PHP - Php Engine Not Picking Up Mysql While Script

Using phpMyAdmin I loaded 6 test records with the id set to auto_increment and it loaded all the data correctly with id # 1-6.

Then from somewhere it got the number 333353  and auto_increments it as the value for the id. So now I have id's 1-6 and 333353, 333354, ect. For every record I add it increments it.

I deleted all but records 1-6 and tried again but it has the last value of 3333xx stored somewhere and increments it. Deleted them again, closed the program, came back and it still does it.

I have 3 staff members that need to pick vacation in a certain order.
Each have X weeks vacation and can pick off a calendar their choice 1-3 weeks  per round.
I'm trying to loop through the DB to find who is next to pick with weeks left

~~~~~~First round of picking~~~~~~
STAFF 1 - PICK ORDER 1 -  3 weeks available
STAFF 2 - PICK ORDER 2 -  5 weeks available
STAFF 3 - PICK ORDER 3 -  3 weeks available

Staff 1 takes 2 Weeks
Staff 2 takes 1 Weeks
Staff 3 takes 3 Weeks

~~~~~~Second round of picking~~~~~~
STAFF 1 - PICK ORDER 1 -  1 weeks available
STAFF 2 - PICK ORDER 2 -  4 weeks available
STAFF 3 - PICK ORDER 3 -  No weeks left

Staff 1 takes 1 Weeks
Staff 2 takes 3 Weeks
Staff 3 Skipped

~~~~~~Third round of picking~~~~~~
STAFF 1 - PICK ORDER 1 -  No weeks left
STAFF 2 - PICK ORDER 2 -  1 weeks available
STAFF 3 - PICK ORDER 3 -  No weeks left

Staff 1 Skipped
Staff 2 takes 1 Weeks
Staff 3 Skipped

~~~~~~~~~~~~
All staff 0 weeks left end

--calendar.php--
$year=2020;
$sql = "SELECT * FROM vac_admin WHERE pick_year='$year';
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
$row_admin = mysqli_fetch_assoc($result);
}
$current_pick_staff = $row_admin['current_pick_staff'];

$sql = "SELECT *  FROM vac_pick_order WHERE  pick_year='$year'  &&  pick_order = '$current_pick_staff'";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
if($row['vac_c_counter'] < 0){
$emp_num = $row['emp_num'];
}ELSE{
??????????????????
goto next staff with weeks > 0
?????Somthing like
if ($current_pick_staff == 3){
	$current_pick_staff = 1;
}ELSE{
	$current_pick_staff++;
}
??????????????????  
}

~<FORM>~~~~~~~~~~~~~~~~~~~~~
Staff with $emp_num can now pick ~~~~~~ $_POST -> $date = XXXX-XX-XX; $num_weeks = X; $emp_num;
~</FORM>~~~~~~~~~~~~~~~~~~~~~

--process.php--
$year = 2020;
$date = $_POST['date'];
$num_weeks = $_POST['num_weeks'];
$emp_num = $_POST['emp_num'];

$sql = "INSERT INTO vac_picks (pick_year,emp_num,date) VALUES ($year,$emp_num,$date)";
$sql = "UPDATE vac_pick_order SET vac_c_counter=vac_c_counter - $num_weeks WHERE emp_num='$emp_num';
$sql = "UPDATE vac_admin SET pick_order=pick_order +1 WHERE pick_year='$year' ;

Then back to calendar.php until all weeks gone.

 

 

 

 

I want to know how to display results from mysql database by filling in a form. but i have found a tutorial which shows typing in a text box which displays results. if i follow this tutorial will it help me to understand and create php coding to display results for my form?

Hi,

Can anyone help me with a script, I have tried a if statement but I cannot get it to work and its driving me mad.

Basically I have a string my website uses to get if the user is logged in what their username is and I want it so when they click a certain link it checks to see if a table all ready exists called their username, if it does it displays a message, if it doesnt it creates the table and if the username is "anonymous" is displays a message.

So in short:

if $username is the same as a table display "Table all ready exists"
if $username="anonymous" display "You must be logged int"
if $username not the same as a table then create table.

