PHP - Storing Variable From Sql Query

Hi there,

I am working on a PHP web form and I have a very simple situation I guess, I have a variable named: $MyVal

When I do

print_r($MyVal);
To see whats inside, I get:

SimpleXMLElement Object ( [0] => 8.23 )
Now I am assigning this variable into a session variable so that I can do calculations with it.

So I assign:

$_SESSION['MySesVal'] = $MyVal;
But after assigning to session variable, when I do my calculations:

$finalValue = $_SESSION['MySesVal'] * 4;

I get 0. So is it because the actual $MyVal variable has some XML stuff as:

SimpleXMLElement Object ( [0] => 8.23 )

So what is the right way to properly assign $MyVal variable to a session variable to do calculations.

Please reply.

All comments and feedbacks are always welcome.

Thank you!

I have a submit.php file that includes the following jQuery code:

<script> [...] request.done(function( json ) { jQuery('input[name=thumbnail-url]').val(json.thumbnail_url); jQuery('input[name=job_title]').val(json.title); jQuery('textarea[name=htmlcode]').val(json.html); [...] </script>

I need to remove

jQuery('textarea[name=htmlcode]').val(json.html);

and pass the "json.html" value into a PHP variable in another php file (functions.php).

The code in the functions.php file is already there.

The thing i need to do (and i am seriously struggling with it) is placing the value of json.html into a $phpvariable that i can then call from the function.php file.

Do you need any more info for this issue?

Thanks.

okay i have table named ZAMJENE with 2 rows only row that i need to setup is named id_zamjena and have values 15 and 16

when i do echo on page it work great but when i press button i get only 16 for both entries

full cde is too long so i paste only important parts

$query_zamjene = mysql_query ("SELECT * FROM zamjene  WHERE id_user_1='$moj_id' AND zamjenjeno='0' ORDER BY id_event_1");

echo "<form method='post' action='event_zamjenjen.php'>";


while ($rows = mysql_fetch_array($query_zamjene))
{
$id_zamjena = $rows['id_zamjena'];
$id_event_1 = $rows['id_event_1'];
$id_event_2 = $rows['id_event_2'];
$id_user_2 = $rows['id_user_2'];

echo "$id_zamjena";   // THIS ROW WORKS CORRECTLY
echo "<input name='id_zamjena' type='hidden' value=$id_zamjena>";  // output of this is always 16

echo "<input name='response' type='submit' value='Prihvati zamjenu' />
<input name='response' type='submit' value='Odbaci zamjenu' />
<br>";

}
echo"</form>";

 

$id = '9';

$MyQuery = "SELECT * FROM bookings WHERE id = '$id'";
$retrieve = mysql_query($MyQuery) or die(mysql_error());
if(mysql_num_rows($retrieve) != 0):
$row = mysql_fetch_assoc($retrieve);
else:
echo 'No Information';
endif;

echo 'Full Name: ' . ($row['fullname']) . '<br><br>';

The above code is working for me however when I change it to what I do...

$id = ($row['id']);

It tells me I have an undefined row variable... What will fix this? Thanks in advance...

I have this variable that I need to get into my database and I just can't think through it. Here is the code.


$sql = "INSERT INTO names (id, name, color, uploaddate) VALUE (NULL,'$master[2]['dbid']','$color','$time')";


I know I need to escape those quotes around dbid somehow, but I can't seam to get it figured out.  If I have to I will set it = to a different variable, but I don't want to do it that way if I don't have to

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

This does not make sense to me, I hope someone can tell me why this query don't work:

Code: [Select]
safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$postdate', $select_map $select_date challenged_info='".$_POST['info']."', status='2' WHERE chalID='".$_GET['challID']."'");;
and this query does work:

