How Can I $_GET a variable?

[co.ador]

Notice: Undefined variable: rmenu in C:\wamp\www\nyhungry\restaurants\shoe1.php on line 76

I know how to global GET an integer

" . (int) $_GET['shoe'] 

But not a variable, which in this case the rmenu value pass undefined being.

How could I global get the $rmenu variable so I can use it's value.

Thank you.

[Dan Grossman]

Please restate your question in coherent English. An example of the problem would be useful as well.

The error you have means you tried to use a variable $rmenu that was not defined before you used it. The solution is to not use variables which you haven't defined.

[co.ador]

It was defined before but the value of $rmenu does not pass

while($rmenu = mysql_fetch_array($regularshoe_set)){
  echo "<li"; 
  if ($rmenu["id"] == $sel_shoe) {
  echo " class=\"selected\""; 
  } 
  echo "><a href=\"shoe1.php?menu=" . urlencode($rmenu["id"]) ."\";>
{$rmenu["shoename"]}</a></li>";  }
  }
 if ($rmenu["id"] == $sel_menu) { // this is line 76 

If you note before line 76 the rmenu value is difined but so how it doesn't pass it to the $rmenu below it.

Sorry about the incoherent English.

[Dan Grossman]

That variable only exists within the loop. Even if it existed outside the loop, $rmenu would simply hold false as that was the last return value of mysql_fetch_array() when there were no more rows in the result set.

[co.ador]

so It doesn't exist.

I will abandon this possibility then.


Thanks though!

[decowski]

That variable only exists within the loop.

This statement is false. Loops run in the scope they are created in so all variables set in a loop do exist outside it.

[...] $rmenu would simply hold false as that was the last return value of mysql_fetch_array() when there were no more rows in the result set.

True, but it's got nothing to do with PHP throwing an undefined variable error. The variable is defined and set to FALSE.

@co.ador: Please paste in more code (especially the code after problematic line).

EDIT: After formatting your code correctly:

  while($rmenu = mysql_fetch_array($regularshoe_set)) {
    echo "<li"; 
  
    if ($rmenu["id"] == $sel_shoe) {
      echo " class=\"selected\""; 
    } 
  
    echo "><a href=\"shoe1.php?menu=" . urlencode($rmenu["id"]) ."\";>{$rmenu["shoename"]}</a></li>";
  }
} // what block of code is this closing?

if ($rmenu["id"] == $sel_menu) { // this is line 76 

it is obvious that you are closing another block of code (after closing the loop) before line 76. Please paste more code before the lines you have pasted.

[co.ador]

Skundbad it is debuged, I debuged it for a while but still showing up the problem. I will show some more detail so you guys can see more details.

after line 76 the fallowing bit of code

if ($rmenu["id"] == $sel_menu) { 

I am trying to place or arrange the bit of code below this paragraph, so it can open the subject and hold on to its submenus and avoid to close the submenus once the users click on a submenu. The problem is that when a user open or click on a subject it opens it's submenus but once users click on a submenu it closes everything and it return to show the subject only closing all the submenus below the subject. If users want to open another submenu other than the one opened again, the users have to click on the subject again to choose another submenu instead of leaving all of the submenus visible and available to the users regardless users clicking on a submenu or not. Is there a format or coding it could be arranged for that? The way it is arranged now, the subject always closes its submenus once a submenu is click on instead of leaving all the submenus visible whether it is clicked on or not.

 echo "<ul class=\"submenu2\">";
 while($rmenu = mysql_fetch_array($regularshoe_set)){
  echo "<li"; 
  if ($rmenu["id"] == $sel_shoe) { 
  echo " class=\"selected\"";
  } 
  echo "><a href=\"shoe1.php?menu=" . urlencode($rmenu["id"]) ."\";>{$rmenu["shoename"]}</a></li>";  }
echo "</ul>"; 

decowski this is the rest of the code before the while loop and what the curly braces below closes is the If statement after the mysql_query, Thank you for the help.

