Displaying form Data

[Lambneck]

Hello,

 

I have a single form that collects a user's resume information. The form consists of four fields: the user's 'Name' and 'Email', as well as a 'Subject' and 'Message'. That information is then stored in a mysql database.

 

What I am trying to do is make it so the 'Subject' information of each user is displayed permanently as an individual link on another page in ascending order by Date as they are posted. Then, the more users that fill out the form the more links the page will display.

The displayed links, when clicked on, would take the viewer to another page that would display the complete form data (Name, Email, Subject, Message) that coincides with the Subject link chosen.

 

My question is how can I separate or gather the 'Subjects' from the other collected form data(Name, Email, Subject, Message) and compile them as links to that corresponding form data? ???

 

A demonstration of what I am trying to accomplish can be found <a href="http://www.eslcafe.com/jobs/wanted/#PostMessage">here</a>.

 

Any and all advice will be greatly appreciated.

 

Thanks.

[writer]

Once you store the data, you can manipulate it however you want using your $query string.

[Lambneck]

Thanks for the advice!

However, I am kinda looking for more information than that because I am pretty new to php.

 

How can I specify just the 'Subject' rather than everything in the table?

I tried this code but its not working out for me:

 

<?php

$table = 'Info';

if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");

$result = mysql_query("SELECT subject FROM {$table}");
if (!$result) {
    die("Query to show fields from table failed");
}

while($row = mysql_fetch_row($result))
{

$subject = $row[4];

        echo "Subject : $subject <br>";
}
mysql_free_result($result);
?>

[uniflare]

<?php

$table = 'Info';

mysql_connect($db_host, $db_user, $db_pwd) or die("Can't connect to database: ".mysql_error());
mysql_select_db($database) or die("Can't select database");

$result = mysql_query("SELECT `subject` FROM `$table`") or die("Query to show fields from table failed: ".mysql_error());

while($row = mysql_fetch_array($result))
{
   echo "Subject : $row['subject'] <br />";
}
mysql_free_result($result);
?>

notice the changes i made:

Used mysql_fetch_array instead.

Changed the echo variable.

 

Also:

Changed the error handling statements for mysql functions to or die().

Added mysql_error() function to the error output (to see what mysql said).

Cosmetically changed the mysql query, don't think you need braces for variables in strings.

 

 

hope this helps,

[writer]

Nice work, uniflare.

[uniflare]

thanks 

[Lambneck]

Thank uniflare!