PHP - Create Array Using Sql Query Result Set

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!

hello
I'm using this code:
Code: [Select]
$query="SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1";to query first row data when I want see it and echo it I recived :
Code: [Select]
Resource id #2can anyone help me ?


thank you

Hello, I am new to the board and PHP. I created a simple store locator app with PHP. It pulls the info from a database. I want to  take the website field from the table, once it is returned from the query, and have it print out a a hyperlink.
Please help.

Thanks, I have posted most of the relevant code.


<?php   if (isset ($stores)) {     if (!empty ($stores)) {       
   echo '<p><strong>' . count ($stores) . ' results were found.</strong></p>';     foreach ($stores as $value) {         
   echo '<p><strong>' . $value['name'] . '</strong><br />';       
   echo $value['address'] . '<br />';       
   echo $value['town'] . ', ' . $value['state'] . ' ' . $value['postal'];     
   echo '&nbsp;<a target="_blank" href="http://maps.google.com/maps?q=',
            $value['address'], ' ', $value['town'], ', ', $value['state'], ' ',
            $value['postal'],'">Map this location</a><br />';       
   echo 'Phone: ' . $value['phone'] . '<br />';
   echo 'Website: ' . $value['website'] .  '<br />'; 
   echo 'Email: ' . $value['email'] . '<br />';   
   echo 'Hours: ' . $value['hours'] ;     
   echo '</p>';       }     }
      else {     
   echo '<p><strong>No results found</strong></p>';   }   }   
   ?>
   

Hello.  This one is driving me up the wall.  It used to work,

I have a sports league database (MsSql) and one of the tables contains team Captain's first name, last name, etc..  To create a unique but simple login, the query searches for the captain's first initial of their first name + lastname + league.  However since there is the remote possibility of two similar usernames (i.e. David Smith) if the last name is found, the query adds an integer to the end of the last name.

So the result would look like this:  DSmithM4A

At the moment, my database is completely blank.  However what is happening when I run the query below is I get: DSmith1M4A

Can someone figure out why this code is not working correctly?  If there is not more than one "Smith" no integer is supposed to be returned.  There is something in the "Captain Count" that I think is not working correctly.

Code: [Select]
    /*
     * Gather captian information
     */
    $strCaptainFirstInitial = substr($arrPost['cpt_first'],0,1);
    $strCaptainLastName = preg_replace("/[^a-zA-Z]/","",$arrPost['cpt_last']);
    $strCaptainName = strtolower($strCaptainFirstInitial.$strCaptainLastName);
    $sqlCaptainSearch = sprintf
    (
        $strCaptainSearch,
        $strCaptainName
    );
    $resCaptainSearch = mssql_query($sqlCaptainSearch,$conDB);
 
    /*
     * Create unique captain login and Password
     */
    $strCaptainCount = (mssql_num_rows($resCaptainSearch) > 0)
        ? mssql_num_rows($resCaptainSearch)
        : "" ;
    $strLeagueTypeInitial = substr($arrLeagueSearch['type'],0,1);
    $strCaptainLogin =
        $strCaptainName .
        $strCaptainCount .
        $strLeagueTypeInitial .
        $arrLeagueSearch['size'] .
        $arrLeagueSearch['division'];

Hello!

I am busy developing a script that returns all users that have the cell "Available" set to 1
but I am unsure how to do this, I am familiar with basic SQL and I have made this:

$sql = "SELECT * FROM `clans` WHERE `available` =1 LIMIT 0 , 30";
if ($result = mysql_query($sql)) {
if (mysql_num_rows($result)) { 
$row = mysql_fetch_assoc($result);
echo ($row['name']);

but it only displays information from the first row found, and I want it to display all the rows found, how can I do this?
Thanks

- GreenFanta

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=306966.0

I have this query to show the current class I'm in:
Code: [Select]
$GetTimeTable = "SELECT * FROM timetable WHERE username='$username' AND Subject_date='$today_date'";
Then I have this to display current class:
Code: [Select]
if($TimeNow >= $Start_time && $TimeNow <= $End_time){
echo "Current Class: ";
echo $Subject;}
How would I find next class that day?

