PHP - Assigning An Id And Showing Results By A Group?

Does anybody know how to show the below php results in a table format?

<?php

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

$fname = $_POST['fname'];
$lname = $_POST['lname'];
$skill = $_POST['skill'];
   
//connect  to the database
$db=mysql_connect  ("127.0.0.1", "root",  "") or die ('I cannot connect to the database  because: ' . mysql_error());
   
//select  the database to use
 $mydb=mysql_select_db("resource matrix");
   

//query  the database table
$sql="SELECT DISTINCT First_Name, Last_Name, l.Resource_ID FROM ((resource l inner join resource_skill ln on l.Resource_ID = ln.Resource_ID) inner join skill n on ln.Skill_ID = n.Skill_ID) WHERE First_Name LIKE '$fname' OR Last_Name LIKE '$lname' OR Skill_Name LIKE '$skill'";
   
//run  the query against the mysql query function
$result=mysql_query($sql);
   
//create  while loop and loop through result set
while($row=mysql_fetch_array($result)){
$First_Name  =$row['First_Name'];
$Last_Name=$row['Last_Name'];
$Resource_ID=$row['Resource_ID'];
   
//display the result of the array
echo "<ul>\n";
echo "<li>" . "<a  href=\"a.php?id=$Resource_ID\">"   .$First_Name . " " . $Last_Name .  "</a></li>\n";
echo "</ul>";
}
}
else{
echo  "<p>Please enter a search query</p>";

}
}

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

So I have 2 queries, that has the potential to return alot of data, foreach loops running. The first shows each group heading, and the foreach nested is calling another query specific to the group heading. The result currently is a lengthy delay in results showing. (Now this may/may not be the most ideal code practise in this instance however... short term solution discussion please).

For Each State
   Read Each States Details from DB
   For Each State Listing
      Display Details
   Next
Next

Is there a command or other where I can say, Display HTML page as it stands while it continues to process the FOREACH loops?

EG:

For Each State
   Read Each States Details from DB
   For Each State Listing
      Display Details
      REFRESH HTML DISPLAYED
   Next
Next

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.

I was wondering how does one go about showing results from SELECT query in columns in a html table.

I have a list of products in a table, and would like to show them on the page in 4 columns. I have done many searches on google to try and find the sulution, but the majority of what im finding instead is about displaying a table from phpmyadmin as a table in html.

If its a large operation to do this, I would be very happy if someone could poiint me in the direction of a tutorial maybe.

Here is the code I have so far to display the products, but for some reason, it only show 1 row instead of all the rows from my table.

Code: [Select]
<?php

$dbhost = "localhost";
$dbuser = "user";
$dbpass = "pass";
$dbname = "dbname";

mysql_connect ($dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

$result = mysql_query("SELECT * FROM mcproducts");
while($row = mysql_fetch_array($result))
 {
  $products_local_id = $row['products_local_id'];
  $productname = $row['product_name'];
  $thumburl = $row['image_from_url'];
  $productlink = $row['product_local_url'];
  $thumbnail = $row['product_image_small'];
  $currencysymbol = $row['product_currency'];
  $price = $row['product_price'];
  $flagicon = $row['product_country_from'];

 }

?>
<html>
<head>
<link href="style/stylesheet.css" rel="stylesheet" type="text/css" />
</head>
<body>

<div class="displaybox">
                <div class="productimage">
                <a href="<?php echo $productlink; ?><?php echo $products_local_id; ?>"><img src="<?php echo $thumburl; ?><?php echo $thumbnail ?>" width="150" height="150"></a>
                </div>
                <div class="productdescription">
                  <div class="pro_name">
      <a href="<?php echo $productlink ?><?php echo $products_local_id; ?>"><?php echo $productname; ?></a>
                  </div>
                  <div class="pro_description">
                  </div>
                  <?php
                  if ($flagicon=="Ireland") {
  $flagicon = "<img src=\"flags/ireland.jpg\">";
  }
  
  elseif ($flagicon=="UK") {
  $flagicon = "<img src=\"flags/uk.jpg\">";
  }  else
  echo "";
  ?>
                  <div class="pro_description"><?php echo $flagicon; ?><?php echo $currencysymbol ?> <?php echo $price ?></div>
                </div>
              </div>
             
</body>
</html>


Many thanks,

DB

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

I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database.  I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window.  I do not know how to correct this.  I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page.  Suggestions on what I need to look for to correct would be great.  I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. 
Thank you for any assistance I may get!!  - Amy -
 
