PHP - Showing Results In A Table

I am writing a CRON job that will execute daily.  First it will identify from a MySql table the date in a field 'FAPforSale_repost35'  If the date is the today date it will then execute commands to delete photo images in a directory, delete the directory, and finally remove the record from the database.   I am on step one which is to build the array of records that match the days date.  When I run the code, there are no errors but I am not getting results even though the records in the test table are set for today.  Below is the select

<?php

 define( "DIR", "../zabp_employee_benefits_processor_filesSm/", true );
 
 require( '../zipconfig.php' );
 require( DIR . 'lib/db.class.php' );
 require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/configRecognition.php' );
 require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/libRecognition/MailClass.inc' );
 
 
 
$todayRepost35 = date("Y-m-d");
echo $todayRepost35;



function repostEndSelect()
 
      {
        
        global $db;
            
    $this->db = $db;
        
            
     $data = $this->db->searchQuery( "SELECT `FAPforSale_IDnumber`,  `FAPforSale_image1`,  `FAPforSale_image2`,  `FAPforSale_image3`,  `FAPforSale_repost35`  FROM  `FAP_forSaleTest` Where `FAPforSale_repost35` = '$todayRepost35' ");
    
    
         $this->FAPforSale_IDnumber = $data[0]['FAPforSale_IDnumber'];
        $this->FAPforSale_image1 = $data[0]['FAPforSale_image1'];
        $this->FAPforSale_image2 = $data[0]['FAPforSale_image2'];
        $this->FAPforSale_image3 = $data[0]['FAPforSale_image3'];
        $this->FAPforSale_repost35 = $data[0]['FAPforSale_repost35'];
        
        echo $this->FAPforSale_IDnumber;
        echo $this->FAPforSale_image1;
        echo $this->FAPforSale_image2;
        echo $this->FAPforSale_image3;
        echo $this->FAPforSale_repost35;
        
    } // ends function...
        
echo( ' Finished...' );

?>

Thanks in advance for any suggestions or direction.  Chapter two will be when I start testing the commands to delete.

Hi   I have 2 tables that I am searching infor for. One holds the phone info and the other the repairs that are available.   I have the code below and it displays on the webpage in a table for the phone information and then another table the repairs but they seem to be separate tables and headers. inbetween them is the phone and the details which separates the repairs but its not in a table.   Hopefully the code below will explain better.   What I would like it the phone details in one table and then a table full of all the repairs (I am not bothered about a new heading for each repair).

$result = mysql_query("SELECT * FROM phone, phonerepairs WHERE phone.model_no='".$_POST['model']."' AND phone.phone = phonerepairs.phone");

while($row = mysql_fetch_array($result))
{
echo '<td width="70" align="center"><img src="'.$row['icon'].'" width="66"></td>';
echo '<td width="100" align="center">'.$row['model_no'].'</td>';
echo '<td width="130" align="center">'.$row['model'].'</td>';
echo '<td width="50" align="center">'.$row['year'].'</td>';
echo '<td width="150" align="center">'.$row['capacity'].'</td>';
echo '</tr>';
echo '</table><br>';
echo '</p>';
echo '<p>';
echo '<div id="pageheading"><h3><div id="title">Repairs we carry out for '.$row['model'].'</div></h3></div>';
echo '<table width="99%" cellpadding="5" cellspacing="0" border="1">';
echo '</p>';
echo '<p>';
echo '<tr>';
echo '<td align="center"><strong>Image</strong></td>';
echo '<td align="center"><strong>Fault</strong></td>';
echo '<td align="center"><strong>Repair</strong></td>';
echo '<td align="center"><strong>Cost</strong></td>';
echo '</tr>';
echo '<tr>';
echo '<td width="70" align="center"><img src="'.$row['icon'].'" width="66"></td>';
echo '<td width="100" align="center">'.$row['fault'].'</td>';
echo '<td width="130" align="center">'.$row['description'].'</td>';
echo '<td width="50" align="center">&pound'.$row['cost'].'</td>';
echo '</tr>';
echo '</table><br>';

Thanks in advance    Martyn

Hi people.

I am creating a carpet website for a good friend of mine as a favour.  I am storing the carpets info in a MYSQL database and am currently trying to relay the info on a page.  I have the array kind of done but it is not producing the results I want.  Let me explain further:

The columns in my database under the table "carpets" a

id (auto incremented)
colour
type
title
price
description
imageloc

------

under the type I have various types of carpet (6-ish).  They a Twist, Striped etc etc...

Heres the php problem.