Many thanks in advance

Jay

hi i am having a problem with a php script. i have 3 files index.php,  home.html.php, form.html.php.
home.html.php is working fine, if has a form that is submitting fine but when it submits i get a problem. maybe someone can help me out please i would be grateful as i am new. the is nothing wrong with anything else mysql database is running fine and there is no problem with connection.

index.php (you can ignore what is commented out)

<?php
 include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php';
 //$_GET['id'] is categoryid
if (isset($_GET['id']))
{
   include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php';

   $id = mysqli_real_escape_string($link, $_GET['id']);
   $sql = "SELECT businessid FROM businesscategory WHERE
          categoryid LIKE '$id'";   
  $result = mysqli_query($link, $sql);
  if ($result !='')
  {
   $businessid = mysqli_fetch_array($result);
  $sql = "SELECT content FROM business WHERE
          id LIKE '$businessid'"; 
  $result = mysqli_query($link, $sql); 
  }
  if ($result !='') 
  {
  $content = mysqli_fetch_array($result);
  include 'form.html.php';
   exit();
  }
} 

/*
   // The basic SELECT statement
   $select = 'SELECT content';
   $from   = ' FROM business';
   $where  = ' WHERE TRUE';

   $id =  mysqli_real_escape_string($link, $_GET['id']);
   if ($category != '') // An owner is selected
   {
      $where .= " AND categoryid='$categoryid";
   }

   $categoryid =  mysqli_real_escape_string($link,
         $_GET['category']);
   if ($categoryid != '') // A category is selected
   {
      $from  .= ' INNER JOIN businesscategory ON id = business';
      $where .= " AND categoryid='$categoryid'";
   }

   $text = mysqli_real_escape_string($link, $_GET['text']);
   if ($text != '') // Some search text was specified
   {
      $where .= " AND content LIKE '%$text%'";
   }

   $result = mysqli_query($link, $select . $from . $where);
   if (!$result)
   {
      $error = 'Error fetching businesses.';
      include 'error.html.php';
      exit();
   }

   while ($row = mysqli_fetch_array($result))
   {
      $businesses[] = array('id' => $row['id'], 'text' => $row['content']);
   }

   include 'form.html.php';
   exit();
} 
 
  */
 
 
 
 
 
 
$result = mysqli_query($link, 'SELECT id, name FROM category ORDER BY name');
if (!$result)
{
   $error = 'Error fetching categories: ' . mysqli_error($link);
   include 'error.html.php';
   exit();
}

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

include 'home.html.php';
?>

form.html.php

<?php include_once $_SERVER['DOCUMENT_ROOT'] .
      '/includes/helpers.inc.php'; ?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
  <head>
  <meta http-equiv="content-type" content="text/html; charset=windows-1250">
  <meta name="generator" content="PSPad editor, www.pspad.com">
  <title>Business</title>
  </head>
  <body>
       it works
       <?php echo $content; ?>
  </body>
</html>

Hello,
I'm new to this website so forgive me if I'm posting in the wrong section.
I just need to know if there is a way that I can retrieve php script that is stored in a database?

for example, in my database in table:
<? echo"Hello World"; ?>

and in my php file, I call that using:
<?
mysql_connect("asdas","asdas","asd");
   
mysql_select_db("asdasda");

$result = mysql_query("select * from somedatabse");

while($r=mysql_fetch_array($result))
{   

   $table=$r["table"];
   
}

 echo"$table";
?>

i know its confusing, i just hope i made sense and that someone understands what im trying to do.

any help will be appreciatedddd
thank you : )

hello guys.

i sitting here at my pc , have used alot days on google and forums.
but i just cant seem to find what i am looking for

so i hope that anyone here got a min to tell me what to do 

okai soo what i am looking for is a php / mysql script that can countdown for a user , when its finish add a number ( to usertabel[2].
and the problem is it need to do it also if the user closes the browser.

i am total lost , have no ide to fix / wihte it
 



Sorry for my spelling i am danish 

TRUNCATE TABLE CubeCart_cats_idx;
INSERT INTO CubeCart_cats_idx (productid, cat_id) SELECT productid, cat_id FROM CubeCart_inventory;

I an running the following in MyPHPAdmin, but need this to be automated in a PHP script, how would I do it ?
all help is appreciated, still new to PHP & MySQL 

many thanks

D

Please help !

Hi guys, I am literally at my wits end with this and I am almost positive its in my code.

This code is from a blackberry app that sends the infor into PHP, that part works flawlessy and was working just fine until I added the code to get the supervisor email from  a different table based off a query.

