5
hi all.i have faced a problem.
input used:
i have a mysql database named form.there is a table named dynamic.
three attributes here
id ( primary key,auto increment)
name.
company.
Expected result:
i want to show the data of the name field and make them a link to go to another page as like as:
mushfiq
mishuk
the two above name will used to go to the page "dynamic2.php" .As for example ,if i click on mushfiq,the url will be "dynamic2.php" & in that show the company where mushfiq works.but
mushfiq
mishuk
is not showed by ..
error codes:
Parse error: syntax error, unexpected $end in c:\wamp\www\dynamic.php on line 33
[ code ]
<!DOCTYPE html PUBLIC
"-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<head>
<title>
</title>
</head>
<body>
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("form");
$result=mysql_query("select name from dynamic") or die(mysql_error());
if(mysql_num_rows($result)==0)
{
echo "database is empty <br>";
} else
{
while(list($id,$name)=mysql_fetch_array($result))
{
?>
<a href="dynamic2.php?id=<?php $id ?>"> <?php echo $name; ?> </a> <br >
<?php
}
?>
</body>
</html>
[code]
may be problem is in that line:
<a href="dynamic2.php?id=<?php $id ?>"> <?php echo $name; ?> </a> <br >
but i m not sure.if anyone can help me, i will be greatful.
