PHP - Read Variables From Mysql Table

Once upon a time needed help on March 12, 2008 this community helped me, I am still a learner tried to help back but all the problems posted here beyond my skills to solve them so I kept quiet.  and Now I need help once again hope I will get it done.

Its PHPPOS developed in Codeigniter This code below is from application\controllers\inventory_summary.php file

Code: [Select]
<?php
require_once("report.php");
class Inventory_summary extends Report
{
function __construct()
{
parent::__construct();
}

public function getDataColumns()
{
return array($this->lang->line('reports_item_name'), $this->lang->line('reports_item_number'), $this->lang->line('reports_description'), $this->lang->line('reports_count'), $this->lang->line('reports_reorder_level'));
}

public function getData(array $inputs)
{
$this->db->select('name, item_number, quantity, reorder_level, description');
$this->db->from('items');
$this->db->where('deleted', 0);
$this->db->order_by('name');

return $this->db->get()->result_array();

}

public function getSummaryData(array $inputs)
{
return array();
}
}
?>

And this partial code below  is from application\controllers\report.php file


Code: [Select]
function inventory_summary($export_excel=0)
  {
    $this->load->model('reports/Inventory_summary');
    $model = $this->Inventory_summary;
    $tabular_data = array();
    $report_data = $model->getData(array());
    foreach($report_data as $row)
    {
      $tabular_data[] = array($row['name'], $row['item_number'], $row['description'], $row['quantity'], $row['reorder_level']);
    }

    $data = array(
      "title" => $this->lang->line('reports_inventory_summary_report'),
      "subtitle" => '',
      "headers" => $model->getDataColumns(),
      "data" => $tabular_data,
      "summary_data" => $model->getSummaryData(array()),
      "export_excel" => $export_excel
    );

    $this->load->view("reports/tabular",$data); 
  }
 
}
?>
Together they get the job done as follow:



Item Name   Item Number   Description   Count   Reorder Level   (Count is quantity in stock)
Item 1                   0002            xyz                    5                  3
Item 2                   0s00            xyz                    5                  3
Item 3                   0005            xyz                    5                  3
Item 4                   0006            xyz                    5                  3


The output I want need to be like this:

Item Name   Item Number   Description    cost price         Count        Reorder Level
Item 1                   0002            xyz                       250                5                  3
Item 2                   0s00            xyz                       120                5                  3
Item 3                   0005            xyz                       300                5                  3
Item 4                   0006            xyz                       500                5                  3
                                                                                   
                                                Total Stock Value  5850  Total Count 20  <-- this doesn't require any formation just needs to be at bottom of all the queries  like shown here.
           

This is what I want mainly >>>>>>>>>>  Total Stock Value (cost price * count foreach and then sum of all results ) and  sum of all count


cost price table and column is different than one used above in the code which is:

databse table is ' receivings_items '
Column is 'item_cost_price'

example php code to fetch is :


Code: [Select]
$sql = "SELECT `item_cost_price` FROM `receivings_items`";
config.php contains all the connection variables to MySQL



Please Help me   

I am having trouble trying to save my image file names to mysql after reading a dir. I am trying to use the file below. Any help would be appreciated. I was originallly getting the files to list but they we not showing up in the database. Now the page shows up blank.

<?php
require("config_0.php");

// Connect to server and select database.
mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect");
mysql_select_db("test")or die("cannot select DB");


$files = glob("images/*.jpg"); 
for ($i=1; $i<count($files); $i++) 
	{ 	
		$num = $files[$i];
		$sql="INSERT INTO images (url) VALUES ('$num')";

			if (!mysql_query($sql))
			  {
			  die('Error: ' . mysql_error());
			  }
		echo '<img src="'.$num.'" alt="random image">'."&nbsp;&nbsp;";
	}
?>

I have a MySQL database with each record of a person who has registered for an event, I am displaying the information on a web page for a user, but he wants to be able to print out all the records in alphabetical order by last name, first name later on so he will have a hard copy of each person who has registered at the table when they arrive.  How can I write each record to a Txt file that he can print out later that will be formated with the record contents along with each fields definition (Ex. Last Name - Smith, First Name - John, etc)?