Code: [Select]
safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$postdate', map1='".$_POST['map1']."', map2='".$_POST['map2']."', map3='".$_POST['map3']."', date1='".$_POST['date1']."', date2='".$_POST['date2']."', date3='".$_POST['date3']."', date4='".$_POST['date4']."', challenged_info='".$_POST['info']."', status='2' WHERE chalID='".$_GET['challID']."'");

Variables in $select_map and $select_date makes the query fail but when I copy/paste this (bold) in query it works:

$select_map = map1='".$_POST['map1']."', map2='".$_POST['map2']."', map3='".$_POST['map3']."',
$select date = date1='".$_POST['date1']."', date2='".$_POST['date2']."', date3='".$_POST['date3']."', date4='".$_POST['date4']."',

am I using the variable in query incorrectly?

i have a small search profile search and need to set $getuname variable to users username

if((trim($country) !== ''&&($state) !== ''&&($city) !== '')){


   $data = mysql_query("SELECT    *  FROM users WHERE users.state='$state' AND users.country='$Country' AND users.city='$city' ORDER BY RAND() DESC LIMIT 1 ");   
Echo "Country city state search";
$getuname = mysql_real_escape_string($_GET['username']);

am i meant to use the get or is that only for forms?

Hi there,
im trying to have a form show up when user clicks "add joke". I need the variable to be retrieved from the url query string. I cant get the form to show up.  I think its either an issue with the GET function at the top or the link down at the bottom. Please help!


<?php
  // If the user wants to add a joke
  
    $_GET['addjoke'] = $addjoke;

  if (isset($addjoke)):
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Type your joke he <BR>
<TEXTAREA NAME="joketext" ROWS=10 COLS=40 WRAP>
</TEXTAREA><BR>
<INPUT TYPE=SUBMIT NAME="submitjoke" VALUE="SUBMIT">
</FORM>

<?php
  else:

    // Connect to the database server
    $dbcnx = @mysql_connect("servername", "username", "password");
    if (!$dbcnx) {
      echo( "<P>Unable to connect to the " .
            "database server at this time.</P>" );
      exit();
    }

    // Select the jokes database
    if (! @mysql_select_db("jhodara2") ) {
      echo( "<P>Unable to locate the joke " .
            "database at this time.</P>" );
      exit();
    }

    // If a joke has been submitted,
    // add it to the database.

$joketext = $_POST['joketext']; 
$submitjoke = $_POST['submitjoke']; 

    if ("SUBMIT" == $submitjoke) {
      $sql = "INSERT INTO jokes SET " .
             "JokeText='$joketext', " .
             "JokeDate=CURDATE()";
      if (mysql_query($sql)) {
        echo("<P>Your joke has been added.</P>");
      } else {
        echo("<P>Error adding submitted joke: " .
             mysql_error() . "</P>");
      }
    }
  
    echo("<P> Here are all the jokes " .
         "in our database: </P>");
  
    // Request the text of all the jokes
    $result = mysql_query(
              "SELECT JokeText FROM jokes");
    if (!$result) {
      echo("<P>Error performing query: " .
           mysql_error() . "</P>");
      exit();
    }
  
    // Display the text of each joke in a paragraph
    while ( $row = mysql_fetch_array($result) ) {
      echo("<P>" . $row["JokeText"] . "</P>");
    }
  
    // When clicked, this link will load this page
    // with the joke submission form displayed.
    echo("<P><A HREF='$PHP_SELF?addjoke=1'>Add a Joke!</A></P>");
  
  endif;
  
?>


see the problem live at http://www.freewaycreative.com/insert2.php

I got my search page error and the server pop up below message:   Notice: Undefined variable: query in /home/tz005/public_html/COMP1687/search.php on line 64
Minimum length is 3   Where should I make a correction in the script? Here is my php script:   <?php  

I have a template page that opens with a GO click from a drop-down in a different page. The variable is being passed through to the URL and in the GET statement. However, my PHP code is producing the following error: Quote

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /data/23/2/100/53/2263053/user/2487001/htdocs/myalaskacenter/events/venues/test_venues-results.php on line 52

