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Old 08-09-2012, 01:19 PM   #1
lhenzg5
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Registered: Aug 2012
Posts: 1

Rep: Reputation: Disabled
function slideSwitchtop() { var $active = $('#slideshowtop IMG.active'); if


This is the error message:
Parse error: syntax error, unexpected ';'..thanks.

// Display Code in Profile
$query = "SELECT * FROM user WHERE user_id='$user'";
$result = mysql_query($query);

while($row = mysql_fetch_assoc($result))

echo "UserID : {$row['user_id']} <br>" .
"Password : {$row['password']} <br>" .
"First Name : {$row['fname']} <br>" .
"Last Name : {$row['lname']} <br>" .
"Age: {$row['age']} <br>" .
"Gender : {$row['gender']} <br>" .
"Course: {$row['course']} <br>" .
"Year Level: {$row['ylevel']} <br>" .
"Email Add : {$row['email']} <br>" .
"Contacts : {$row['contact']}; <br>" .

?>
 
Old 08-19-2012, 10:42 PM   #2
byannoni
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Registered: Aug 2012
Location: /home/byannoni
Distribution: Arch
Posts: 128

Rep: Reputation: 36
This might not solve the problem, but:
Code:
"Contacts : {$row['contact']}; <br>" . // There should probably be a ';' instead of a '.' here
 
  


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