PHP - Proper Displaying Of A Mysql Query Result Set

Hi Everyone.

I get the following data from a result from database search

<?php echo $rsjobs['CompanyURL'];?>  It prints out the url but when I try to turn it into a clickable link it does not work

Here is the coding I have so far.


<a href="<?php echo $rsjobs['CompanyURL'];?>"><?php echo $rsjobs['CompanyURL'];?></a>


 What am I doing wrong ? please help.

$sql = "SELECT option_id, id FROM bets WHERE win_status = '1' AND updated_at BETWEEN '$today 13:00:00.000000' AND '$tmrw 08:00:00.000000'";
$result = mysqli_query($connt, $sql);

foreach($result as $value)
{
	$value[] = arary_multisort();
	$oid = $value['option_id'];
	$gid = $value['id'];
     var_dump($value);
     //$gim = var_export($value);
}


foreach($result as $key => $value) {

	$info = array($value);

}
mysqli_close($connt);
$fl = array($oid, $gid);
print_r($fl);

array (
  'option_id' => 'game1a',
  'id' => '155670450',
)array (
  'option_id' => 'game2h',
  'id' => '386887537',

Either I can return 1 row and have the data sorted nicely or I can return an array but I can only seem to access the last array. Been stuck on this bit for atleast 2 hrs sadly. How can I rename the array? Some days I might have 100 results. Today we have two winners xP I really need to ctrl+z sublime cuz I tried what feels like everything. Ancy to move to the next part of this code MATH if you can kindly assist.

Yall have been extremely helpful in my last two posts 2/2 going on 3. Is there a place I can properly contribute lite silver :3

Hi guys, 

I'm starting to get back into coding for a hobby and wondering is there a better way of doing these php and mysql calls? they all work but not sure if it was the efficent way or is there others that everyone else uses?

 

//connection for user details lookup
$sql0 = "SELECT uid, fullname, emailaddress, created_at FROM MK_users WHERE uid={$_SESSION['id']}";
$result0 = $conn->query($sql0);

//connection for user against role acceess
$sql2 = "SELECT role_uid, role_bid, role_access, access_created_at FROM MK_role_access WHERE role_uid={$_SESSION['id']}";
$result2 = $conn->query($sql2);

 while($row2 = $result2->fetch_assoc()) {
    $_SESSION['role_bid'] = $row2["role_bid"];
  }

//connection for role access against business lookup
$sql3 = "SELECT businessname, account_type, website, main_address, logo FROM MK_baccounts WHERE bid={$_SESSION['role_bid']}";
$result3 = $conn->query($sql3);

while($row3 = $result3->fetch_assoc()) {
  
    $_SESSION['businessname'] = $row3["businessname"];
  }
 

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>";
}
}
?>

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!

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 am building a product search engine using MySql and PHP (Jquery / AJAX).
As i am getting the search results i wondered is it possible to count the instances of a category?
For example:
If i search for "Dell Laptops" i get 400+ results back because it has lots of laptops and accessories.
What i am aiming for it to do is while looping through look at the categories and if its the first instanc eof it i want it to begin a count. until it has finished looping and then i can display it.
Results: 436 [ Laptops(127) - Accesories (244) - Components(65)  ]

Im not even sure where to start here so any advice will be helpful

Here is the code im using to generate the search:
   $id = str_replace(" ", "%", "$id");
   $q = "SELECT * FROM products WHERE prod_name LIKE '%$id%' order by fee asc LIMIT 20";

Hello all,   Am trying to calculate some data and display the result of the current month in a chart form. i just don't know how to get the current month. this is not working..

<?php
include('mysql_connect.php');

$date = 'MONTH(CURRENT_DATE())';
$status = 'paid';
//mysql_select_db("hyprops", $con);

$query = mysql_query("SELECT sum(amount) 'amount',  department FROM requisition WHERE status = '$status' AND  date='$date' ");

$category = array();
$category['name'] = 'department';

//$series1 = array();
//$series1['name'] = 'actual';

$series2 = array();
$series2['name'] = 'amount';

//$series3 = array();
//$series3['name'] = 'Highcharts';


while($r = mysql_fetch_array($query)) {
    $category['data'][] = $r['department'];
   // $series1['data'][] = $r['actual'];
    $series2['data'][] = $r['amount'];
  //  $series3['data'][] = $r['highcharts'];   
}

$result = array();
array_push($result,$category);
//array_push($result,$series1);
array_push($result,$series2);
//array_push($result,$series3);


print json_encode($result, JSON_NUMERIC_CHECK);

mysql_close($con);
?> 

Please how can i solve this issue?   Thanks in advance

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=330221.0

Hello all.   I am using a premade login script which has some very powerful function built in, the one I am looking at using is as follows:

  /** 
    * query - Performs the given query on the database and 
    * returns the result, which may be false, true or a 
    * resource identifier. 
    */ 
   function query($query){ 
      return mysqli_query( $this->connection, $query); 
   } 

The issue is, displaying the results.  I call is via the following code:

$database->query('SELECT LastName, Firstname FROM Patrons WHERE LastName = "Delude"');

But I have no idea how to display it.  I have attempted to echo it out, but I get the following error in the logs:   PHP Catchable fatal error:  Object of class mysqli_result could not be converted to string    

Hi all !,   I am stuck on the following piece of code which does not give an error nor does it give a result. ( i.e. it gives 0 num_rows which should be > 1).    If, however, I execute the query in phpmyadmin by simply substituting the values of $pp,$ll and $room_no in the query it gives the correct result.      Please can someone tell me what I may be doing wrong here.  Thanks !