$regularshoe_set = mysql_query($query, $connection);
  if(!$regularshoe_set){
  die("Database query failed:" . mysql_error());} 
if ($shoesubject["id"] == $sel_shoe['id'])  {   
while($rmenu = mysql_fetch_array($regularshoe_set)) {
    echo "<li"; 
  
    if ($rmenu["id"] == $sel_shoe) {
      echo " class=\"selected\""; 
    } 
  
    echo "><a href=\"shoe1.php?menu=" . urlencode($rmenu["id"]) ."\";>{$rmenu["shoename"]}</a></li>";
  }
} //That carly brace is closing the first If then statement at the top of the code.

if ($rmenu["id"] == $sel_menu) { // this is 

line 76

[decowski]

You have a logical error in your code:

if ($shoesubject["id"] == $sel_shoe['id'])  {
  while($rmenu = mysql_fetch_array($regularshoe_set)) {
  // [...]
  }
}

if ($rmenu["id"] == $sel_menu) { 

So only if the first condition is met you set the $rmenu variable but you're trying to use it even if the condition is not met. You need to rethink what you're doing here.

Sorry I can't help more but I can't understand the purpose of this code.

[co.ador]

if ($rmenu["id"] == $sel_shoe) 

It is not sel_menu is $sel_shoe

[decowski]

This is irrelevant. I'm talking about $rmenu variable, because that's what PHP is complaining about being not set. Look again at the snippet:

if ($shoesubject["id"] == $sel_shoe['id'])  {
  while($rmenu = mysql_fetch_array($regularshoe_set)) {
  // [...]
  }
}

if ($rmenu["id"] == $sel_menu) { 

You only set $rmenu variable if the first condition is met but then try to use the variable outside the condition (thus it may be not set).

[Dan Grossman]

Define a variable before the loop, set its value within the loop, and after the loop is over, it will hold the value from the last row.

[co.ador]

ok let me try difine a variable with that name before the loop starts

$regularshoe_set = mysql_query($query, $connection);
  if(!$regularshoe_set){
  die("Database query failed:" . mysql_error());} 
  $rmenu = mysql_fetch_array($regularshoe_set) 
// I am not sure If I should define or call $rmenu with a mysql_fetch_array?
//another thing is that before the while loop I have an If statement.
//I am not sure if I should place this variable in between the if statement and the while loop thought it was better to locate it here. 
  echo "<ul class=\"submenu\">";
  if ($shoesubject["id"] == $sel_shoekind['id'])  {
while($rmenu = mysql_fetch_array($regularshoe_set)){
  echo "<li"; 
  if ($rmenu["id"] == $sel_shoe) {
  echo " class=\"selected\""; 
  } 
  echo "><a href=\"shoe1.php?menu=" . urlencode($rmenu["id"]) ."\";>
{$rmenu["shoename"]}</a></li>";  }
  }
 if ($rmenu["id"] == $sel_menu) { // this is line 76 

[Dan Grossman]

Doing that will throw away the first row of the result set, since you advanced the row pointer twice before getting inside your loop by calling mysql_fetch_array twice.

$lastrow = array();

while ($rmenu = mysql_fetch_array($regularshoe_set)) {
  // other code
  $lastrow = $rmenu;
}

//now you can use $lastrow 

[co.ador]

Thank you Dan Grossman!!!

I will keep trying tomorrow morning

[skunkbad]

One thing you should know about variables is that they can be set, unset, and not set at all. Being set means that the variable has a value of some sort applied to it. Unsetting a variable makes the variable lose all value, and a variable without a value is not set.

When debugging your php applications, you will see errors like the one you saw:

Notice: Undefined variable: rmenu in C:\wamp\www\nyhungry\restaurants\shoe1.php on line 76

and this error tells you that the rmenu variable was not set. If you thought you had set it, you were wrong.

Sometimes, programatically speaking, you will not know if a variable is set or not, and in that case, you can check if it is set or not, and do what is necessary.

Tonight, after reinstalling my operating system, wamp, and all of my programs, one of my applications wasn't working, and it was hard to track down, because of the complex nature of the framework I was using. In the end, my error came down to a problem similar to yours. My script was checking for a server variable - $_SERVER['HTTP_USER_AGENT'] - and it didn't exist. I used the php function isset() to fix my problem like this:

if(isset($_SERVER['HTTP_USER_AGENT'])){
			$ie6 = "MSIE 6.0";  
			$browser = $_SERVER['HTTP_USER_AGENT'];  
			$browser = substr("$browser", 25, 8);
			if($browser == $ie6){
				$this->template->javascripts[] = 'js/supersleight-min.js';
			}
		}

I just thought you might need to know this, because I had a hard time learning php, and you seem sort of new.