5
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
Expand|Select|Wrap|Line Numbers
- <!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>
<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.
