PHP - Shall I Use Mysqli_stmt_free_result($stmt) Or Not ?

Hello ,   I am trying to update a record in my table if the particular value exists in the rows of a table. But my if-else not working properly. I am not getting where i am going wrong.   here is my table authorization, where i define authorization % with min and max.  authorisation.JPG   16.83KB   0 downloads   after adding line items, before display i will check whether any one of the line item discount lies within this min and max and also checks whether authorization is required (required='Yes'/'No')  for that discount in my authorization table.  If authorization required then i will update that order number as authorized.    here is my line items table    line_items.JPG   20.37KB   0 downloads   I am doing like this

while($row=mysql_fetch_array($query))
{
$dis1 = "SELECT auth_id, auth_min, auth_max, required FROM sales_authorisation";
$dis2 = mysql_query($dis1) or die (mysql_error());

while($d1 = mysql_fetch_array($dis2))
{
	$min = $d1['auth_min'];
	$max = $d1['auth_max'];
	$req = $d1['required'];
	//echo $req;
	
if( ($min <= ($row['discount'])) && ($max >= ($row['discount'])) && ($req='Yes'))
{
$auth = "UPDATE orders SET authorise='No' WHERE order_id=".$order_id."";
echo "hello";
}
else
{

$auth = "UPDATE orders SET authorise='Yes' WHERE order_id=".$order_id."";

}
$auth1 = mysql_query($auth) or die (mysql_error());

}

?> 

<tr> 
<td><?php echo $counter++; ?></td>
<td><?php echo $row['itemname']; ?> - <?php echo $row['uom']; ?></td>
<td><?php echo $row['description']; ?></td>
<td><?php echo $row['quantity']; ?></td>
<td><?php echo number_format($row['selling_price'],2); ?></td>
<td><?php echo $row['discount']; ?> %</td>
<td><?php echo $row['tname']; ?>-<?php echo $row['rate']; ?> %</td>
<td><?php echo $row['freight']; ?></td>
<td><?php echo number_format($row['total'],2); ?></td>
 
 </tr> 

While loop is for displaying line items for the order.   But my if condition doesn't works . Not getting how to do it. please suggest       

Hi,

I have two scripts using almost identical code.
One of them works and finds the field (email) - the PRIMARY KEY (email)
and the other doesn't (user_key) - not indexed

Is this because PHP 7 needs to use only indexes in the WHERE part of SQL Query?

Here are the two scripts - first one works

$sql = 'SELECT access,fname FROM clients WHERE email=?';

	    $stmt = $connect->prepare($sql);

	    $stmt->execute([$email]); 

	    $data_exists = ($stmt->fetchColumn() > 0) ? true : false;

	    if ($data_exists) { //   account found.

	        $row = $stmt->fetch();

	        $access = $row['access'];

	        $fname = $row['fname'];

	        }

	    else {

	        $err_msg = 'Invalid Email Address and/or Password.';

	        $email = '';

	        $pass = '';

	        require_once ("login_fm.php");

	        exit;

	        }

	[/PHP]

This second one drops through to the ERROR page.

 

$sql = 'SELECT email,fname,lname,confirm FROM clients WHERE user_key=?';
    $stmt = $connect->prepare($sql);
    $stmt->execute([$the_key]); 
    $data_exists = ($stmt->fetchColumn() > 0) ? true : false;
    if ($data_exists) { //   Account found.
        $row = $stmt->fetch();
        $email        = $row['email'];
        $name         = $row['fname'].' '.$row['lname'];
        $confirm     = $row['confirm'];
        
        $_SESSION['auth'] = "yes";
        $_SESSION['email'] = $email;
        $_SESSION['name'] = $name;
        
        $sql = 'UPDATE clients SET confirm = ?,log_count = log_count+1,last_date=? WHERE email=?';
        $stmt= $connect->prepare($sql);
        $stmt->execute(['y',$today_time,$email]);
     
        require_once("index.php");   //  SUCCESSFUL LOGIN: THIS LOADS THE START PAGE  
        exit;    
        }  // end if
    else {
        $err_page_message = "ERROR - Account not found $the_key : $name";
        require_once("err.php"); 
        exit;
        }  // end else  

 

May be it's a different reason - like I'm too tied !

On the second one. I could rewrite it to just try the update and fail-over to the ERROR page,

But it would be nice to know why it isn't working.

 

It's quite simple, I just want to check to see if a row exists with the condition given, that's all.

I never had this problem the old mysql but the PDO seems a bit tricky or I'm just using the wrong code lol.

 

Thanks

Edited June 15, 2019 by David-London

Code: [Select]
$params = array();
$type=array();
$fragments = array();

while (list($chiave,$val) = each($_POST)){
if($val!=""){
$fragments[] = $chiave." = ?";
//echo $chiave;
if($chiave=="ritiro" or $chiave=="data")
$params[] = normalToDbDate($val);
else
$params[] = $val;
if($chiave=="legale")
$type[]='i';
else
$type[]='s';
}
}

$prova=implode("",$type);
foreach($params as $param)
echo "<p>".$param."</p>";
//echo "val(".count($params).")-->".implode(',',$params);
//echo "type(".strlen($prova).")-->".implode("",$type);
$sql = $db->prepare("..... AND ".implode(" AND ", $fragments));
array_unshift($params,$prova)
call_user_func_array( array( $sql, 'bind_param' ),$params);
//$sql->bind_param($prova,$params);
$sql->execute();
     
hi all, with the tructure above i've tried to implement the cration of a dynamic query for a search form. i debugged with echoes and che numbers of bind_param and the those in the prepared statement match in sense that count($params)=strlen($prova). having said that i get this error:

Code: [Select]
Warning: call_user_func_array() expects parameter 2 to be array, integer given in
and no results are shown to me even when the filter match.

Hi all !   I have a piece of code he

$result = display_all(fcon, $var1, $var2);

function display_all( // defined in another file
$query = "SELECT one, two, three four, index1, index2 FROM numbers WHERE index1 = $var1 LIMIT 0, 1"; $result = mysqli_query($fcon, $query); return ($result); )

and then I use the returned variable $result to display the value as follows:-

		while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
		{
			echo "<tr>";
			echo "<td>".$one."</td>";
			echo "<td>".$two."</td>";
			echo "<td>".$three. "</td>";
			echo "<td>".$four. "</td>";
			echo "<td>".$index1. "</td>"; 
                        echo "</tr>";
               }    

and this displays the n rows of data returned.       Now I have started using prepared statements and the function is now 

function display_all{

      $query = "SELECT one, two, three four, index1 FROM numbers WHERE index1 = ?
                LIMIT 0, 10";
      $stmt = $conn->prepare($query);
      $stmt->bind_param('i',$var)
      if($stmt->execute())
      {
          $stmt->bind_result($one, $two, $three, $four, $index1);
          $stmt->store_result();
      }
return($stmt);
}

However the returned $stmt object is unable to display the n rows of data since it shows null values. I assume that this is not the right way to use the $stmt object to display data. I must be missing something. So I request you guys to help me with this.   Thanks loads.
Edited by ajoo, 23 December 2014 - 11:24 AM.