I have all the images there and ready on the page and want to link them all with linking commands. i.e:

the main page is at: http://www.ircdirect.co.uk/FTPServers/supremecarpets/index.php

the carpets results page is at: http://www.ircdirect.co.uk/FTPServers/supremecarpets/carpetlist.php

on the index.php page I want to link them so that for example:

If he looks for a striped carpet he clicks "striped" and is shows carpetlist.php but with the striped carpets displayed.

example link: <A HREF="carpetlist.php?type=striped">IMAGE HERE</A>

I have tried various coding on the carpetlist.php and cant seem to get it to work.

here is the snippets:

<?php
// Make a MySQL Connection
$query = "SELECT * FROM carpets GROUP BY type"; 
 
$result = mysql_query($query) or die(mysql_error());


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


echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100% />";

echo "<TR />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/topleft.png /></TD />";

echo "<TD BACKGROUND=images/main/search/top.png /> </TD />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/topright.png /></TD />";

echo "</TR />";

echo "<TR />";

echo "<TD BACKGROUND=images/main/search/left.png />";
echo " ";
echo "</TD />";

echo "<TD BACKGROUND=images/main/search/bg.png />";

echo "<FONT FACE=VERDANA SIZE=1 />";
echo $row['title'];

echo "</TD />";

echo "<TD BACKGROUND=images/main/search/right.png />";

echo " ";

echo "</TD />";

echo "</TR />";

echo "<TR />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/bottomleft.png /></TD />";

echo "<TD BACKGROUND=images/main/search/top.png /> </TD />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/bottomright.png /></TD />";

echo "</TR />";

echo "</TABLE />";

echo "<BR />";

}
?>

It displays ALL the results which is not what I want.  I want only striped carpets, blue carpets etc....

Help is needed and VERY much appreciated!

Kind Regards,
Ian

i have built pages that paginate with 10 rows per page (some pages show more but for the moment i want to focus on this particular page)

//Define Some Options for Pagination
$num_rec_per_page=10;
if (isset($_GET["page"])) { $page  = $_GET["page"]; } else { $page=1; }; 
$start_from = ($page-1) * $num_rec_per_page;

$results = mysql_query("SELECT * FROM `ecmt_memberlist` WHERE toonCategory='Capital' AND oldMember = 0 ORDER BY CONCAT(MainToon, Name) LIMIT $start_from, $num_rec_per_page") or die(mysql_error());

	$results_array = array();
	while ($row = mysql_fetch_array($results)) {
    	$results_array[$row['characterID']] = $row;
}

The above sets the variables for the pagination and below the results are echo with 10 rows per page   then i show the pagination links:

<?php 
$sql = "SELECT * FROM `ecmt_memberlist` WHERE toonCategory='Capital' AND oldMember = 0 ORDER BY CONCAT(MainToon, Name)";

$rs_result = mysql_query($sql); //run the query
$total_records = mysql_num_rows($rs_result);  //count number of records
$total_pages = ceil($total_records / $num_rec_per_page); 
?>


      <table width="100%" border="0" cellspacing="0" cellpadding="0">
        <tr>
          <td align="center"><div style="width:100%; text-align:center;">
          <ul class="pagination">
    <li class="selected">
	<?php
echo "<a href='capitalmember.php?page=1'>".'«'."</a> ";?> 
</li>
<?
for ($i=1; $i<=$total_pages; $i++) { 
            echo "<li><a href='capitalmember.php?page=".$i."'>".$i."</a></li> "; 
};
?>
<li class="selected">
<? 
echo "<a href='capitalmember.php?page=$total_pages'>".'»'."</a> "; // Goto last page
?></li>
  </ul></div>
    <?php
  $pageNr = $page;   // Get Current Page Number
$from = $pageNr * $rowsPerPage;  // 3 * 10 = 30  // 3 * 10 = 30
$to = $from + $rowsPerPage;      // 30 + 10 = 40
echo $pageNr;
/* Result: From page 30 to 40 */
  ?></td>
        </tr>
      </table>

this works great and shows me 10 results per page,   what i need to work out and work on next is:   echo the number of results above the records (for example: "showing records 1 to 10" and then on page 2 "showing records 11 to 21" and page 3 "showing records 22 to 32"   how can i work out the maths for this echo?   i was thinking along the lines of;

    <?php
$pageNr = $page;   // Gets Current Page Number
$from = $pageNr * $rowsPerPage;  // 3 * 10 = 30  
$to = $from + $rowsPerPage;      // 30 + 10 = 40
// Now Show Results
echo $from; // Echo from
echo $to // Echo to
  ?>

but i'm still working on this..   then i need to look at shortening the amount of page links on the page if for example i have 500 records it shows me links 1 to 50 and eats up the page....   Appreciate any help and light onto my problems.   Many thanks
Edited by jacko_162, 11 January 2015 - 05:43 PM.

