syntax problem? not sure what i did wrong

votum76

I'm new to php and sql I have a background in C++ though
I typed some code to input data into a mysql database and that works perfectly
but I can't query the database and get it out
here is my code

<html>
<body>
<?php
$db = mysql_connect("localhost", "root","");
mysql_select_db("learndb",$db);
$results = mysql_query("Select * FROM personnel");
$num=mysql_numrows($result);
mysql_close();
echo ($num);
$x = 0;
while ($x < $num){
$first=mysql_result($result,$i,"firstname");
$last=mysql_result($result,$i,"lastname");
$nick=mysql_result($result,$i,"nick");
$email=mysql_result($result,$i,"email");
$salary=mysql_result($result,$i,"salary");
echo ("$first $last nick: $nick email: $email Salary: $salary <hr> ");
++x;
}
?>
</body>
</html>
when I go to my broswer and run it I get this error
Parse error: parse error, unexpected T_STRING, expecting T_VARIABLE or '$' in c:\program files\apache group\apache\htdocs\get1.php on line 18
that's the line that I echo the information out
I'm throughly confused help is very much appreciated

Kudose

I am new to this too, but I would use print instead of echo on line 18.

Also, it should be x++; instead of ++x;

votum76

tryed that with print and x++; instead of ++x; just changes error slightly
to
Parse error: parse error, unexpected T_INC in c:\program files\apache group\apache\htdocs\get1.php on line 19

dnszero

You left out the dollar sign:

++$x;

votum76

I now get a new error
<html>
<body>
<?php
$db = mysql_connect("localhost", "root","");
mysql_select_db("learndb",$db);
$results = mysql_query("Select * FROM personnel", $db);
$num_rows = mysql_num_rows($result);
mysql_close();
$x = 0;
while ($x < $num_rows){
$first=mysql_result($result,$i,"firstname");
$last=mysql_result($result,$i,"lastname");
$nick=mysql_result($result,$i,"nick");
$email=mysql_result($result,$i,"email");
$salary=mysql_result($result,$i,"salary");
echo "$first $last nick: $nick email: $email Salary: $salary ";
$x++;
}
?>
</body>
</html>
error msg is
I really don't understand this because I have the same mysql_num_rows function in other file without a problem
Notice: Undefined variable: result in c:\program files\apache group\apache\htdocs\get1.php on line 7

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\get1.php on line 7

dnszero

You have a typo in this line:
$num_rows = mysql_num_rows($result);

It should be:
$num_rows = mysql_num_rows($results);