Thank you for your help. please try to provide more details. I am a newbie.

Thank you for your help. please try to provide more details. newbie & lt; html & gt; & lt; head & gt; & lt; metahttp-equiv = & quot; content-Type & quot; content = & quot; text/html & quot; cha helps you. please try to provide more details.



Handle_posting


$ First_name = $ _ POST ['first _ name'];
$ Last_name = $ _ POST ['last _ name'];
$ Email_address = $ _ POST ['email _ address'];
$ Sex = $ _ POST ['Sex '];
$ Posting = nl2br ($ _ POST ['posting']);
$ Name = $ first_name. ''. $ last_name;
Print "Thank you, $ name for your posting:

$ Posting

Our team will send the lastest meassage to $ email_address.

"
$ Name = urlencode ($ name); // use urlencode for link
$ Email_address = urlencode ($ _ POST ['email _ address']);
Echo"

Clickhereto continue.

";
?>


The following error occurs: Parse error: syntax error, unexpected '$ name' (T_VARIABLE) in C: \ Program Files \ Apache Software Foundation \ Apache2.2 \ htdocs \ handle_posting.php on line 18.

------ Solution --------------------
1. try to use the delimiter when outputting html
2. the quotation marks are incorrectly used.

PHP code

    
      Handle_posting
  $ Posting
Our team will send the lastest meassage to $ email_address.
Html; $ name = urlencode ($ name); // use urlencode to link $ email_address = urlencode ($ _ POST ['email _ address']); echo"
Clickhereto continue.
";?>
------ Solution --------------------

"Add a semicolon, dumb!


";

In addition, use echo output.
------ Solution --------------------

The following error occurs: Parse error: syntax error, unexpected '$ name' (T_VARIABLE) in C: \ Program Files \ Apache Software Foundation \ Apache2.2 \ htdocs \ handle_posting.php on line 18.

This error is almost the case where a semicolon is missing or a quotation mark is missing.
------ Solution --------------------
Less semicolon