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=330251.0

Our product is running on LAMP(PHP) architecture. For MySQL operation PDO library has been used.

We are in the process of moving the product to a new Linux server.
On the new server we are facing MySQL query output related issue. At the end of the post I have listed the server configuration details of both the servers and the sample PHP program with the output.

It works fine in server 1 but when the same code moved to Server 2, the output is not correct.

Would appreciate any points to solve this problem

The following piece of code is tested and output is also given.

<?php

/*** mysql hostname ***/
$hostname = 'localhost';

/*** mysql username ***/
$username = 'ram';

/*** mysql password ***/
$password = 'tellmehow';

try {
       $dbh = new PDO("mysql:host=$hostname;dbname=inhouse", $username, $password);

       $sql = "SELECT first_name,last_name,user_id,email FROM users LIMIT 1";
       $stmt = $dbh->prepare($sql,array(PDO::MYSQL_ATTR_USE_BUFFERED_QUERY => TRUE));
       $stmt->execute();
       $result = $stmt->fetch(PDO::FETCH_OBJ);
       $stmt->closeCursor();
       /*foreach($result as $key=>$val)
       {
               echo $key.' - '.$val.'<br />';
       }*/
       print_r($result);
       $dbh = null;
   }
   catch(PDOException $e)
   {
       echo $e->getMessage();
   }
?>


Output for server 1:

stdClass Object
(
   [first_name] => super
   [last_name] => admin
   [user_id]=>1
   [email] => ram@greynium.com
)

output for server 2:

stdClass Object
(
   [first_name] => super
   [last_name] => admin
   [users] => ram@greynium.com
)

On server 2 I am not getting the output for field 'user_id' from the table 'users'.

For the field 'email' from 'users' table, it is displaying the name of the table itself (users). Not sure what is wrong.

The configuration of the servers are given below
Server 1
--------
OS Details:
Linux 2.6.9-5.ELsmp

PHP Details:
PHP 5.2.3 (cli) (built: Aug 28 2007 11:48:30)
Copyright (c) 1997-2007 The PHP Group
Zend Engine v2.2.0, Copyright (c) 1998-2007 Zend Technologies
   with Suhosin v0.9.22, Copyright (c) 2007, by SektionEins GmbH

PDO details:
PDO
PDO support => enabled
PDO drivers => sqlite2, sqlite, mysql
PDO Driver for MySQL, client library version => 5.0.37
PDO Driver for SQLite 3.x => enabled



Server 2
---------
OS Details:
Linux 2.6.18-8.el5

PHP Details:
eAccelerator requires Zend Engine API version 220060519.
The Zend Engine API version 220090626 which is installed, is newer.
Contact eAccelerator at http://eaccelerator.net for a later version of eAccelerator.

PHP 5.3.2 (cli) (built: Jul  6 2010 17:57:31)
Copyright (c) 1997-2010 The PHP Group
Zend Engine v2.3.0, Copyright (c) 1998-2010 Zend Technologies


PDO Details:
PDO
PDO support => enabled
PDO drivers => mysql, sqlite, sqlite2
PDO Driver for MySQL => enabled
PDO Driver for SQLite 3.x => enabled

I am using html5 and all the pages have h2 headings as titles.  They all show up in google search results fine.  The only thing that doesn't show up fine are some of the descriptions under each heading. Some meta descriptions show up fine under the correct heading title, while rest shows ALL the heading tags AS description under each search result.     For eg.   Cars www.mywebsite.com/category?id=5&name=cars

when I do a "SELECT * FROM system_disks WHERE system_id = 'aNumber'" and their is more than one result with that system_id, I get more than one result obviously.

I essentially need two results in one row. I've took a screen shot of what it is doing and what I need it to do.

The blurred row is a different system_id

This is what the while ($row = mysql_fetch_array($query)) is currently doing:



This is what I would like it to do:


Sorry if this is a dumb question, I've been coding the past 48 hours and my brain is fried 

Hi,
I'm quite new to this and I'm trying to get this to line up in a table with 3 columns (unlimited rows)
I have searched and tried but I'm not having any luck.

