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Need help with strange error - Parse error: unexpected $end

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#1 johnqpublic24

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Posted 16 July 2003 - 03:25 PM

Thanks for looking - I\'m trying to create a field where people can enter existing records in a database. When the PHP page first loads, they are prompted to enter the number of the record they want to change. When they enter their record, the page reloads and displays the fields of that record in text boxes. The user then has the option of changing the data and clicking the Update button.

I can pull up the record fine, but the code that processes the revised information gives me a strange error message:

Parse error: parse error, unexpected $end in c:public_htmlupdate.php on line 113

Where line 113 is the last line of the page. When I pull out the if (isset($_POST[\'submit\'])): block the code works fine, but with no update when submitted. Does anyone have any ideas what I\'m doing wrong?

<html><head><title> View Client </title></head><body><?php  // Connect to the database server  $dbcnx = @mysql_connect(\'localhost\');  if (!$dbcnx) {    die( \'<p>Unable to connect to the \' .         \'database server at this time.</p>\' );  }  // Select the jokes database  if (! @mysql_select_db(\'online\') ) {    die( \'<p>Unable to locate the \' .         \'database at this time.</p>\' );  }?><?phpif (isset($_POST[\'submit\'])):  $dbcnx = mysql_connect(\'localhost\');  mysql_select_db(\'online\');$Status = $_POST[\'Status\'];  $sql = \"UPDATE data SETStatus=\'$Status\' WHERE SessionID=\'$results\'\";  if (@mysql_query($sql)) {    echo(\'<p>data added</p>\');  } else {    echo(\'<p>Error adding new data: \' .         mysql_error() . \'</p>\');  }?><?phpif (isset($_GET[\'record\'])):$record = $_GET[\"record\"];   echo(\"$record\");      echo(\"<form action=\'$_SERVER[PHP_SELF]\' method=post>\");  echo(\'<table border=1 cellpadding=5><TR bgcolor=\"#e5e5e5\"><TD>SessionID</TD><TD>Student Number</TD><TD>First Name</TD><TD>Last Name</TD><TD>Email</TD><TD>Status</TD></TR>\'); 		$result = @mysql_query(\"SELECT SessionID, StudentNumber, FirstName, LastName, Email, Status FROM data WHERE SessionID=\'$record\'\");while ($row = mysql_fetch_array($result) ) {	$SessionID=$row[\'SessionID\'];$StudentNumber=$row[\'StudentNumber\'];$FirstName=$row[\'FirstName\'];$LastName=$row[\'LastName\'];$Email=$row[\'Email\'];$Status=$row[\'Status\'];		echo(\'<TR bgcolor=\"#e5e5e5\"><TD>\' .$SessionID . \'</TD><TD><input type=text name=StudentNumber value=\' .$StudentNumber . \'></TD><TD><input type=text name=FirstName value=\' .$FirstName . \'></TD><TD><input type=text name=LastName value=\' .$LastName . \'></TD><TD><input type=text name=Email value=\' .$Email . \'></TD><TD><input type=text name=Status value=\' .$Status . \'></TD></TR>\');			}    echo(\'<input type=submit name=submit value=submit></form>\');?><?phpelse: // Allow the user to enter a new author?><form action=\"<?=$_SERVER[\'PHP_SELF\']?>\" method=\"get\"><p>Look up a client (using session ID #):<br />Email: <input type=\"text\" name=\"record\" size=\"20\" maxlength=\"255\" /><br /><input type=\"submit\" value=\"lookup\"></p></form><?php endif; ?></body></html>

#2 holiks

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Posted 16 July 2003 - 04:44 PM

...maybe because of incorrect use of \'if\' statement?
//do something
} else {
//do something else
..is wrong
..but you did use it right the second time which is maybe why that worked..
if (isset($_GET[\'record\'])):
//do something
else: // Allow the user to enter a new author
..use the first if statement u mentioned like u did the second
__________________________________________________________[br]$php = array('..a stack', '...once you pop', '...you don't stop');

#3 johnqpublic24

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Posted 16 July 2003 - 05:03 PM


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