	$fcon = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');

				$pp = "(ms.level = 'Beginner' || ms.level = 'Intermediate')";	  
				$ll = 'ms.diff <= 7';
				$room_no = 4;

				$query = "SELECT md.Member_reg_id, md.fname, md.lname, md.email, md.cell, 
						ms.level, ms.diff, ms.score,  
						r.ID_Status 
						FROM register as r
						JOIN member_detail as md ON r.ID = md.Member_reg_id  
						JOIN memstatus as ms On r.ID = ms.ID  
						WHERE r.CENTERCODE = ? AND r.ID_Status ='A' AND ? AND ?
						ORDER by level, diff, score DESC";
		
				$stmt=$fcon->prepare($query);
				$stmt->bind_param('iss',$room_no,$pp,$ll);
				if(!$stmt->execute()) die('Failed to execute the query'.$fcon->error);
				else
				{
					echo "Executed";
					$stmt->bind_result($Member_reg_id,$fname,$lname,$email,$cell,$level,$diff                                               ,$score,$ID_Status);
					$numrows = $stmt->num_rows;
					$stmt->store_result();
//					echo $numrows;
					while($stmt->fetch())
					{
						echo "<br>".$fname.' '.$lname;
						echo "<br>".$level;
						echo "<br>".$diff;
						echo "<br>".$score;
						echo "<br>".$cell;
						echo "<br>".$email;
					}
				}

Edited by ajoo, 03 January 2015 - 08:00 AM.

Hi everyone. I have this code that suppose to divide the query result 3 columns in a row. Like this :
Shoes   Hats   Jeans
Shorts  Shirts  Belts
But for some reason this one doesn't work. It gave me everything in a single row. Like this: Shoes Hats jeans Shorts Shirts Belts. I noticed that before I put this two 2 lines of code, it's working fine. It didn't give me any error messages or anythings. Can someone point me out what did I did wrong?

This is two lines I put in:
Code: [Select]
$vuong_query = tep_db_query("select categories_id from " . TABLE_GENDER_CATEGORIES . " where parent_id = 1");
 while ($product_check = tep_db_fetch_array($vuong_query)) {
This is the whole piece of code:
    Code: [Select]
  <tr>
    <td><table width = "100%" border="1" bordercolor="maroon">
      <tr>
<?php
$vuong_query = tep_db_query("select categories_id from " . TABLE_GENDER_CATEGORIES . " where parent_id = 1");
 while ($product_check = tep_db_fetch_array($vuong_query)) {

$categories_querys = tep_db_query("select c.categories_id, cd.categories_name, c.categories_image from " . TABLE_CATEGORIES . " c, " . TABLE_CATEGORIES_DESCRIPTION . " cd where c.categories_id = '" . $product_check['categories_id'] . "' and c.categories_id = cd.categories_id and cd.language_id = '" . (int)$languages_id . "' order by sort_order, cd.categories_name");
    

    $number_of_categories = tep_db_num_rows($categories_querys);

    $rows = 0;
    while ($categories = tep_db_fetch_array($categories_querys)) {
      $rows++;
      $cPath_new = tep_get_path($categories['categories_id']);
      $width = (int)(100 / MAX_DISPLAY_CATEGORIES_PER_ROW) . '%';
      echo '         <td align="center" class="smallText" width="' . $width . '" valign="top"><a href="' . tep_href_link(FILENAME_DEFAULT, $cPath_new) . '">' . tep_image(DIR_WS_IMAGES . $categories['categories_image'], $categories['categories_name'], SUBCATEGORY_IMAGE_WIDTH, SUBCATEGORY_IMAGE_HEIGHT) . '<br>' . $categories['categories_name'] . '</a></td>' . "\n";
      if ((($rows / MAX_DISPLAY_CATEGORIES_PER_ROW) == floor($rows / MAX_DISPLAY_CATEGORIES_PER_ROW)) && ($rows != $number_of_categories)) {
echo '       </tr>' . "\n";
echo '       <tr>' . "\n";
      }
    }
}
    ?>
      </tr>
            </table>
          </td>
        </tr>
Thanks anyone for helping me out.
Vanvoquan

Hello,
So I have a content management site set up, but I'm stuck on one part. I have a query. Code: [Select]
Select `email` FROM author WHERE author.id IN(SELECT authorid FROM job WHERE id = '$_POST[id]')
I'm a beginner in PHP and SQL. Basically I need the query to run, it will come out with one result and I need that to be where the email is sent to.

This is the existing code I have for sending out the email $subject = " Job Application Confirmation";

$message = "Hello $_POST[name] , \r\rYou, or someone using your email address, has applied for the job: $text Using the Resource Locator at Prahan.com. Keep this as future reference. \r\r Full Name: $_POST[name] \r Location: $_POST[location] \r Email: $_POST[email] \r Additional Information: $_POST[info] \r Job ID: $id  \r Job Name: $text  \r \r Expect a response shortly. \r\r Regards, Prahan.com Team";

$headers = 'From: Prahan.com Team <noreply@prahan.com>';

    'Reply-To: noreply@prahan.com' . "\r\n" .

    'X-Mailer: PHP/' . phpversion();

$to1      = 'pratik@prahan.com';

mail($to1, $result, $message, $headers);


Thanks in Advance!

Hi there I am new to this forum and not much of a php expert. I hope you can assist me. My problem is as follows: I have a column consisting of several values (in this case: sceience, art, humanistics and music) I would like to use the following mysqli query in order to count the appearances of each value in this column:     $query = "SELECT SUM (IF(department = 'science', 1, 0)) AS science, SUM(IF(department = 'art', 1, 0))-> AS art FROM new2"   I have tried lots of ways to loop the results but couldn't get the it working. Can anyone suggest any idea?   Thanks   Hanan