PHP - Show Every 3rd Entry From Table

Newbie here, I am actually a math teacher working on creating a simulated store front for my students where they keep up with inventory, budgeting etc.

Everyday my students will have to look at how much product is in stock, how much product was sold, and then input how much should be ordered for the next day.

I am working on an answer key that will be used to validate students input but the 'quantity' and 'sold' fields will be relative to the most recent post. Right now the code below pulls all the rows under 'quantity', but I need it to only pull the most recent post.

I have done my best to search the forum and try to find how to do this, but admittedly I am not even really sure what to be searching for. Any help would be great.



Code: [Select]
<?php
$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("myDatabase", $con);

$result = mysql_query("SELECT * FROM myTable");

while($row = mysql_fetch_array($result))
  {
 $quantity = $row['quantity'];
 $sold = $row['numbersold'];
 $needed = $row['remaining'];
 echo $needed-($quantity-$sold);
  }

mysql_close($con);
?>

Hello there,   I would be really grateful if someone could advice please, how do I find this pesky entry in the 'db' table? This is what it says,   "Found an entry in the 'db' table with empty database name; Skipped." MySQL version is 5.5.40   I am pretty much stuck on it. Does it mean the table's name is db? Plus, I am not sure I understand the wording of it too. The empty database name cannot exist by default as it won't allow you to create a database without a name in the first place. I do not even know how and where to begin to look for it because honestly speaking I do not quite understand the phrase itself, because it does not make much sense to me. I would really highly appreciate it if someone could explain and suggest how do I find the... "whatever it says" :-) and to fix it. Many thanks in advance!   P.S. I know this warning may easily and safely be ignored as it is of minor inconvenience and not that important the more so as MySQL is working fine but my curiosity is just running wild for two reasons: 1. I do not really understand what this warning message is trying to tell me and 2. How do I find and adjust it to turn off the warning.   Simple regular things as running myisamchk -r *.MYI on where MySQL is did not help as this warning is actually out of scope of what myisamchk can fix.   And yes, I've Googled it long enough but couldn't seem to come up with anything reasonable solutions.   Thankful for any suggestions / pointers / assistance / comments at all.
Edited by Klaipedaville, 29 October 2014 - 06:17 AM.

Hi all,
firstly apologies as this is a cross post from another forum and we have hit a block.. I am hoping that opening this up to another set of gurus we can get a resolution.

What I am trying to achieve is this...
I have 2 tables Main and FinancialYear. Main holds all data which I use a form to post the data to it..(all works fine). I use this code to create a drop down in the insert.php form. again this works.
Code: [Select]
<tr><td>Financial Year: xxxx/xxxx</td><td>
<!-- pulls the data from the table variable to populate the dropdown menu -->
   <?php
      $database = 'Projects_Main';
      $fintable = 'FinancialYear';
         if (!mysql_connect($db_host, $db_user, $db_pwd))
               die("Can't connect to database 'cos somethin' is wrong");
         if (!mysql_select_db($database))
             die("Can't select database");
      $result = mysql_query("SELECT FinancialYear_id, FinancialYear FROM {$fintable} order by FinancialYear");
      $options="";
         while ($row=mysql_fetch_array($result)) 
            {
               //$id=$row["FinancialYear_id"];
               $thing=$row["FinancialYear"];
               $options.="<OPTION VALUE=\"$thing\">".$thing.'</option>';
            }
   ?>
      <SELECT NAME="FinancialYear">
      <OPTION VALUE=0>Choose</OPTION>
      <?=$options?>
      </SELECT>
   </td></tr>

What I have done is built another form which list all records in the database and creates an update url for every record that passes the field Project_id where i use $_get to retrieve the Project_id to retrieve the relevant data into the update.php form.

I am able to populate the form with all the correct information BUT I am looking to introduce some dropdowns to aid updating the data and provide consistency to the data. .

