PHP - Php Not Displaying Mysql Result - So Strange?? Please Help

Hello -

Suppose I had a MySQL result set that was something like this:
ITEM: COLOR: ball red ball blue book red book green book black
If I were to use the code:
Code: [Select]
do {
   echo $row_myQuery['item'] . " - " . echo $row_myQuery['color'] . "<br />";
} while ($row_myQuery= mysql_fetch_assoc($myQuery));

Then I end up with this:

Code: [Select]
ball - red
ball - blue
book - red
book - green
book - black
MY QUESTION:

What I'd like to know is, how could I re-write that php code so that I could get an output like this?:

Code: [Select]
ball - red, blue
book - red, green, black
Thanks!

For some reason the line commented // executes unexectedly does exactly that

Any ideas why?
if ($_POST['user'] == null){
$errors[] = 'An Interviewer ID is Required';
}elseif ($_POST['task'] != 'receipting' || $_POST['task'] != 'dataentry'){
$errors[] = 'Invalid Task'; // EXECUTES UNEXPECTEDLY
}
if (!empty($errors)){
print_LoginForm($errors);
}

Code: [Select]
echo "</b><select name='task'>\n";
echo "<option value='receipting'>Receipting</option>\n";
echo "<option value='dataentry'>Data Entry</option>\n";
echo "</select>\n";

i have it set up so the user picks the year and session and then the database shows all the games in that specific year and session and then displays them with a link to their photo galleria and also shows the record(wins - losses - ties) but the way i have it know its not displaying the first result here is the code
Code: [Select]
<?php
 //declares $y as year played and $s as session
 $y = ' ';
 $s = ' '; 
 
 //handles submition of form for picking year and session
 if (isset($_POST['submit'])) {
 $data = mysql_real_escape_string($_POST['session']);
 $exploeded = explode(",", $data);
 $y = $exploeded[0];
 $s = $exploeded[1];
 }

  //Query for the selection menu
    $squery = "SELECT DISTINCT(Year_Played), Sessions FROM pinkpanther_games";
    $sresults = mysql_query($squery) or die("squery failed ($squery) - " . mysql_error());
   
    //checks $y and $s to see if their is a vaule
    if ($y == ' '){$y = '20112012';}
    if ($s == ' '){$s = '2';}
   
    //Query for getting record and for getting all the appropriate info for displaying the games
    $query = " SELECT (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Tie' ) AS 'Tie', 
         (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Win') AS 'Win',
         (SELECT COUNT(id) FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s AND Win_Loss = 'Loss') AS 'Lose',
         Opponent AS Opponent, Score AS Score, Gallery_no AS Gal, Win_Loss AS Record FROM pinkpanther_games WHERE Year_Played = $y AND Sessions = $s";
    $results = mysql_query($query)  or die("Query failed ($query) - " . mysql_error());
    $row_a = mysql_fetch_assoc($results);   
    //Set record variables
    $wins = $row_a['Win'];
    $loss = $row_a['Lose'];
    $tie = $row_a['Tie'];

    //creates from and selection menu for year and session
echo "<tr><td colspan='2'><form method='post' action=''><select name='session' class='txtbox2'>";
while ($srow = mysql_fetch_assoc($sresults)){
echo "<option value='" . $srow['Year_Played'] . "," . $srow['Sessions'] . "'>" . $srow['Year_Played'] . " Session " . $srow['Sessions'] . "</option>";
}
echo "</select><input type='submit' name='submit' value='Go'  class='txtbox2' /></form></td></tr>";
//displays  record
echo "<tr><td colspan='2'>" .$wins . " - " . $loss . " - " . $tie . "</td></tr>";
echo "<tr></tr>";
//displays games and scores in appropriate year and session
while ($row = mysql_fetch_assoc($results)){
$opp = $row['Opponent'];
$score = $row['Score'];
$gal = $row['Gal'];
$wlt = $row['Record'];
//styles games acording to if win lose or tie
if ($wlt == 'Win'){
echo "<tr><td class='win'>";
echo "<a  href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>";
echo "</td><td class='win'>";
echo   $score . "";
echo "</td></tr>";
}
if ($wlt == 'Loss'){
echo "<tr><td class='loss'>";
echo "<a  href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>";
echo "</td><td class='loss'>";
echo   $score . "";
echo "</td></tr>";
}
if ($wlt == 'Tie'){
echo "<tr><td class='tie'>";
echo "<a  href='galleries.php?g=$gal'" . $gal . " >" . $opp . "</a>";
echo "</td><td class='tie'>";
echo   $score . "";
echo "</td></tr>";
}
}
?>

i cannot figure out why my first result is not displaying properly it successfully shows all results but me first
myurl.com/other.php?id=2 now showing any thing kindly help to solved this problem! 

it seems like problem is in variable $http$ids but dunno how to combine them to get result $Ids always b numeric number

<?php
ini_set ("display_errors", "1"); 
error_reporting(E_ALL);

$http = "myurl.com/other.php?id=" ;

$conn=odbc_connect('greeting','','');
if (!$conn)
  {exit("Connection Failed: " . $conn);}
$sql="SELECT * FROM mytable";
$rs=odbc_exec($conn,$sql);
if (!$rs)
  {exit("Error in SQL");}
echo "<table border= 1><tr>";
echo "<th>ID</th>";
echo "<th>Like</th>";
echo "<th>title</th></tr>";
while (odbc_fetch_row($rs))
  {
  $ids=odbc_result($rs,"ID");
  $title=odbc_result($rs,"title");
  echo "<tr><td>$ids</td>";
  echo "<td><fb:like href=\"$http$ids\" layout=\"button_count\"  show_faces=\"false\" width=\"100\" font=\"tahoma\" colorscheme=\"dark\"></fb:like> </td>";
  echo "<td>$title</td></tr>";
  }
echo "</table>";
odbc_close($conn);
?>

Hi!

I want the image of my site to be clickable so that it shows the next image contained in the result set of the sql query.

Code: [Select]
<?php
require_once('includes/connection.inc.php');
$conn = dbConnect();
$sql = 'SELECT filename FROM images ORDER BY image_id DESC LIMIT 1';
$msg = '';
if(!$sql) { echo "Something went wrong with the query. Please try again."; }
$result = $conn->query($sql) or die(mysqli_error());
if(mysqli_num_rows($result) == 0) {
header('Location: empty_blog.php');
$conn->close();
} else {
$row = $result->fetch_row();
$image = $row[0];
}
?>
<?php include('header.php'); ?>
<div class="content main">
<img id="image" src="images/<?php echo $image; ?>"/>
</div>
<?php include('includes/footer.inc.php'); ?>

All the filenames of the images are contained correctly in the $result variable, all I need to do now is to fetch the next image in the set. How would I go about this? Any help is greatly appreciated!