Problem with prepared statements

PHP
1

Hi all,
I have some problems running prepared statements in a function. I’m getting some warnings which I’m not sure what they mean. Anyway here is my code.

include_once 'login.php'; 
  
function read()  
{  
   $parameters = array();  
   $results = array();  
   $username = 'iTenzo';

   $mysql = new mysqli($server,$user,$pass,$db) or die('There was a problem connecting to the database');  
   $stmt = $mysql->prepare('SELECT firstname, lastname FROM users WHERE username=?') or die('Problem preparing query');  
   $stmt->execute();  
  
   $meta = $stmt->result_metadata();  
  
   while ( $field = $meta->fetch_field() ) {  
  
     $parameters[] = &$row[$field->name];  
   }  
  
   call_user_func_array(array($stmt, 'bind_result'), $parameters);  
  
   while ( $stmt->fetch() ) {  
      $x = array();  
      foreach( $row as $key => $val ) {  
         $x[$key] = $val;  
      }  
      $results[] = $x;  
   }  
  
   return $results;  
  
}  
  
$results = read();

Here are the warnings I’m getting:


Warning: mysqli::mysqli() [mysqli.mysqli]: [2002] No such file or directory (trying to connect via unix:///var/mysql/mysql.sock) in /Users/tenzo/Sites/prepare_function.php on line 11

Warning: mysqli::mysqli() [mysqli.mysqli]: (HY000/2002): No such file or directory in /Users/tenzo/Sites/prepare_function.php on line 11

Warning: mysqli::prepare() [mysqli.prepare]: Couldn't fetch mysqli in /Users/tenzo/Sites/prepare_function.php on line 12
Problem preparing query

Thanks

2

Where is “$server,$user,$pass,$db” coming from?

3

include_once ‘login.php’;
at the top

4 

include_once 'login.php'; 
  
function read()  
{  
echo "<p>server: $server</p><p>user: $user</p><p>pass: $pass</p><p>db: $db</p>";
  
}  
  
$results = read();
5

this is what “login.php” contains:

<?php

//database settings
$server = 'localhost';
$user =  'root';
$pass = 'password';
$db = 'login';

?>

I’m not sure what you mean oddz…

6

What he is trying to tell you is that the variables such as $server are declared outside of the function, so the function cannot find them.


$server = 'EEK';

function read()  {  
echo "<p>server: $server</p><p>user: $user</p><p>pass: $pass</p><p>db: $db</p>";
  
}  
  
$results = read();

On my localhost system this creates 4 massive errors because I am running with error_reporting turned on - not on my live server you understand!

Whats more with xDebug installed those errors are bright orange and spill out even more information;


Notice: Undefined variable: server in C:\\var\\www\\html\\code\\shell\\xxx.php  : eval()'d code on line 6
Call Stack
#	Time	Memory	Function	Location
1	0.9665	146488	{main}( )	..\\index.php:0
2	0.9691	199264	eval( ''$server = \\'EEK\\'; function read() { 
echo "

server: $server

user: $user

pass: $pass

db: $db
";}$results = read(); 
0.9691	199480	read( )	..\\index.php(492) : eval()'d code:10

So the lesson is not really give your functions access to the variables it needs, but learn how to debug your own code for the sake of your own sanity and effectiveness.

The term you are looking for is variable scope , it is not complicated - have a read and you should be able to work it out, if not come back and ask here again.

7

yes, that’s so true. As soon as I resolved the scope problem everything fell into place. Of course it is scope! man, I feel so stupid…

many thanks ODDZ/CUPS!

8

Glad you got it sorted, don’t worry many of us have done the same as you, and we felt bad as well.

All I ask is, just dont forget how it feels, and sometime in the future, if you get time, come back here and “pass it on”.

9

Thank you Cups