What happens is the information gets posted three times, The first time the emails are triggered and the post happens succefully, both the user and the supervisor get the properly formatted emails.

This happens almost immediately, works great. Then almost immediately, the supervisor will get two more emails, with the same information except missing the orignal users email, presumably the user would have gotten two more emails as well if the email address was available. The mysql DB gets all 3 records with the second two missing the user email.

Here is my code:


<?php
$dbServer1='contractor.emcee.tv';
$dbUser1='******';
$dbPass1='******';
$dbName1='purchaseorders';
$link = mysqli_connect("$dbServer1", "$dbUser1", "$dbPass1") or die("Could not connect post");
mysqli_select_db($link,"$dbName1") or die("Could not select database your_db_name1");
$job = $_GET['jobnumber'];
$vend = $_GET['vendor'];
$cat = $_GET['category'];
$amt = $_GET['amount'];
$note = $_GET['note'];
$email = $_SERVER['HTTP_RIM_DEVICE_EMAIL'];
$sql = mysqli_query($link,"insert po (job,vend,cat,amount,user,notes,date) values ('$job','$vend','$cat','$amt','$email','$note',NOW())");
if (!$sql){
print("FAILED");
}
else
{ print("SUCCESS");
$query = "SELECT MAX(id) AS PO from po";
   $result = mysqli_query($link, $query) Or die(mysqli_error($link)) ;
    while ($row = mysqli_fetch_assoc($result))
 {
$po = $row['PO'];
}
$query2 = "SELECT supervisoremail FROM job WHERE jobnum = $job";
$result2 = mysqli_query($link, $query2)Or die(mysqli_error($link));
while ($row2 =mysqli_fetch_assoc($result2)){
$sup = $row2['supervisoremail'];
}
$subject = "Purchase Order Request";
$supsubject = "FYI Purchase Order Completion";
$body = "Your purchase order request has been completed. \n \n Your PO number is: $po \n \n Job Number:$job \n Vendor:$vend \n Category:$cat \n Amount:$amt \n Items: \n $note \n \n A copy of this email has been sent to your supervisor.";
$supbody = "A purchase order request has been completed. \n \n PO number $po was issued to $email for the following: \n \n Job Number:$job \n Vendor:$vend \n Category:$cat \n Amount:$amt \n Items: \n $note \n \n Please address this if this request is in error.";
mail($email, $subject, $body);
mail($sup, $supsubject, $supbody);
}
?>

Hello. i am writing an API for lua... that sends POST requests to a php script. this php script is malfunctioning. below is the code and error.

Hey guys! I'm not that good with PHP, and I really need to finish this script. I'm having a hard time finding the problem..it keeps saying Error, query failed and I've also checked the database info, connectivity info and even tried a connection and database selection verification to see if that was working. Is there any other error you guys can find in the script that may be causing the message?

Edit: The files I want to be able to upload should be pdf's, txts, docs, etc. Would that be possible with this script?

Thanks!!

Code: [Select]
<form method="post" enctype="multipart/form-data">
<table width="350" border="0" cellpadding="1" cellspacing="1" class="box">
<tr>
<td width="246">
<input type="hidden" name="MAX_FILE_SIZE" value="104857600"></td>
<td width="80">&nbsp;</td>
</tr>
</table>
<p>Archivo:
  <input name="userfile" type="file" id="userfile" />
</p>
<p>Nomb
  <label for="name"></label>
 <input type="text" name="nombrelista" id="nombrelista" />
</p>
<p>Password (para luego borrar si necesario):
  <input type="text" name="pass" id="pass" />
</p>
<p>
  <input name="upload" type="submit" class="box" id="upload" value=" Upload " />
</p>

<?php
if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$filePass = $_POST['pass'];
$nombreLista = $_POST['nombrelista'];



$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

$host_db = "localhost";
$user_db = "melkien_rudesa";
$pass_db = "jorlan2407";
$base_db = "melkien_rudesa";



$link = mysql_connect($host_db, $user_db, $pass_db);
if (!$link) {
    die('Could not connect: ' . mysql_error());
mysql_close($link); 
}

mysql_select_db($base_db, $link) or die(mysql_error());



if(!get_magic_quotes_gpc())
{
   $fileName = addslashes($fileName);
}


$query = "INSERT INTO 'upload'('nombrelista', 'name', 'pass', 'size', 'type', 'content' ) VALUES ('$nombreLista', $fileName', '$filePass', $fileSize', '$fileType', '$content')";


mysql_query($query) or die('Error, query failed'); 


echo "<br>File $fileName uploaded<br>";

mysql_close($link); 