Hi again, i'm wondering how can i make Sort (for example a-z) an column in php reading mysql informations.

Here's the script:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<html>

<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<link rel="stylesheet" href="table.css" type="text/css">
</head>
<body style="margin: 0 0 0 0;">
<?
include("passwd.php");
@$start = $_GET["start"];
if($start =='')
    $start =0;
include("lib.php");
$link = mysql_connect($host,$username,$password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db($db, $link);
if (!$db_selected) {
    die ("Can't use $db : " . mysql_error());
}
//total number of records in the table
$res = mysql_query("SELECT * from `$table`");
@$rows = mysql_num_rows ($res);

$result = mysql_query("SELECT * from `$table` limit $start,10");
if (!$result) {
    die('Invalid query: ' . mysql_error());
}
 echo "<p align=center class = 'menu'> Ocitana tabela: $table </p>";
$cols = mysql_num_fields($result);
$records = mysql_num_rows ($result);
 echo "<table align='center' width='1200' >";
 echo "<tr bgcolor='BBCCDD' class='menu'>";
 for ($i = 0; $i < $cols;$i++)
 {
    echo "<td align='center'>".mysql_field_name($result,$i)."</td>";
 }
 echo "</tr>";
 while ($row = mysql_fetch_array($result, MYSQL_NUM))
 {
    echo "<tr bgcolor='F6F6F6' class='normal'>";

   foreach ($row as $value)
   { echo "<td align='center'>".$value ."</td>";
   } 


   echo "</tr>";
 }
 $end = $start + $records;
 echo "<tr  align = 'center' bgcolor = 'BBCCDD' class='menu'><td colspan=$cols> Prijema $start  do  $end  od $rows </td></tr>";
 echo "<tr  align = 'center' class='mylink'><td colspan=$cols> ";



    if($start != 0)
    {
    $prev = $start - 10;
    echo "<a  href='tabela.php?start=$prev'> Prethodna </a> ";
    }
    if($start<$rows-10)
    {
    $next = $start + 10;
    echo "<a href = 'tabela.php?start=$next'>Sledeca</a> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;";
    }
     echo "</td></tr>";
     echo "</table>";
 ?>
</body>

</html>

Hello, I try to get website speed of some website, but i can read only ''domain.com'' i can't read website files like css , js ... why ? i use proxies for this job. here is the php code:

$options = array(
'useragent' => "Firefox (+http://www.firefox.org)", // who am i
'connecttimeout' => 120, // timeout on connect
'timeout' => 120, // timeout on response
'redirect' => 10, // stop after 10 redirects
'referer' => "http://www.google.com",
'proxyhost' =>'85.25.8.14:80'
);
$response = http_get("http://solve-ict.com/wp-content/themes/ict%20theme/js/jquery-1.7.1.min.js", $options , $info);

but it works fine with http://domain.com/ , but with files css or js it gives 404, using some free proxy servers available ? Thanks.

Hi All! I've written up a script for my website. It\ is basically a virtual job quest. My queries are all correct it just isn't registering the variable for the session. It is $-SESSION[theid']. I want to be bale to use it in my table but I get an error. How do I write this in my SQL query for it to work. The page (when no errors), doesn't show my data. Here is my ocde:

Code: [Select]
<?php
session_start();
include("config536.php");
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css" />
</head>

<?php
if(!isset($_SESSION['username'])) {
echo "<ubar><a href=login.php>Login</a> or <a href=register.php>Register</a></ubar><content><center><font size=6>Error!</font><br><br>You are not Logged In! Please <a href=login.php>Login</a> or <a href=register.php>Register</a> to Continue!</center></content><content><center><font size=6>Messages</font><br><br></center></content>";
}

if(isset($_SESSION['username'])) {
echo "<nav>$shownavbar</nav><ubar><img src=/images/layout/player.gif><a href=status.php>$showusername</a>.......................<img src=/images/layout/coin.gif> $scredits</ubar><content><center><font size=6>Basic Quests</font><br><br>";