Code: [Select]
<body><font color="006600">
 <div style="background-color:#f9f9dd;">
  <fieldset>
<h1>Asset Entry Results</h1>
<?php
  // create short variable names
  $tag=$_POST['tag'];
  $serial=$_POST['serial'];
  $category=$_POST['category'];
  $status=$_POST['status'];
  $branch=$_POST['branch'];
  $comments=$_POST['comments'];

  if (!$tag || !$serial || !$category || !$status || !$branch) {
     echo "You have not entered all the required details.<br />"
          ."Please go back and try again.";
     exit;
  }

  if (!get_magic_quotes_gpc()) {
    $tag = addslashes($tag);
    $serial = addslashes($serial);
    $category = addslashes($category);
    $status = addslashes($status);
    $branch = addslashes($branch);
    $comments = addslashes($comments);
  }

  //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory');
  $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database");

  $query = "insert into assets values
            ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')";
  $result = $db->query($query);

  if ($result) {
      echo  $db->affected_rows." asset inserted into Inventory.";
  } else {
     echo "An error has occurred.  The item was not added.";
  }

  $db->close();
?>
  </fieldset>
 </div>
</body>

Hi all,

I can't seem to designate an array key by using a variable and I was wondering if this is possible. I'm looking to do something like this:

Code: [Select]

<?php
$key = "apple";
$arr = array($key => "fruit");
?>
Any suggestions would be appreciated!

I need to assign unknown array keys and their values to strings.

The code I am using is:
Code: [Select]
$cart = $_SESSION['cart'];
$items = explode(',',$cart);
$qty = array();


foreach ($items as $item) {
$qty[$item] = $qty[$item] + 1;
}
print_r($qty);
The keys change and are not ordered. I do not know what the key or it's value may be.  So as an example, I might have:
Code: [Select]
Array ( [2] => 5 [4] => 8 ) or in a different scenario, I might have:
Code: [Select]
Array ( [1] => 6 )
What those numbers represent is a product id (the key), and the quantity (the value of the key).

Can someone help me to assign these to a string?  I have been reading all morning, and not making any progress. 

Thank you!

hi there,
i'm trying to put a message in the footer of a page which welcomes a person who is logged in with his/her name, using sessions of course; when i place this:
Code: [Select]
$username = $_SESSION['valid_user'];
in the footer before the echo:
Code: [Select]
echo "You are logged in as $username";
but the session is also needed before the footer to use the username for other things , such as -checking his credit- so if place in footer the footer shows name in browser but checking credit would not happen as the assignment is at the buttom.

IF i place the assignment above at the top of the file:

everything works for the user and checking credit ..etc...but the footer is not there...

cud not put this any clearer..sorry...hope if someone cud help...thanks

I just have a quick question, what happens if I don't assign a visibility element to methods?

In other words, if I just have:

[php]
class Article_model
{
   public $data;

   function index()
   {
      return $data;
   }
[php]
...instead of declaring public, private, static, and so on before the function. Will php just consider it public?

Hey everyone. I have been trying to assign dates to image so I could sort them, and also display the date they were uploaded. But I need some help. I have a script that uploads an image to a file directory, and also stores the file directory and the upload date in a database. And I also have a script that displays the images in the file directory but I cannot find a way to assign the right date to the right image.
Here is the script that displays the images:

Code: (php) [Select]
if (isset($_POST['submit'])) {
// fill an array with all items from a directory
$dir = "thumbs/";

$handle = opendir($dir);
while ($file = readdir($handle)) {
$files[] = $file;

// This removes the full stops from the array
unset($files[0]);
unset($files[1]);
               
}   
            foreach($files as $fileimg)
{

echo "<img src='". $dir . $fileimg ."'/>\n";
                echo "$date";
}
       
}
and here is the script for calling the date from the database:

Code: [Select]
$query = mysql_query("SELECT imgaddress, uploaddate FROM thumbs WHERE imgaddress='" . $dir . $file . "'");

                if (mysql_num_rows($query) == 1) {
                     $row = mysql_fetch_assoc($query);

                     $date = $row['uploaddate'];
If I use the database statement before the foreach, then it does not get the right file directory, but if I call it inside the foreach, then I cannot use krsort() to sort the array.

Any help would be greatly appreciated, thanks