}



?>


</form>

Hi,
Im just in the middle of creating an update script for my mysql database but don't know why it's not working. p.s. I'm a little new to PHP, but know quite a bit, it's probably something really small.. *facepalm*

Here's the script:

the form (update.php)
<?
// Connect to the database
$link = mysql_connect('###', '###', '###');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('###', $link);

$id = $_GET['id'];

// Ask the database for the information from the links table
$query="SELECT * FROM orders WHERE id='$id'";
$result = mysql_query("SELECT * FROM orders");
$num=mysql_numrows($result);
mysql_close();

$i=0;
while ($i < $num) {
$name=mysql_result($result,$i,"Name");
$location=mysql_result($result,$i,"Location");
$fault=mysql_result($result,$i,"Fault");

?>

<form action="updated.php" method="post">
<input type="hidden" name="ud_id" value="<? echo "$id";?>">
Name: <input type="text" name="ud_name" value="<? echo "$name"?>"><br>
Location: <input type="text" name="ud_location" value="<? echo "$location"?>"><br>
Fault: <input type="text" name="ud_fault" value="<? echo "$fault"?>"><br>

<input type="Submit" value="Update">
</form>

<?
++$i;
}
?>

------------------------------------------------------
(processor) updated.php

<?php
// Connect to the database
$link = mysql_connect('###', '###', '###');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('###', $link);

$query="UPDATE orders SET Name='" . $_POST['ud_name'] . "', Location='" . $_POST['ud_location'] . "', Fault='" . $_POST['ud_fault'] . "' WHERE $id='" . $_POST['ud_id'] . "'";
echo $query;
$checkresult = mysql_query($query);
if ($checkresult) echo '<p>update query succeeded';
else echo '<p>update query failed';

mysql_close();
?>


------------------------------------------------------

Every time I want to update, it comes up with:

UPDATE orders SET Name='TEST', Location='TEST', fault='jbjh' WHERE =''
update query failed

Any help would be appreciated.

Hey everyone,

I'm currently working on a friends online script and i have a slight problem that i need help with.

Basically the code first searches "TBL_Friends" to see if you have any friends added. If it returns results it then turns your friends ID's into a variable. It then searches "TBL_Users_Online" to see if any body is logged based on the friend's ID it returned before. The first bit of the code works and it retrieves all the friends i got added. The second half is odd, if i have one or two friends added it will show that one is online. If i have more then three friends added it returns no results. I know my code is a bit sloppy and probably not the best way of writing it, im still learning PHP.

Anyways this is the code, any help is appreciated.

Code: [Select]
<?php
$FriendsOnline = mysql_query("SELECT Sender_ID FROM TBL_User_Friends WHERE Reciever_ID = $UserID");
while($fo=mysql_fetch_array($FriendsOnline))
{
     $FriendsOnlineID = $fo[Sender_ID];

     $FriendsOnlineNumber = mysql_query("SELECT * FROM TBL_Users_Online WHERE User_ID = $FriendsOnlineID");
     $FriendsNumber = mysql_num_rows($FriendsOnlineNumber);

     echo $FriendsNumber;
}
?>
$SenderID = Friends ID
$Reciever_ID = User ID
$UserID = User ID

Hi, I am trying to convert the register & login script from mysql to mysqli. I have converted the easy parts and have the connection to the database, but the following functions all need changing and I can't work out the correct solution mainly due to the deprecation of mysql_result()

The code that needs updating is
 

<?php
function user_count() {
	return mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `active` = 1"), 0);
}

function users_online() {
	return mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `logged_in` = 1"), 0);
}

function change_profile_image($user_id, $file_temp, $file_extn) {
	$file_path = 'images/profile/' . substr(md5(time()), 0, 10) . '.' . $file_extn;
	move_uploaded_file($file_temp, $file_path);
	mysql_query("UPDATE `users` SET `profile` = '" . mysql_real_escape_string($file_path) . "' WHERE `user_id` = " . (int)$user_id);
}

function has_access($user_id, $type) {
	$user_id 	= (int)$user_id;
	$type 		= (int)$type;
	
	return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `user_id` = $user_id AND `type` = $type"), 0) == 1) ? true : false;
}

