Undefined variable error in Php

[Capers]

Hello friends,

I am getting an error message whenever i am trying to run any query in php.I don't know whats wrong with it.I am getting the following error message as follows


PHP Notice: Undefined variable: array in C:\xampp\htdocs\mysql_class_2.php

[kelfro]

Sure i will help you out with this is it possible for you to provide me the code so that i can give you a perfect solution for it.

[Capers]

Following are the code on which i am getting the error message

<?php
 class mysql {
 

  function Connect($host, $name, $pass, $db){
 
  $connection = mysql_connect("$host","$name","$pass");
  mysql_select_db("$db", $connection);
 
  }//ends the connection function
 

  function Close(){
 
  mysql_close($this->connection);
 
  }//ends the close function
 

  function FetchRow($query){
  $rows = mysql_fetch_row($query);
  return $rows;
  }
 

  function FetchArray($query){
  $array = mysql_fetch_array($query);
  return $array;
  }
 
  function FetchNum($query){
   $num = mysql_num_rows($query);
  return $num;
  }
 

  function Query($sql){
  $query = mysql_query($sql) or die(mysql_error());
  return $query;
  }//ends the query function
 

   function FetchAssoc($query2){
   if($query2 and !is_bool($query2)){
  $array = mysql_fetch_assoc($query2);
         }
  return ($array);
         }

 /*
  function TanimotoSimilarity($query,$input_arrays) {
             $i=0;
             $row= array();
             while( $row=mysql_fetch_assoc($query) )
        {
 
      $first[$i]=(count(array_intersect($row,$input_arrays)))/(count($row)+count($input_arrays)-count(array_intersect($row,$input_arrays)));
      $i++;
 
        }
 
         return($first);
 
 
} */

 
}//ends the class
 
 ?>

[kelfro]

Try to use the following code instead of your

$x = 1;
$y = 2;
// Because our conditional will return false, it will skip the creation of $xy.
if ( $x === $y )
{
  $xy = $x + $y;
}
// Therefore, we are now trying to return a variable that doesn't exist.
return $xy;
 
$x = 1;
$y = 1;
// Now that the if conditional will return true, it sets the variable and we can
// successfully return it.
if ( $x === $y )
{
  $xy = $x + $y;
}
return $xy; 

And with that you need to use else statement to return a NULL value for your $array.