$startjob = $_POST['submit'];
$jobq = "SELECT * FROM jobs WHERE username='$showusername'";
$job = mysql_query($jobq);
$jobnr = mysql_num_rows($job);

if($jobnr == "0") {
?>
<form action="<?php echo "$PHP_SELF"; ?>" method="POST">
<input type="submit" name="submit" value="Start Job"></form>
<?php
}
if(isset($startjob)) {
$initemidq = "SELECT * FROM items ORDER BY RAND() LIMIT 1";
$initemid = mysql_query($initemidq);
while($ir = mysql_fetch_array($initemid)) {
$ids = $ir['itemid'];
}
mysql_query("INSERT INTO jobs (username, item, time, completed) VALUES ('$showusername', '$ids', 'None', 'No')");
$wegq = "SELECT * FROM items WHERE itemid='$ids'";
$weg = mysql_query($wegq);
while($wg = mysql_fetch_array($weg)) {
$im = $wg['image'];
$nm = $wg['name'];
$id = $wg['itemid'];
}
$_SESSION['theid'] = $id;

echo "<font color=green>Success! You have started this Job!</font><br><br>Please bring me this item: <b>$nm</b><br><br><img src=/images/items/$im><br><br><br>";
echo $_SESSION['theid'];
}

if($jobnr == "1") {
$finish = $_POST['finish'];
$okgq = "SELECT * FROM items WHERE itemid='$yes'";
$ok = mysql_query($okgq);
while($ya = mysql_fetch_array($ok)) {
$okname = $ya['name'];
$okid = $ya['itemid'];
$okimage = $ya['image'];
}
echo "Where is my <b>$okname</b>?<br><br><img src=/images/items/$okimage><br><br><br>";
echo $_SESSION['theid'];
?>
<form action="<?php echo "$PHP_SELF"; ?>" method="POST">
<input type="submit" name="finish" value="I have the Item"></form>
<?php
}
}

if(isset($finish)) {
$cinq = "SELECT * FROM uitems WHERE theitemid='$_SESSION[theid]'";
$cin = mysql_query($cinq);
$connr = mysql_num_rows($cin);
if($connr != "0") {
echo "<font color=green>Success! You have the item.</font>";
} else {
echo "<font color=red>Error! You do not have my item!</font>";
}

}
?>
. I basically just want to know how I can set this session as a variable. Also..I have a user login on every page and I want to be able to destroy JUST THE "theid" session and NOT the username session. How would I do that too? thanks for the help in advance!

I have a table and the structure is Code: [Select]
ID, UID, Site, Uname, PassI already have this <?php

$result = mysql_query(sprintf("SELECT * FROM Logins WHERE UID = %d", $_COOKIE['UID_WatsonN']));
//check login table against cookie
if(($num = mysql_num_rows($result)) > 0){
mysql_close();
?>
<table border="0" cellspacing="2" cellpadding="2">
<tr>
<th>ID</th>
<th>UID</th>
<th>Site</th>
<th>Uname</th>
<th>Pass</th>
</tr>
<center>
<?php
$i=0;
while ($i < $num) {

$f1=mysql_result($result,$i,"ID");
$f2=mysql_result($result,$i,"UID");
$f3=mysql_result($result,$i,"Site");
$f4=mysql_result($result,$i,"Uname");
$f5=mysql_result($result,$i,"Pass");
?>

<tr>
<td><?php echo $f1; ?></td>
<td><?php echo $f2; ?></td>
<td><?php echo $f3; ?></td>
<td><?php echo $f4; ?></td>
<td><?php echo $f5; ?></td>
</tr>

<?php
$i++;
}
}
?>

which only gets the data from that one person. What is in the table is usernames and passwords for diffrent sites and i want to take all the usernames and passwords and put them into variables to pass on to the login form.

Hey Guys,

I have to insert some data in MySQL but it wont work  .

Please have a look.

