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I have been staring at this code for hours now and I cannot figure out where my mistake is. I know this syntax error usually comes up because of some missing or out of place curly brace or some issue with single/double quotes and I'm not sure there is one anywhere in my code. I am only trying to fix my syntax right now so I can get the code to completely compile. Any help would be much appreciated. Here is my code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <title>Add to and Read from the Database</title> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" /> </head> <body> <?php function print_form() { echo <<<END <form action="$_SERVER[PHP_SELF]" method="post"> <h3>Please put your comments below.</h3> <input type="hidden" name="stage" value="process" > <p>Name:</p> <input type="text" size="30" name="WholeName" /> <p>Comment:</p> <input type="text" size="200" name="Comment" /> <input type ="submit" value ="Submit" > </form> END; } function process_form() { print "<p>adding comment...</p>"; $Name = $_POST['WholeName']; $Comment = $_POST['Comment']; if( preg_match("^[a-zA-Z]+$", $Name)) { if( preg_match("^[a-zA-Z0-9]_\-\'[.?!]+$", $Comment)) { $sql = "insert into comments1 values ( '$Name', '$Comment')"; $result = mysql_query($sql) or die("Mysql query failed"); } else { print "invalid name"; } } else { print "invalid characters"; } } $db = mysql_connect("", "", ""); if (!$db) { print "Error - Could not connect to mysql"; exit; } $er = mysql_select_db(""); if (!$er) { print "Error - Could not connect to comments1 database"; exit; } if (isset($_POST['stage']) && ('process' == $_POST['stage'])) { process_form(); } else { print_form(); } ?> </body> </html>
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  • 14
    It's good practice to state where exactly the error is being reported (i.e., Line number and column number). Commented Dec 6, 2011 at 12:44
  • this is indeed "too localized" question Commented Dec 6, 2011 at 12:48
  • Just a note to save yourself a lot of hair-pulling: These kind of errors can easily be tracked down by using a good IDE that inspects your code. Commented Dec 6, 2011 at 12:52
  • FWIW, the line in question is $Name = $_POST['WholeName']; codepad.viper-7.com/NEJqgK Commented Dec 6, 2011 at 12:53
  • It's telling me that the error is occurring on line 32. Can you recommend a good IDE for me to use? Commented Dec 6, 2011 at 12:54

4 Answers 4

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50

I stumbled on this question as I had the same error. Mine was due to a slightly different problem and since I resolved it on my own I thought it useful to share here. Original code with issue:

$comment = "$_POST['comment']";

Because of the enclosing double-quotes, the index is not dereferenced properly leading to the assignment error. In my case I chose to fix it like this:

$comment = "$_POST[comment]";

but dropping either pair of quotes works; it's a matter of style I suppose :)

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1 Comment

In my case it was a stray ` backtick. Geany's syntax highlighter didn't notice it. But Joe's syntax highlighter caught it. Initially your answer helped but I was really fighting an imaginary enemy all along ;-)
42

Your problem is that you're not closing your HEREDOC correctly. The line containing END; must not contain any whitespace afterwards.

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1 Comment

And it must not contain any whitespaces BEFORE the END; either, as I just found out when trying to format my code for readability.
4

You have extra spaces after END; that cause the heredoc not terminated.

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0

Might be a pasting problem, but as far as I can see from your code, you're missing the single quotes around the HTML part you're echo-ing.

If not, could you post the code correctly and tell us what line is causing the error?

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