Can any one help?
Thank you

Code: [Select]
<?php 
echo '<div class="resultados_sub_cat">';

foreach($this->subcats as $key => $subcat) {

$subcat->link = JRoute::_('index.php?option=classcliff&view=list&catid='.$subcat->id."&Itemid=".$this->Itemid);
if ($key != 0)
echo ' - ';
echo '<a href="'.$subcat->link.'">'.$subcat->name.'</a>';

} 
?>

I have three tables: events, orderdetails & orders.

First I query orderdetails to find all the records that match the EventID:

$query1 = SELECT * FROM orderdetails WHERE EventID = $_SESSION['EventID'];

This returns 4 records.  These 4 records have a field called DetailOrderID which is the foreign key for orders.OrderID.  Next I need to query the results of the first query to find all the records in the orders table that match up.  For example:

SELECT * from orders where $query1.DetailOrderID = orders.OrderID.

How would I go about doing this?  I'm head down the temporary table solution but wanted to through this one out for discussion before I invest too much time.

Ive gotten some results user selcts check box on first page
The php page will say which brackets it falls between example price is between 100-300 say

I dont seem to be able to populate a table with the data in the database:S

Code: [Select]
$row_number= 0;

while ($row = mysql_fetch_array($result3))
    {
If  (($row["price"] = $price_low) && ($row["price"] <= $price_high)) //price
If (($row["storage"] = $storage_low) && ($row["storage"] <= $storage_high)) //storage
if (($row["Processor "] = $processor_low) && ($row["Processor"] <= $processor_high))  //Processor
{ $row_number++;

?>

<tr>

<td align="center"><?php print $row["Computer_Price"]; ?> </td>
<td align="center"><?php print $row["Computer_Storage"]; ?> </td>
<td align="center"><?php print $row["Computer_ProcessorSpeed"]; ?> </td>

</tr>
</table>
Thats wat im usign atm the price_low and price_high are what sets the low and high price for the search

  i think this is more of a HTML problem, but it is in my PHP code so i will ask anyway!
if i have the search code:


 if (count($error) < 1) {
      $searchSQL = "SELECT id, placing, racedate, horseid FROM race WHERE ";
     
      // grab the search types.
      $types = array();
      $types[] = isset($_GET['placing'])?"`placing` LIKE '%{$searchTermDB}%'":'';
      $types[] = isset($_GET['racedate'])?"`racedate` LIKE '%{$searchTermDB}%'":'';
      $types[] = isset($_GET['horseid'])?"`horseid` LIKE '%{$searchTermDB}%'":'';
     
      $types = array_filter($types, "removeEmpty"); // removes any item that was empty (not checked)
     
      if (count($types) < 1)
         $types[] = "`placing` LIKE '%{$searchTermDB}%'"; // use the body as a default search if none are checked
     
          $andOr = isset($_GET['matchall'])?'AND':'OR';
      $searchSQL .= implode(" {$andOr} ", $types) . " GROUP BY `horseid` ORDER BY `racedate`"; // order by title.

      $searchResult = mysql_query($searchSQL) or trigger_error("There was an error.<br/>" . mysql_error() . "<br />SQL Was: {$searchSQL}");
     
      if (mysql_num_rows($searchResult) < 1) {
         $error[] = "The search term provided {$searchTerms} yielded no results.";
      }else {
         $results = array(); // the result array
         $i = 1;
echo '<table border="1"><tr><th>Num</th><th>Placing</th><th>Race Date</th><th>Horse ID</th></tr>';
while ($row = mysql_fetch_assoc($searchResult))
{
   echo "<tr><td>$i</td>
                    <td>{$row['placing']}</td>
                    <td><a href=\"viewrace.php?date=" . urlencode($row['racedate']) . "\">{$row['racedate']}</a></td>
                    <td>{$row['horseid']}</td></tr>";
   $i++;
}
echo '</table>';

how do i get the 'horseid' to appear at the header of the table rather than in a column?

thank you for all the brilliant help this site provides!!

My db is called localDB, and the table in that db is called MonthlySales. I want to display the table in a table in php. But to be able to filter the table by the column year, so if 2000 is selected only the values with 2000 are shown. Is there also a better way to populate the drop down menu?

