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jade5357
Forum Newbie
Posts: 5 Joined: Thu Apr 30, 2009 4:25 am
Post
by jade5357 » Fri May 01, 2009 4:03 am
well i have this code :
Code: Select all
$sql1=mysql_query("SELECT personFirstName, personLastName, personMiddleInitial
FROM person where personid='$pid'");
$result1 = mysql_fetch_array($sql1);
$lastname = result1['personLastName'];
$firstname = result1['personFirstName'];
$mi = result1['personMiddleInitial'];
and all i get is a parse error at this line
Code: Select all
$lastname = result1['personLastName'];
$firstname = result1['personFirstName'];
$mi = result1['personMiddleInitial'];
can somebody help me with this.. I can't really figure out why.. thanks in advance
Last edited by
Benjamin on Fri May 01, 2009 4:26 am, edited 1 time in total.
Reason: Changed code type from text to php.
requinix
Spammer :|
Posts: 6617 Joined: Wed Oct 15, 2008 2:35 am
Location: WA, USA
Post
by requinix » Fri May 01, 2009 4:23 am
Variables start with a $.
jade5357
Forum Newbie
Posts: 5 Joined: Thu Apr 30, 2009 4:25 am
Post
by jade5357 » Fri May 01, 2009 4:25 am
*sigh* very simple indeed... sorry.. i forgot to write the '$' sign again..
thank you very much for the reply