.
Here is my code: Code: [Select]
<?php
//Assigns the venueCode tothe variable with a more convenient name
$venueURL = $_GET['VenueCode'];
@$DB = mysqli_connect('server','username','password','database');
if (mysqli_connect_errno()){
echo '</blockquote><br/><br /> &nbsp; &nbsp; &nbsp; Sorry, this webpage is temporarily unavailable.<br /> &nbsp; &nbsp; &nbsp; <a href="http://alaskapac.centertix.net/">Click here to search for events.</a>';
} else {
 ?>
<?php
   $Query = "SELECT Events.EventTitle, DATE_FORMAT(Performance.startDateTime, '%W, %M %e, %Y') AS startDate, DATE_FORMAT(Performance.startDateTime, '%h:%i %p') AS startTime, Events.EventID, Events.thumb, Events.ShoWareEventLink, Events.tagline, Performance.category_id, Promoters.Presenter, Promoters.website, Events.startDATE, Events.endDATE, Events.EventOnSaleDate, venues.VenueName FROM ((Events LEFT JOIN Performance ON Events.EventID = Performance.EventID) LEFT JOIN Promoters ON Events.PromoterCode = Promoters.PromoterCode) LEFT JOIN venues ON Performance.VenueCode = venues.VenueCode WHERE venues.VenueCode=".$venueURL." AND Events.group_id=1 AND Performance.category_id!=2 AND Performance.category_id!=5 AND Performance.category_id!=7 AND Performance.category_id!=8 AND Events.EventOnSaleDate IS NOT NULL AND (Performance.PerfType='Public Event' OR Events.EventID='79') AND Performance.endDateTime >= now()-INTERVAL 1 DAY AND Events.PublishDate <= now() AND Events.startDATE IS NOT NULL ORDER BY Performance.startDateTime";
   $Result = mysqli_query($DB,$Query);
   $NumResults = mysqli_num_rows($Result);
?>
<?php
 if($NumResults=0){
    echo "<p class='submenu'>$NumResults Performances</p>";
 }

while ($Row = mysqli_fetch_assoc($Result)){
$eventtitle = $Row['EventTitle'];
$eventDate = $Row['startDate'];
echo '<p><a href="' . $Row['ShoWareEventLink'] . '"><img src="https://alaskapac.centertix.net/UPLImage/' . $Row['thumb'] . '" alt="' . $Row['EventTitle'] . '" title="' . $Row['EventTitle'] . '" align="left" border="0"><span class="Heading3_blue">' . $Row['EventTitle'] . '</span>';
if($Row['FreeEvents']==TRUE){
echo '<img src="/images/free.gif" alt="Free Event" title="Free Event" width="80" height="80" align="right" border="0"></a><br />';
} elseif ($Row['EventOnSaleDate'] <= date("Y-m-d g:i a")){ /** IF ONSALEDATE<=NOW**/
echo '<img src="/images/logos/ctx/BUY_Tickets_gold.gif" alt="Buy Tickets" title="Buy Tickets" width="85" height="32" align="right" border="0"></a><br />';
} else {/** IF ONSALEDATE!<NOW**/
echo '<img src="/images/logos/ctx/AvailableSoon.png" alt="Available Soon" title="Available Soon" align="right" border="0"></a><br /><i>Tickets available ' . date("l, F j, Y", strtotime($Row['EventOnSaleDate'])) . ' at ' . date("g:i a", strtotime($Row['EventOnSaleDate'])) . '.</i>'; /** +ONSALEDATE **/
}
echo '<br />Presented by <a href="' . $Row['website'] .'" target="_blank">' . $Row['Presenter']  . '</a>';
echo '<br />'.$Row['startDate']. ' at ' .  $Row['startTime'].' - '.$Row['VenueName'];
if ($Row['FreeEvents']==TRUE){
echo '<br /><br />'.$Row['BriefDescription'];
}
    echo '<br /><br /></p><hr>';
}
?>
<?php
mysqli_free_result($Result);
mysqli_close($DB);
}
?>
Here is my sample Results page: http://www.myalaskacenter.com/events/venues/test_venues-results.php?VenueCode=ACH

Can anyone point out how to write a MySQL query with a PHP variable in the WHERE clause. I've tried {} {'xx'} and () and it still doesn't work.

