Re: [PHP] Converting code to a function

[Simon Garner]

Clayton,

The problem is that some of your variables, e.g. $HOSTNAME, $DB_USER, etc.,
are not global variables, and so are not accessible from inside the
function.

I recommend making those settings define()s instead of variables, but if you
keep them as variables then you just need to add:


<?php
    function viewdoc ($docid)
    {
        global $HOSTNAME, $DB_USER, ...;
        ...
    }
?>


But to do them as constants instead, do:

<?php
    define("HOSTNAME", "myhost.net");
    define("DB_USER", "myuser");
    define("DB_PASS", "jksajfeioe");

    function viewdoc ($docid)
    {
        ...
        $dbh = mysql_connect(HOSTNAME, DB_USER, DB_PASS);
        ...
    }
?>


Hope this helps.


Cheers

Simon Garner


----- Original Message -----
From: "Clayton Dukes" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, February 28, 2001 3:51 PM
Subject: [PHP] Converting code to a function


Hi everyone :-)

I need help, how can I convert the following code into a function so that I
can just place it in a global include file and call it using
viewdoc($docid); whenever necessary?

-------START-------
if ($docid == "") {
          echo "Fatal Error - DocID not supplied";
          exit;
  }

  $dbh = mysql_connect("$HOSTNAME","$DB_USER","$DB_PASS");
  mysql_select_db("$DATABASE");

  $sth = mysql_query("SELECT docdata FROM documents WHERE docid like
                  '$docid'");

  $row = mysql_fetch_array($sth);

  if ($row[0] != "") {
          echo "DocID: $docid\n\n";
        echo (wraptext($row[0],72));

  } else {
          echo "Fatal Error - Can't fetch docid";
  }
-------END-------

I tried to do this:
function viewdoc($docid) {
...PASTE FROM ABOVE...
}

But then I get:

Warning: Supplied argument is not a valid MySQL result resource in
/home/www/websites/gdd/inc/funcs.inc on line 76
Fatal Error - Can't fetch docid

when I call it from a web page using:

viewdoc($docid);


What have I done wrong?



TIA!
Clayton Dukes




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