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Following webmonkey and it doesn't work

Posted: Sun Dec 01, 2002 9:21 pm
by dawho9
I am really new to PHP/MySQL. I am familar with straight HTML, but have never tried connecting to a database. So I was following an example on webmonkey (http://hotwired.lycos.com/webmonkey/99/ ... rogramming) and am having problems. I think it has something to do with this global variable thing, but I cannot seem to figure out what it all is saying. There is too much techno talk. Here is the code I am trying.

Code: Select all

<?php
$db = mysql_connect("localhost", "root");
mysql_select_db("mydb",$db);
// display individual record
if ($id) {
   $result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);
   $myrow = mysql_fetch_array($result);
   printf("First name: %s\n<br>", $myrowї"first"]);
   printf("Last name: %s\n<br>", $myrowї"last"]);
   printf("Address: %s\n<br>", $myrowї"address"]);
   printf("Position: %s\n<br>", $myrowї"position"]);
} else {
    // show employee list
   $result = mysql_query("SELECT * FROM employees",$db);
    if ($myrow = mysql_fetch_array($result)) {
      // display list if there are records to display
      do {
        printf("<a href="%s?id=%s">%s %s</a><br>\n", $PHP_SELF, $myrowї"id"], $myrowї"first"], $myrowї"last"]);
      } while ($myrow = mysql_fetch_array($result));
    } else {
      // no records to display
      echo "Sorry, no records were found!";	
    }
}
?>
A simple explaination of why "if ($id)" does not work and the correct syntax would be great. The error I am getting is:

Notice: Undefined variable

dawho
?>

Posted: Sun Dec 01, 2002 9:24 pm
by hob_goblin