Code: [Select]
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db('Projects_Main')or die("cannot select DB");

// get value of id that sent from address bar
$Project_id=$_GET['Project_id'];
//define vars
$FinancialYear=$_POST['FinancialYear'];
// other vars defined here also.. about 30

// Retrieve data from database
$sql="SELECT * FROM Main WHERE Project_id='$Project_id'";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
?>
//update_record_ac.php posts the data to the dbase.
<form name="form1" method="post" action="update_record_ac.php">
<center>
<table>
<tr><td><b>Store Details<b></td></tr>
<tr><td>Financial Year:</td><td>
// takes the data from $rows and present to form
<input name="FinancialYear" type="text" id="FinancialYear" value="<?php echo $rows['FinancialYear']; ?>">

// this is where I need to create the drop down.. see my other comments in the post.....

</td></tr>


the financialYear table consists if the following;
financialyear_id - pri, auto inc. ---- data format is 2010/2011, 2011/2012....
financialyear

the main table contains 30 fields .. won't list em all...
Project_id - pri, auto inc
financialyear

I need the drop down to pull the data from the financialyear table and then to present or focus on the currently stored data...
so if the store value in the table Main is 2010/2011 if Ii was to select the update url in the list_record.php it will pull all the relevant data into update_record.php form. the financialyear field in the form should be a dropdown with all the financial years listed but the 2010/2011 is selected or focused. I still need to be able to change the entry and post this back to the table Main.....

So the dropdown contains the list of years from the financialyear table but when the record is pulled from table main the year that is stored in table Main should be highlighted in the dropdown and I should be able to select a new record and post back to the table Main..


any thoughts...

please don't slate for the cross post, I haven't sanatised the data at any stage. I know i'm open to injection attacks. and yes my code is a little dirty... all these will be rectified as i finalise the process and ensure the consept works.

Thanks for taking the time to read and hopefully you are able to understand the requirement and are able to assist.

thanks
Balgrath

Alright,  wasn't quite sure how to summarize this in the title,  but I want to:

Check if a user status is "active" or not based on the UserName input.

I have a table witch holds:

Code: [Select]
VarChar Username      Var CharPassWord     int Active
 Ted                        TedsPW                1
something like the above(assuming it formatted correctly.

In my php script I will want to input a variable for Username to check for:

inputUN in this example would be "Ted".

$UserNameToCheck = $_GET['inputUN'];

Then I want to check for that UserName in the database,  if it exists,  I want pull the value for the "Active" field for just that UserName and echo it.

Thanks for any help.
 

I'm trying to have a running total of the number of views an image gets. My test file works perfectly every time. The actual file, image_win.php, does not and after much testing can't see what's wrong. This files STRANGENESS: the first two views count fine, with 1 added the total in the field image_visit. But every time after that 2 is added to the total in image_visit. Very bizarre.

Thanks for taking a look -Allen
Code: [Select]
<?php
////////TEST IMAGE_WIN //////////THIS WORKS
include_once "scripts/connect_to_mysql.php";
include_once "scripts/paths.php";
$art_id = 372;
$QUERY="SELECT user_id FROM artWork WHERE art_id = '$art_id'";
$res = mysql_query($QUERY);
$num = mysql_num_rows($res);
if($num >0){

while($row = mysql_fetch_array($res)){ 
     $owner_id = $row['user_id'];
    }
}
mysql_query("UPDATE userInfo SET image_visit = image_visit +1  WHERE user_id = '$owner_id'"); 
?>

<?php 
//////// image_win.php  /////////THIS DOES NOT WORK
$user_id=$_SESSION['user_id'];
include_once "scripts/connect_to_mysql.php";
include_once "scripts/paths.php";
$image_link = $_GET['image_link'];
$art_id = $_GET['art_id'];
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<meta name="viewport" content="width=device-width, height=device-height, target-densityDpi=device-dpi">
<title>Image Window</title>
</head>
<body bgcolor="#000000" leftmargin="0">
<div  align="center" >
<img src="slir/w640-h535/<?php echo $path.$image_link.$postCat; ?>" />
<?php
$QUERY="SELECT user_id FROM artWork WHERE art_id = '$art_id'";
$res = mysql_query($QUERY);
$num = mysql_num_rows($res);
if($num >0){
while($row = mysql_fetch_array($res)){ 
     $owner_id = $row['user_id'];
    }
}
mysql_query("UPDATE userInfo SET image_visit = (image_visit +1)  WHERE user_id = '$owner_id' ");

?>

MOD EDIT: code tags added.