<?php

// to values are set to empty

$vatsim="";
$ivao="";

// values from form in other page are set
if(isset($_POST["pilotid"])) $pilotid=$_POST["pilotid"];
if(isset($_POST["network"])) $network=$_POST["network"];
if(isset($_POST["vid"])) $vid=$_POST["vid"];
if(isset($_POST["pilot"])) $pilot=$_POST["pilot"];

// if value is that copy data in this value, otherways in that value
if ($network == "IVAO") { $ivao="$vid";}
if ($network == "VATSIM") { $vatsim="$ivao";}

// connect db
include(dbconnect.inc.php);

// first sql to update some data in one table
$sql = "UPDATE `360283`.`jos_users` SET `IPS` = \'1\' WHERE `jos_users`.`id` = \'$pilotid\'";
$result1 = mysql_query($sql);

// 2nd sql to insert some data in other table
$sql2 = "INSERT INTO `360283`.`IPS_Pilots` (`ID`, `Name`, `Hours`, `Flights`, `LastFlight`, `IVAO`, `VATSIM`, `Enabled`, `Rating`) VALUES ('$pilotid', '$pilot', NULL, NULL, NULL, '$ivao', '$vatsim', '1', '0');";
$result2 = mysql_query($sql2);

// sql to check if it was succesful
$sql3 = "SELECT * FROM `IPS_Pilots` WHERE `ID` = '$pilotid'  LIMIT 0, 30 ";
$result3 = mysql_query($sql3);
$num3 = mysql_numrows($result3);


// echo succesfull or not
if (!$num3) { echo "Sorry, but I failed to apply this pilot."; } else { echo "Pilot succesfully applied."; }



?>

Thanks

Hi,

I'm only just getting started with php and mysql but I like to go off the deep end.

something is wrong with my script.

it seems to work up untill it comes to the UPDATE part. It wont actually update the table and the mysql_affected_rows bit returns -1 I've searched for an explanation and can't find one.

Can someone tell me what I'm doing wrong

Thanks.

//connect to mysql
$con = mysql_connect("xxx","yyy","zzz");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
//use the right database
mysql_select_db("foo_bar");

//select all cancelled tickets
$status_query = "SELECT * FROM jos_vm_orders where order_status = 'X'";
$status_result = mysql_query($status_query) or die(mysql_error());

while ($newArray = mysql_fetch_array($status_result, MYSQL_ASSOC)) {
                $order_id = $newArray['order_id'];


$quantity_query = "SELECT * FROM jos_vm_order_item where order_id = $order_id";
$quantity_result = mysql_query($quantity_query) or die(mysql_error());

        while ($newArray2 = mysql_fetch_array($quantity_result, MYSQL_ASSOC)) {
                $product_quantity = $newArray2[product_quantity];
                $order_item_sku = $newArray2[order_item_sku];

echo ("$order_id cancelled $product_quantity tickets sku was $order_item_sku \n");

mysql_query("UPDATE LOW_PRIORITY jos_vm_product SET product_in_stock = product_in_stock + $product_quantity
WHERE product_sku = $order_item_sku AND STR_TO_DATE($order_item_sku, '%Y-%d-%m')
$quantity_query = "SELECT * FROM jos_vm_order_item where order_id = $order_id";
$quantity_result = mysql_query($quantity_query) or die(mysql_error());

        while ($newArray2 = mysql_fetch_array($quantity_result, MYSQL_ASSOC)) {
                $product_quantity = $newArray2[product_quantity];
                $order_item_sku = $newArray2[order_item_sku];

echo ("$order_id cancelled $product_quantity tickets sku was $order_item_sku \n");

mysql_query("UPDATE LOW_PRIORITY jos_vm_product SET product_in_stock = product_in_stock + $product_quantity)
WHERE product_sku = $order_item_sku AND STR_TO_DATE($order_item_sku, '%Y-%d-%m')
>= CURDATE()");
echo ("$product_quantity tickets released into $order_item_sku \n");
echo ("Affected rows ");
echo (mysql_affected_rows());
echo ("\n");

}
}
mysql_free_result($status_result);
mysql_free_result($quantity_result);
mysql_close($con);
?>

