mark + 1;


$Fail = $Mark + 0;

  
$sqlb = "SELECT * FROM Answers WHERE qid = ?";
$stmtb = $conn->prepare($sqlb); 
$stmtb->bind_param('s', $q);
$stmtb->execute();
$resultb = $stmtb->get_result();

$rowb = $resultb->fetch_assoc();

$Answer = $rowb['Answer'];

$Pick = $_POST['ans'];

if($Pick ==  $Answer)

{
  
  
    
  $sqld = "UPDATE Students SET Mark=? WHERE Email = ?";
$stmtd = $conn->prepare($sqld); 
$stmtd->bind_param('ss', $Correct,$_SESSION['login']);
$stmtd->execute();
$resultd = $stmtd->get_result();  


   echo "correct" ;
}

}
else

{
    
    echo 'Ate';
echo "<script type='text/javascript'>document.location.href='home.php?q='.$next.';</script>";
}

}//post submit

?>

The else statement doesnt run. so it keeps reloading the same page.   

  Also i want to ask why i keep getting the value of 2 for my $Correct variable stored on the database into column Mark for a single correct question instead of 1. What could be wrong?

Lastly, i dont want the script to run if the browser is edited by the user by changing the value of $q from the browser, because i am using get method, is there a way to do that.

Please be nice with your comments. Thanks in advance.!!!

   In my php web program I'm trying to execute a bash shell script on the web server passing to it some parameters gathered within the program.  Ideally I wouldl ike to run this script as another user.  However, I get only a return code of "1" and no other error messages that I can see.

I tried using system and passthru as well and I'm sure the script is executable by anyone.  I've also tried the sticky bit on the target shell script.

Can someone/anyone please provide some gotcha pointers about doing this?  As can be expected, after playing with this for a week now I'm extremely frustrated and nearly exhausted my resources (except for this forum).

Ideas/suggestions please?

Thanks 

i'm creating a small app for someone who wants to control the content of each page from a DB.

effectively it'd be a template file with header, footer, etc all pre-built, and the content area supplied from a TEXT field in a MySQL table - basically the same idea wordpress uses.

this is all fine, but a couple pages would require some php - e.g., a list of events or users or articles, whatever, that are managed in different tables.

i can do this with eval - something like:

Code: [Select]
function render_content($string) {
ob_start();
eval("?>$string<?php ");
$returns = ob_get_contents();
ob_end_clean();
return $returns;
}
but i wonder if there's a better way.  i can probably limit whatever code needs to be executed in include files, so i thought maybe include some arbitrary tag and use regexp to parse it out...  maybe modeled after a conditional comment, e.g.,

Code: [Select]
<!--[include]some-file.inc.php-->
// or even...
<include>some-file.inc.php</include>

but, again, not thrilled with the approach, and wondered if anyone had a better idea.

i should probably mention that it's not going to be a "content or include" setup - it probably won't be one or the other, exclusively, and is likely to be a mix on those pages that require it - the php might need to appear before, after, or in the middle of whatever arbitrary markup the user happens to supply, e.g.

Code: [Select]
<h1>This is a list of stuff</h1>
<p>Some explanation lorem ipsum dolor sit ahmet.</p>
<?php include('some-file.inc.php'); ?>
<em>But this caveat applies to the above list.</em>
<div>
  Something totally unrelated.
  <img src="pic.jpg" />
</div>
not sure I explained that very well, but hopefully the concept comes across.

TYIA

I have a PHP application where I am trying to execute a PHP routine when the user clicks on a link to go to another page.  The routine needs to execute before I use a header location command to send the user to the next location.  I can't do this all with JavaScript, but it could be part of the solution if necessary.  I'd prefer to just use PHP if I could, but I'm not sure that this is possible.

Can anyone share with me a brief example of how I could do this?

Thanks!

In a PHP file [php_one.php], I wrote some PHP code that contained a form.  I used ECHO to write the HTML inside of the PHP tags.

On another PHP page [php_two.php], I wanted to show the code that was used to create the first page [php_one.php].

My first effort provided an interesting result, since the PHP code was readable.  However, all HTML code was visible in its executed form.

There was an actual input field and a submit button (not the code used to produce them) in the context of my coding.

I have tried using:

file_get_contents("FILE")
$array = file("FILE")
readfile("FILE")

All results were similar and carried the same problem.

How can I view code so that even the HTML is visible as code?

Looking to INSERT a row in a data table and keeping the User on the current page. 

Will it work like what I have below, or do I need to send it to its own page then return the User back to where he started?

What I have below isn't inserting a row.  No errors showing.  

function bookmark_add () {

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

include("/home2/csi/public_html/resources/con.php");


	$query = "INSERT INTO a_player_bookmark (bookmark)
				VALUES ('1')";
	
	$results = mysqli_query($con,$query);
	echo mysqli_error($con);


}

	$output = '<form name="bookmark_add" method="post" action="" class="bookmark-plus">';
    $output .= '<span><input type="submit" name="Bookmark"></span>';
	$output .= '</form>';
	
	return $output;	



}

 

Hi,

I am using cpanel, where i am want to run a php file, to delete the old records,

i used cronjob in cpanel to do this.

In time columns i have made : * * * * * to execute each and every minute the script..

In the command textbox i have made: /home/server25/public_html/auto_cancel.php


in the email box: admin_email@gmail.com


I am getting Permission denied to my email address:
The error report am getting is :

/bin/sh: /home/server25/public_html/auto_cancel.php: Permission denied

Please help me to solve this.
Thanks,

Hi All

Basically I have 4 fields in my database

image2, image3, image4, image5

I have an image uploader to upload the files but I need what I need is if image2 is empty put in image3, if image3 is empty put in image4

so far I have this


//so first is the query to find the right row which gives the variable $row

//now here is what im doing

if($row['image2']!==""){
 //run this code
}

elseif($row['image3']!==""){
//run this code
}

elseif($row['image4']!==""){
//run this code
}

elseif($row['image5']!==""){
//run this code
}


Now the problem I'm having is the code that is always executing is the first one of $row['image2']

even if image2 has a value in the database it still runs that code and over writes the field in the database. It never moves on to field 3 or 4.  Can someone please tell me if I've done something wrong.


I'm not storing the image in the database just the name.

Thanks

Danny