vaiable access

[robcrozier]

Ok, imp having a bit of a problem accessing a variable that is initialized at the top of the script and then its value changed within an if statement further down the script:
[code]

if($row){
$form_1_valid = true;
$id = $row[0]; // this is the variable in question, it’s an auto incremented id field from the database
$form_2_display = true;
}
[/code]

Now I try to access the $id variable near the bottom of the script and it contains no value???

Any help would be greatly appreciated!

[kenrbnsn]

Is this assignment in a function? Can you post the rest of the script?

Ken

[robcrozier]

No, i'm simply assigining a variable with a random value at the top of the script like this:
[code]
$id = a;
[/code]

Next, if the form on the page is submitted, the vlues are validated and if validation is ok, the value of id is changed to an integer value, like so (the $id variable is changed near the bottom on this piece of code:
[code]
//validate the first form

if (isset($_POST['submit'])) {


  //  check for a username

  if (eregi ("^[[:alnum:]_-]{1,50}$", stripslashes(trim($_POST['username']))))
{

$u = trim($_POST['username']);

  } 

  else {

    $u = false;
    echo '<p><font color = "red" size = "-2" face="Arial, Helvetica, sans-serif"> >>> Please enter a valid username.</font></p>';
}

// check for a password

if (eregi ("^[[:alnum:]]{1,50}$", stripslashes(trim($_POST['password']))))
{

$p = trim($_POST['password']);

} 

else {


  $p = false;

  echo '<p><font color = "red" size = "-2" face="Arial, Helvetica, sans-serif"> >>> Please enter a valid password.</font></p>';

    }

// If everything's OK

  if ($u && $p)

  {

// attemt to find a match for the users dataials in the database

$query = "SELECT id, username, password from users
              WHERE username='$u'
                AND password='$p'";

$result = mysql_query ($query);


$row = mysql_fetch_array ($result, MYSQL_NUM);

                if($row){

// set variables                                                                             $form_1_valid = true;
$id = $row[0]; //THIS IS WHERE MY VARIABLE IS CHANGED!!!
$form_2_display = true;

}
else
{

//display error message
echo "<font color='red' face='arial' size='-2'>invalid username and password, pleas try again</font>";
}
}

}
[/code]

Next, im trying to access the changed variable within another if statement further down the script, like this:
[code]if($result)
{
//insert ok
echo 'id = '.$id;


// delete old cookie, new one issued upon 1st login
setcookie("username", "", time()-60*60*24*365);
setcookie("password", "", time()-60*60*24*365);

}
[/code]

[wildteen88]

Your query might be failing. So change this:
[code]$result = mysql_query ($query);[/code]
To this:
[code]$result = mysql_query ($query) or die("error with query:<br />\n" . mysql_error());[/code]
If you get any errors post them here in full. Do not post part of the error message. It is important you post the error as-is.

Also I'd suggest you change this:
[code]$row = mysql_fetch_array ($result, MYSQL_NUM);

if($row){[/code]
To this:
[code]if(mysql_num_rows($result) > 0)
{
    $row = mysql_fetch_array ($result, MYSQL_NUM);[/code]