Solved

Parse error: parse error, unexpected $ in...

Posted on 2007-04-10
7
233 Views
Last Modified: 2007-04-10
Can't figure this one out, I've been looking at this for 2 hours.

http://www.austinmetrobaseball.com/gamechange.txt
0
Comment
Question by:EddieShipman
7 Comments
 
LVL 24

Expert Comment

by:glcummins
ID: 18885131
Can you post the full error message? All we received was "parse error, unexpected $ in..."
0
 
LVL 29

Accepted Solution

by:
TeRReF earned 500 total points
ID: 18885180
Change these lines:

  if ( isset($_POST["gameid_1"]) ) {
    $sql = "SELECT h.email AS hemail, v.email AS vemail ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "WHERE g.gamenum = ".$_POST["gameid_1"];
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($hemail1, $vemail1)=$db->fetch_row($query);
    }
    $sql = "SELECT gamenum, ";
    $sql = $sql. "       DATE_FORMAT(gamedate, \"%a %b %d  %l:%i%p\"), ";
    $sql = $sql. "       v.teamname AS visitor, ";
    $sql = $sql. "       h.teamname AS home, ";
    $sql = $sql. "       f.field_name ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "JOIN _fields AS f ON (f.id = g.fieldid )";
    $sql = $sql. "WHERE g.gamenum =".$game_num1;
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($gamenum1, $gamedate1, $visitteam1, $hometeam1, $field_name1)=$db->fetch_row($query);
  }
  if ( isset($_POST["gameid_2"]) ) {
    $sql =  "SELECT h.email AS hemail, v.email AS vemail ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "WHERE g.gamenum = ".$_POST["gameid_2"];
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($hemail2, $vemail2)=$db->fetch_row($query);
    }
    $sql = "SELECT gamenum, ";
    $sql = $sql. "       DATE_FORMAT(gamedate, \"%a %b %d  %l:%i%p\"), ";
    $sql = $sql. "       v.teamname AS visitor, ";
    $sql = $sql. "       h.teamname AS home, ";
    $sql = $sql. "       f.field_name ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "JOIN _fields AS f ON (f.id = g.fieldid )";
    $sql = $sql. "WHERE g.gamenum =".$game_num2;
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($gamenum2, $gamedate2, $visitteam2, $hometeam2, $field_name2)=$db->fetch_row($query);
    }



into:

  if ( isset($_POST["gameid_1"]) ) {
    $sql = "SELECT h.email AS hemail, v.email AS vemail ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "WHERE g.gamenum = ".$_POST["gameid_1"];
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($hemail1, $vemail1)=$db->fetch_row($query);
    }
    $sql = "SELECT gamenum, ";
    $sql = $sql. "       DATE_FORMAT(gamedate, \"%a %b %d  %l:%i%p\"), ";
    $sql = $sql. "       v.teamname AS visitor, ";
    $sql = $sql. "       h.teamname AS home, ";
    $sql = $sql. "       f.field_name ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "JOIN _fields AS f ON (f.id = g.fieldid )";
    $sql = $sql. "WHERE g.gamenum =".$game_num1;
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($gamenum1, $gamedate1, $visitteam1, $hometeam1, $field_name1)=$db->fetch_row($query);
    }
  }
  if ( isset($_POST["gameid_2"]) ) {
    $sql =  "SELECT h.email AS hemail, v.email AS vemail ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "WHERE g.gamenum = ".$_POST["gameid_2"];
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($hemail2, $vemail2)=$db->fetch_row($query);
    }
    $sql = "SELECT gamenum, ";
    $sql = $sql. "       DATE_FORMAT(gamedate, \"%a %b %d  %l:%i%p\"), ";
    $sql = $sql. "       v.teamname AS visitor, ";
    $sql = $sql. "       h.teamname AS home, ";
    $sql = $sql. "       f.field_name ";
    $sql = $sql. "FROM games AS g ";
    $sql = $sql. "JOIN teams AS v ON ( v.id = g.visitorid ) ";
    $sql = $sql. "JOIN teams AS h ON ( h.id = g.homeid ) ";
    $sql = $sql. "JOIN _fields AS f ON (f.id = g.fieldid )";
    $sql = $sql. "WHERE g.gamenum =".$game_num2;
    $query = $db->query($sql);
    if ($db->num_rows($query) > 0) {
      list($gamenum2, $gamedate2, $visitteam2, $hometeam2, $field_name2)=$db->fetch_row($query);
    }
  }
0
 
LVL 27

Expert Comment

by:yodercm
ID: 18885194
Well, for one thing, this line looks strange to me...

  $db = new dbstuff;

I don't think that's correct php syntax.
0
Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

 
LVL 26

Author Comment

by:EddieShipman
ID: 18885432
[quote]$db = new dbstuff;[/quote]

That is a db access class that I got from phpclasses.org.
0
 
LVL 26

Author Comment

by:EddieShipman
ID: 18885437
You didn't have to post all that code, all you had to do was say that I needed
a closing } for the if here:

if ( isset($_POST["gameid_1"]) ) {
  .
  .
  if ($db->num_rows($query) > 0) {
    list($gamenum1, $gamedate1, $visitteam1, $hometeam1, $field_name1)=$db->fetch_row($query);
  }
}
0
 
LVL 26

Author Comment

by:EddieShipman
ID: 18885463
[quote]Can you post the full error message? All we received was "parse error, unexpected $ in..."[/quote]
This is the EXACT message sent:

Parse error: parse error, unexpected $ in /home/austin1/public_html/gamechange.php on line 282
0
 
LVL 29

Expert Comment

by:TeRReF
ID: 18885817
Sometimes it's easier to just post some more code instead of explaining where the closing }'s need to be added :)
0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Introduction HTML checkboxes provide the perfect way for a web developer to receive client input when the client's options might be none, one or many.  But the PHP code for processing the checkboxes can be confusing at first.  What if a checkbox is…
Build an array called $myWeek which will hold the array elements Today, Yesterday and then builds up the rest of the week by the name of the day going back 1 week.   (CODE) (CODE) Then you just need to pass your date to the function. If i…
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.
The viewer will learn how to create a basic form using some HTML5 and PHP for later processing. Set up your basic HTML file. Open your form tag and set the method and action attributes.: (CODE) Set up your first few inputs one for the name and …

862 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

24 Experts available now in Live!

Get 1:1 Help Now