Check if variable exists....

[Aes]

I'm working with a dynamic site using PHP and MySQL (what else would I use? ). Say, for example, I have a page named about.php.

And depending upon what an id variable is when passed to the page (i.e. http://host.com/about.php?id=x), it displays the content of the database where the id=x.

But, if no ID is passed to the page (i.e. http://host.com/about.php), I need to assign a value to $id.

So how would I check if the variable exists? That way if it does, the content will be displayed, and if it doesn't, a value wil be assigned to it and that content will be displayed.

if ($id == '')
{
$id="x";
}
// and
if ($id == "")
{
$id="x";
} 

Both of those don't work. As always, thanks for your help.

[Justin Sampson]

Try the isset() function:

if(!isset($id))
{
$id = "1";
} 

[Aes]

That doesn't work either.

[Justin Sampson]

I don't know why that wouldn't work, and I don't see why your 2 examples wouldn't work either.

Are you sure it's not somthing else?

[johnn]

Try
if(empty($id))

[DR_LaRRY_PEpPeR]

do you maybe have register_globals off in php.ini?

[Aes]

That didn't work either. It seems like all of those examples should work!

[delemtri]

Try:   if (!$id) { $id=1; }

[TWTCommish]

Alright, something's up, because ALL of the below should work:

if (!$id) 

if (!isset($id)) 

if (empty($id)) 

If those aren't working, then something else is going on...perhaps the id value is coming from somewhere else, like a cookie? If you're look for the id value in the querystring (and only from the querystring), replace $id with $HTTP_GET_VARS["id"], to ensure that you're looking at the exact variable that you want.

If that doesn't work, then something might be up with your host, because it ought to work.

[Aes]

We appear to be 0 for 5 now.

[Aes]

Here's the code if it helps:

<?php
// check to see if $id contains a value
if (!$id)
{
    $id="2";
}
else
{
    // connect to the main content database
    $maincnx = @mysql_connect("localhost","root");
    if (!$maincnx)
    {
        echo("<p>There was an error in connecting to the database: " . mysql_error() . "<br />\n" .
        "A message has been sent to the webmaster.  Please try again later.</p>");
        exit();
    }
    // select the appropriate database
    $mainselect = @mysql_select_db("main",$maincnx);
    if (!$mainselect)
    {
        echo("<p>There was an error in selecting the database: " . mysql_error() . "<br />\n" .
        "A message has been sent to the webmaster.  Please try again later.</p>");
        exit();
    }
    // extract the content and title from the database
    $content = @mysql_query("select id, content, title, date from maincontent where id='$id'");
    if (!$content)
    {
        echo("<p>There was an error in displaying the content for this page: " . mysql_error() . "<br />\n" .
        "A message has been sent to the webmaster.  Please try again later.</p>");
        exit();
    }
    $results = mysql_fetch_array($content);
    $contentid = $results["id"];
    $contenttitle = $results["title"];
    $contenttext = $results["content"];
    $contentdate = $results["date"];
    // assign the values to a temporary array, reassign to variables, and display on the page
    // site template
    include("header.php");
    include("toppage.php");
    echo("$contenttext");
    include("footer.php");
}
?>

[TWTCommish]

If those outside problems do not exist (IE: if you're using $HTTP_GET_VARS[] or if you're sure that the variable isn't being set through a form, cookie, or session variable of some sort), then something's up with the PHP installation/webhost.

[Aes]

I'll inquire with my host. Thanks.

[Aes]

Well, I would like to apologize. I just realized something. If $id isn't set it sets it to 2 and then forgoes the rest of the script because I have satisfied the if clause and so it skips the else clause.

I'm so stupid!!!!!!

[TWTCommish]

D'oh! I only glanced over your large block of code, so I missed it to. Whoops! Hehe. No problem, glad it's working for you now.