I have run into a Parse Error while I was coding a web site for my father-in-law. It is a basic error, yet I can't seem to find the problem. below is the error, and the code leading up to the error.

Parse error: syntax error, unexpected T_IS_EQUAL, expecting ',' or ')' on line 6

<?php
// Page information will go here
if ($_GET['page'] == 'company') {
    if (isset($_GET['area'] == 'staff')) {
$query = "SELECT * FROM staff";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());

From that last line, it goes into echoing HTML to form a table, and create my pages. I have used these lines of code in so many pages, why won't it work. On a scale of 1-5, I'm about a 4 on PHP, so maybe I am missing something? Could it be something farther in my code?

Inyour example above, you are not closing your if statements:

PHP Code:

<?php
  
  // Page information will go here
  if ($_GET['page'] == 'company')
  {
    if (isset($_GET['area'] == 'staff'))
    {
      $query = "SELECT * FROM staff";
      $result = mysql_query($query) or die(mysql_error());
      $row = mysql_fetch_array($result) or die(mysql_error());
    }
  }
  
?>

I have checked the statement, I still recieve the error. Here is more of the code.

 
<?php
// Page information will go here
if ($_GET['page'] == 'company')
{
 if (isset($_GET['area'] == 'staff') ) 
 {
 $query="SELECT * FROM staff";
 $result = mysql_query($query) or die(mysql_error());
 $row = mysql_fetch_array($result) or die(mysql_error());
 echo ("HTML IS GOING HERE"); }
 }
 
 elseif (isset($_GET['area'] == 'testimonials') ) 
 {
 $query="SELECT * FROM testimonials";
 $result = mysql_query($query) or die(mysql_error());
 $row = mysql_fetch_array($result) or die(mysql_error());
 echo ("HTML IS GOING HERE");
 
 }elseif (isset($_GET['area'] == 'contact') ) 
 {
 $query="SELECT * FROM contact";
 $result = mysql_query($query) or die(mysql_error());
 $row = mysql_fetch_array($result) or die(mysql_error());
 echo ("HTML IS GOING HERE");

 }else {

This code just continues the same coding, just different values. Any ideas where the problem is?

your brackets on the second if statement are misplaced

should be
if( isset($_GET['area']) =='staff')

Jamie

Thank you, Jamie. Looks like that was the problem, throughout my whole page of code. I can't believe I misplaced a ). Thanks!

Originally Posted by JamieLewis

your brackets on the second if statement are misplaced

should be
if( isset($_GET['area']) =='staff')

Jamie

Oops, missed that!

You will probally be better off using Switch for that lot of code. it will use alot less code, and make it a bit easier to read and understand.

If you want any help, i could probally help you

Originally Posted by j.edney

Thank you, Jamie. Looks like that was the problem, throughout my whole page of code. I can't believe I misplaced a ). Thanks!

And if I had a beer for everytime I forgot the d*** semicolon we would have a real labor day party here.

I prefer Vodka

Originally Posted by JamieLewis

your brackets on the second if statement are misplaced

should be
if( isset($_GET['area']) =='staff')

Jamie

Why use isset?

Should be just

if($_GET['area'] == 'staff')

i think its to check that $_GET['area'] is staff??

Originally Posted by dansgalaxy

i think its to check that $_GET['area'] is staff??

You don't need isset for that, wouldn't that code be incorrect? isset returns true or false.

If we use $_GET['area'] == 'staff' without isset that should be just fine.

isset is only used to check if it's set, not what the content is.