Here is the code



<?php
ini_set('display_errors',1);
error_reporting(E_ALL|E_STRICT);

 include ("include/connect.php");
 include ("include/session.php");

$username = $session->userinfo['username'];

$result = mysql_query("SELECT email FROM customer WHERE user = {'$username'} ");

while($row = mysql_fetch_array($result))
  {
$custemail = $row['email'];
  }

echo "Session username: " . $username . "";
echo "Session customer email: " . $custemail . "";
?>



So I'm trying to show the email address for a record that matches the username of the user logged in.

I really appreciate the help.

Hi i have created a query where i want some columns to be retrieved. i get these 2 errors and i dont understand why.   Notice: Undefined variable: db Fatal error: Call to a member function query() on a non-object in      this is my query

function get_products_all() { 

$connect = get_connected_db();

try {
	  $results = $db->query("                                  ***** error on this line****
                SELECT  id, product, description, price, picc
                FROM products ORDER BY id DESC
				");
		
    } catch (Exception $e) {
        echo "Data could not be retrieved from the database.";
        exit;
    }

    $result = $results->fetchALL(PDO::FETCH_ASSOC);
	
	return $result;
}

what am i doing wrong?

Hi,
My issue here is that I cant get my query string variables to ONLY feed into if/else statement  on my secondary page. I include my secondary page (fine.php) from my index page. My query string variables keep being fed back into my original if/ese statement on index.php.

here is the if/else on index.php (these links work fine):

<?php
if($_SERVER['QUERY_STRING']=='/index.php' || $_SERVER['QUERY_STRING']=='')
{
include 'port.php';
}

elseif (isset($_GET['pos'])){
include 'pos.php';
}

elseif (isset($_GET['web'])){
include 'web.php';
}

elseif (isset($_GET['fine'])){
include 'fine.php';
}

        else {include '404.php';}






here is the if/else on my secondary page (fine.php). These links are supposed to alert the if/else in the next table cell. However, they instead alert the if/else in index.php.


<td><br/>
<a href="?backset"><img src="fine/thumbs/x-backset.jpg" border="0"></a><br/><br/> 
<a href="?backside"><img src="fine/thumbs/x-backside.jpg" border="0"></a><br/><br/> 
<a href="?bannerprint"><img src="fine/thumbs/x-bannerprint.jpg" border="0"></a><br/><br/> 
<a href="?chopu"><img src="fine/thumbs/x-chopu.jpg" border="0"></a><br/><br/> 
</td>
<td><br/>
<div id="DivPiece" align="left">
<?PHP 

if (isset($_GET['backset'])){
       include 'fine/backset.php';
}

elseif (isset($_GET['backside'])){
     include 'fine/backside.php';
}

elseif (isset($_GET['bannerprint'])){
     include 'fine/bannerprint.php';
}

elseif (isset($_GET['chopu'])){
include 'fine/chopu.php';
}

?>
</div>
</td>



How can I get the links on the secondary page to only alert the if/else statement on that page, and BLOCK the if/else statement on index.php from seeing them? I still want to use the query string though.

Thanks!
[/quote]

Hi, In some cases I need to execute  a default query , so I am storing the SELECT statement in a variable like this and executing it:

Code: [Select]
$default_query="SELECT * from user where userid='$userid'";
$user_query=mysql_query($default_query);
The above code returns an empty results.
But if I execute it without the variable it works fine.
Code: [Select]
$user_query=mysql_query("SELECT * from user where userid='$userid'");
 Am I syntatically wrong somewhere?