My script looks like:

Code: [Select]
<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' >
     
    <select name="value">
        <option value="all">All</option>
        <option value="2000">2000</option>
        <option value="2001">2001</option>
    </select>
     
    <input type='submit' value = 'Filter'>

</form>

   
    <?php
    $link = mysql_connect('localhost', 'root', 'root');
    if (!$link) {
    die('Could not connect: ' . mysql_error());
    }

 $db_selected = mysql_select_db('localDB', $link);
    if (!$db_selected) {
        die (mysql_error());
    }

// process form when posted 
if(isset($_POST['value'])) 
{ 
if($_POST['value'] == '2000') 
{  
$query = "SELECT * FROM MonthlySales WHERE Year='2000'";   
}   
elseif($_POST['value'] == '2001') 
{   
$query = "SELECT * FROM MonthlySales WHERE Year='2001'";   
} 
else 
{   
$query = "SELECT * FROM MonthlySales";   
}   
$sql = mysql_query($query);   
 
while ($row = mysql_fetch_array($query))
{  
$Id = $row["Id"];  
$ProductCode = $row["ProductCode"];  
$Month = $row["Month"];  
$Year = $row["Year"]; 
$SalesVolume = $row["SalesVolume"]; 
 
echo "<tr>";
   echo "<td>" . $row['Id'] . "</td>";
   echo "<td>" . $row['ProductCode'] . "</td>";
echo "<td>" . $row['Month'] . "</td>";
echo "<td>" . $row['Year'] . "</td>";
echo "<td>" . $row['SalesVoulme'] . "</td>";
   echo "</tr>";
} 
mysql_close($con);  
} 
?>

Hi guys,

I am working with an old script at the moment, there is one page which just will not populate the table results.
I have tried running multiple debugging commands but the only one it flags is the line displaying Quote

last;

saying it's not a used function.
If I comment out this line, no errors are produced but the results do not enter the table.
Can anyone shed some light on this please, I've spent hours and hours and banging my head against a brick wall would probably be more constructive right now.

Many thanks indeed for any help or advice.

<?php

mysql_connect("localhost", "$db_user","$db_upwd") or die ("Error - Could not connect: " . mysql_error()); 
mysql_select_db("$database");

$query="select host,count(*) from badc_mis_prog group by host";
$result = mysql_query($query) or die ("Error - Query: $query" . mysql_error());

$count=0;
$hosts=array();
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    $html_hlname=$row[0];
    $html_hlname=preg_replace("/</","&lt;",$html_hlname);
    $html_hlname=preg_replace("/>/","&gt;",$html_hlname);
            array_push($hosts, $html_hlname,$row[1],0);
    $count++;
        }

$query="select host,count(*) from badc_mis_prog where reported=1 group by host";
$result = mysql_query($query) or die ("Error - Query: $query" . mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    $html_hlname=$row[0];
    $html_hlname=preg_replace("/</","&lt;",$html_hlname);
    $html_hlname=preg_replace("/>/","&gt;",$html_hlname);
            for ($i=0; $i<($count*3); $i+=3) {
        if ($hosts[$i] == $html_hlname) {
    $hosts[($i+2)]=$row[1];
   last;
}
    }
        }

for ($i=0 ; $i<(($count-1)*3); $i+=3){
    for ($j=$i+3 ; $j<($count*3); $j+=3){
        if ($hosts[($i+1)] < $hosts[($j+1)]){
            $temp=array();
    $temp[0]=$hosts[$i];
    $temp[1]=$hosts[($i+1)];
    $temp[2]=$hosts[($i+2)];
    $hosts[$i]=$hosts[$j];
    $hosts[($i+1)]=$hosts[($j+1)];
    $hosts[($i+2)]=$hosts[($j+2)];
    $hosts[$j]=$temp[0];
    $hosts[($j+1)]=$temp[1];
    $hosts[($j+2)]=$temp[2];
                }
    }
}
print "<br><br><br><center><table border=\"1\">\n";
print "<tr><td>Host Name</td><td>Hosted</td><td>Reported</td><td>Ratio H/R</td></tr>\n";

for ($i=0; $i<($count*3); $i+=3) {
   if ($hosts[($i+1)]<15){ break;}
           printf ("<tr><td>&nbsp; %s &nbsp;</td><td> %d </td><td> %d </td><td>%.1f %%</td></tr>\n",$hosts[$i],$hosts[($i+1)],$hosts[($i+2)],(($hosts[($i+2)]/$hosts[($i+1)])*100));
}

print "</table></center>\n";


?>

Hi

I have the following code:

Code: [Select]
$result = mysql_query("SELECT * FROM xbox_games order by gameid");
$row = mysql_fetch_assoc($result);
$id = $row["gameid"];
$title = $row["gametitle"];
$cover = $row["cover"];

                                <?php echo $title;?>

Which displays only the first result from my database.

How can i change this to display all the results either as a list, or in a table?

Thanks