So, I'm working on a quiz system. The text and choices are stored in a MySQL database. I haven't gotten to the choices yet, I'm still having trouble with the text. Here's my code:
Code: [Select]
$querytext = "SELECT text FROM quiz id = '$prob'";
$result = mysql_query($querytext) or die(mysql_error());
echo $result;
Yes, I'm already connected and stuff, that's just the snippet. I made sure that $prob is 1.
Here's the MySQL Error I'm getting:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= ''' at line 1

I've looked all over the web and through the MySQL/PHP section of a 991-page PHP book. What am I doing wrong?
I bet that I really, really failed epicly this time, but I never catch my epic fails and have to ask questions about them in order to fix the problem. So please reply

Ok so I have this script which will take a propery formated .csv file and upload to my database. here's the part i need to know if it's possible. basically it puts a group of codes into the database but i need a field that will specify which user the code pertains to from another table. The most logical way to reference the user is with an id code; but short of having the user reference an id code to put in the csv file, is there anyway I can make a query that will INSERT INTO and upload the csv? here's my current code...

<?php

$tmpName = $_FILES['song_codes']['tmp_name'];

include('database.php');

if($sc != ""){

mysql_query('LOAD DATA LOCAL INFILE \''.$tmpName.'\' INTO TABLE music_codes FIELDS ENCLOSED BY "\"" TERMINATED BY "," LINES TERMINATED BY "\n" ;') or die('Error loading data file.<br>' . mysql_error()); 

Trying to get this done:

Page_1 has many external links, when certain links are clicked i do not want user to go directly to page, but rather go to a special add page_2 where user must click a second time on the link to finally get there.

The add page_2 must show on screen the name of the initial link from page_1, it must change accordingly with the link it came from page_1,once on page_2 the hyperlink redirects outside the site.

So far i am thinking give an id to the div or "<a href..." on page_1 then somehow have page_2 detect that id and fill in the variable for the final external link. Other wise is there a way to detect a url from incoming?

I guess a similar example is how some domain name sellers landing page will indicate the name of the site. Such as "Thisdomain.com" is for sale. same landing page but the name changes according to the domain that was typed.

$var = @$_GET['q'] ;
$trimmed = trim($var);
$table = @$_GET['field'];
$query="SELECT * FROM contacts WHERE @'table' contains @'trimmed' order by id";
$result=mysql_query($query);
$num=mysql_numrows($result);


Why wont this work?

Zacron

Hi there,   I have a table in a MySQL database where I keep a list of user privileges. I am trying to create variables where the name of variable matches the privileges in the table.   This is also known as variable variables (I think).   EDIT (17/07/2014 04:02 PM): This might be a better way to describe what I'd like, so if the value from the table is admin_panel I'd like to dynamically create a variable with that name.   I have created a code so far, but all I seem to be getting is a list of Notice errors telling me that the variable is undefined. (I have supplied a list of errors a bit further down the post).   Here is the code:

<?php
$host = "localhost";
$account = "***";
$password = "****";
$dbname = "****";

$connect = mysql_connect($host,$account,$password) or die("Unable To Connect");
$db = mysql_select_db($dbname,$connect) or die("Unable To Select DB");

$perm_query = "SELECT * FROM `privileges`";
$permission_query = mysql_query($perm_query);

while($row = mysql_fetch_array($permission_query))
{
$rows[] = $row;
}

foreach($rows as $row)
{
${$row['privilege']};
}
?>

The list of errors:   Notice: Undefined variable: admin_panel in C:\xampp\htdocs\DynamicVariables.php on line 20
Notice: Undefined variable: create_user in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: edit_user in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: delete_user in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: create_group in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: edit_group in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: delete_group in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: view_log in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: log_settings in C:\xampp\htdocs\DynamicVariables.php on line 20

Notice: Undefined variable: password_change in C:\xampp\htdocs\DynamicVariables.php on line 20   Thanks
Edited by chrisrulez001, 17 July 2014 - 10:05 AM.

Hi Everyone,

I'm trying to convert my MySql data into variables so I can use them on my website. For example, if I want to convert a city or even first or last name into a variable. How would I go about doing that? Thanks everyone!