function activate($email, $email_code) {
	$email 		= mysql_real_escape_string($email);
	$email_code = mysql_real_escape_string($email_code);
	
	if (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email' AND `email_code` = '$email_code' AND `active` = 0"), 0) == 1) {
		mysql_query("UPDATE `users` SET `active` = 1 WHERE `email` = '$email'");
		return true;
	} else {
		return false;
	}
}

function user_exists($username) {
	$username = sanitize($username);
	$query = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE `username` = '$username'");
	return (mysql_result($query, 0) == 1) ? true : false;
}

function email_exists($email) {
	$email = sanitize($email);
	return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'"), 0) == 1) ? true : false;
}

function user_id_from_username($username) {
	$username = sanitize($username);
	return mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"), 0, 'user_id');
}

function user_id_from_email($email) {
	$email = sanitize($email);
	return mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `email` = '$email'"), 0, 'user_id');
}

function login($username, $password) {
	$user_id = user_id_from_username($username);
	
	mysql_query("UPDATE `users` SET `logged_in` = 1 WHERE `user_id` = $user_id");

	$username = sanitize($username);
	$password = md5($password);
	
	return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `password` = '$password'"), 0) == 1) ? $user_id : false;
}

?>

And here is what the converter gave me:

function user_count() {
return mysql_result(mysqli_query($GLOBALS["___mysqli_ston"], "SELECT COUNT(`user_id`) FROM `users` WHERE `active` = 1"), 0);
}

function users_online() {
return mysql_result(mysqli_query($GLOBALS["___mysqli_ston"], "SELECT COUNT(`user_id`) FROM `users` WHERE `logged_in` = 1"), 0);
}

function change_profile_image($user_id, $file_temp, $file_extn) {
    $file_path = 'images/profile/' . substr(md5(time()), 0, 10) . '.' . $file_extn;
    move_uploaded_file($file_temp, $file_path);
    mysql_query("UPDATE `users` SET `profile` = '" . mysql_real_escape_string($file_path) . "' WHERE `user_id` = " . (int)$user_id);
}

function has_access($user_id, $type) {
    $user_id     = (int)$user_id;
    $type         = (int)$type;
    
    return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `user_id` = $user_id AND `type` = $type"), 0) == 1) ? true : false;
}

function activate($email, $email_code) {
    $email         = mysql_real_escape_string($email);
    $email_code = mysql_real_escape_string($email_code);
    
    if (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email' AND `email_code` = '$email_code' AND `active` = 0"), 0) == 1) {
        mysql_query("UPDATE `users` SET `active` = 1 WHERE `email` = '$email'");
        return true;
    } else {
        return false;
    }
}

function user_exists($username) {
    $username = sanitize($username);
    $query = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE `username` = '$username'");
    return (mysql_result($query, 0) == 1) ? true : false;
}

function email_exists($email) {
    $email = sanitize($email);
    return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'"), 0) == 1) ? true : false;
}

function user_id_from_username($username) {
    $username = sanitize($username);
    return mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"), 0, 'user_id');
}

function user_id_from_email($email) {
    $email = sanitize($email);
    return mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `email` = '$email'"), 0, 'user_id');
}

function login($username, $password) {
    $user_id = user_id_from_username($username);
    
    mysql_query("UPDATE `users` SET `logged_in` = 1 WHERE `user_id` = $user_id");

    $username = sanitize($username);
    $password = md5($password);
    
    return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `password` = '$password'"), 0) == 1) ? $user_id : false;
}

?>

Please could someone point me in the right direction here?     Also my site works perfectly well with MySQL, do I have to convert it to MySQLi?   Many Thanks   Paul If anyone knows how to solve this, it would be much appreciated. I already have a website template and would prefer to continue with mysqli instead of PDO.

Many Thanks

Paul

Hello,

I know i'm new, i hope it wont stop you guys from putting 5 minutes of your time to help me out.
I need a script to get some data from mysql and put this into a table.

i have a mysql database called "denora"
In there is a table called "chan"
In table chan are the columns: chanid (unique number), channel, currentusers, mode_ls

I would like the script to make a html table that lists the top10 channels (most "currentusers" on top)
And when "mode_ls" is "Y" it must not be shown (secret channel)

My knowledge of php and mysql arent that great so i really hope someone takes the time to help me uit

Kind regards
Stefan