Hi all,

Just a pretty basic question that I can't find the asnwer to:

I have a MySQL query of the following form:
Code: [Select]
$result = mysql_query (" SELECT $cols FROM tablename   WHERE.......");
Where $cols is as follows:
Code: [Select]
$cols = firstname, lastname, $price1, $price2, etc.
($price1 and $price2 are variable column names that contains prices)

How do I find the lowest value for $price1 and/or $price2 out of the results set of this query?

Thanks!

soo i get info from a form and i want to search for it in my db .. if i find user_name AND user_password the $result returns true .. otherwise false ..

<?php require_once("../includes/connection.php"); ?>
<?php
$user_name=$_POST["user_name"];
$user_password=$_POST["user_password"];
?>
<?php
$query = "SELECT *
          FROM users
          WHERE  '{$user_name}' = user_name
          AND '{$user_password}'= user_password
          ";
$result = mysql_query($query,$connection);
if($result==false){ 
redirect("../index.php");}
elseif($result==true){
redirect("../welcome.php");
}
?>
<?php //close connection!
mysql_close($connection);
?>


but the problem is that it always returns TRUE !!!!
whats the problem here exactly ???

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.

I´ve reached to

    <?php require_once('Connections/conexxion.php'); ?>
    <?php
    mysql_select_db($database_conexxion, $conexxion);
    $query_consulta = "SELECT venta,compra,taller,regula_mas,regula_menos,movimiento.id_item,sum(compra+regula_mas-venta-taller-regula_menos)  as stock, cilindro, esfera FROM movimiento  join item on item.id_item=movimiento.id_item join rx on rx.id_rx=item.id_rx join cilindro  on cilindro.id_cil=rx.id_cil  join esfera  on esfera.id_esf=rx.id_esf GROUP BY movimiento.id_item ORDER BY esfera desc, cilindro desc";
    $consulta = mysql_query($query_consulta, $conexxion) or die(mysql_error());
    $row_consulta = mysql_fetch_assoc($consulta);
    $totalRows_consulta = mysql_num_rows($consulta);
    ?>
    <!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" />
    <title>test</title>
    </head>
     
    <body>
     
       <?php
    echo "<table align=center>";
     
    $columnes = 10; # Número de columnas (variable)
     
    if (($rows=mysql_num_rows($consulta))==0) {
      echo "<tr><td colspan=$columnes>No hay resultados en la BD.</td></tr> ";
    } else {
      echo "<tr><td colspan=$columnes>$rows Resultados </td></tr>";
    }
     
    for ($i=1; $row = mysql_fetch_row ($consulta); $i++) {
    $resto = ($i % $columnes); # Número de celda del <tr> en que nos encontramos
    if ($resto == 1) {echo "<tr>";} # Si es la primera celda, abrimos <tr>
       echo "<td>$row[1]</td>";
    if ($resto == 0) {echo "</tr>";} # Si es la última celda, cerramos </tr>
    }
    if ($resto <> 0) { # Si el resultado no es múltiple de $columnes acabamos de rellenar los huecos
    $ajust = $columnes - $resto; # Número de huecos necesarios
    for ($j = 0; $j < $ajust; $j++) {echo "<td>&nbsp;</td>";}
    echo "</tr>"; # Cerramos la última línea </tr>
    }
    mysql_close($connexion);
    echo "</table>";
    ?>
     
     
     
    </body>
    </html>
    <?php
    mysql_free_result($consulta);
    ?>

it "works", but results doesn´t start at number 1 but number 2   anyways, The intent of this table is to show the stock of a product, a lens, that has a range of Rx, we use to see this range this way, and show the stock this way would be very productive, but I can´t figure out how to achive it, any sugestion??  
Edited by gralfitox, 16 December 2014 - 06:45 AM.

I have a table korisnici in SQLite with INTEGER field aktivan that can have only 0 or 1 value (CHECK constraint).

Field aktivan has value 0 but PHP returns value 1, why? Is this a bug?

This is PHP code that I am running:

 $sql = "SELECT ime, aktivan FROM korisnici WHERE lower(ime) = '" . $ime . "'" . " AND sifra = '" . $_POST["sifra"] . "'";
                
$result = $db->query($sql);
$row = $result->fetchArray(SQLITE3_ASSOC);

$row['aktivan'] = 1 but in table the value is 0.

When I run same query in DB Browser for SQLite I get correct value 0.

Is this a bug?

This topic has been moved to CSS Help.

http://www.phpfreaks.com/forums/index.php?topic=353608.0