How to link a record to a detailed report

[Geli19]

Hello everyone.

I have build a simple MS Access database to keep track of volunteer info (contact, address, phone etc). I have a report that displays each record in a list:

record1. Last Name, First Name, phone, email
record2. Last Name, First Name, phone, email
etc...

What I want to accomplish is to have either the Volunteer's name in this record be a Link or add a link itself to the recod that when you click it it takes you to an individual detailed report that will display more accurate information about the volunteer.

I know I can create a button and link it to a different form/report/query... but I am not sure this will work as intended since the report does not show all the volunteers rather the ones that meet a specific criteria, therefore Ineed this to be available for every single record regardles of the report or query that is being run.

I built a form that displays all the detailed information of the volunteers and you can scroll through it to view them.... is there a way the link can actually point to this form, but the form shows the actuall volunteer selected instead of the standard default first volunteer?

Any help and ideas will be greatly appreciated.
Thanks!

[fobos]

yes, since i know php any mysql, i would use that. just display that data on one page have have them echo a link that uses a keyword or id to specifically display the info on that person. Ex

Members.php
<?php
//database connection
// select statement for database

//from w3schools.com with some modding
<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM Persons");

while($row = mysql_fetch_array($result))
  {
  echo "<a href='members2.php?id=".$row['id']."'>".$row['name']."</a>";
  }

mysql_close($con);
?>

Members 2.php
<?php
$id = $_POST['id'];
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM Persons WHERE id = $id");

//took out while so it doesnt repeat your output.
$row = mysql_fetch_array($result)
?>
  //display the info of what you want to display
 <table><tr><td><?php $row['name'] ?></td></tr></table>
 <?php

mysql_close($con);
?>

Hope this helps

[Geli19]

Thanks a lot!!! I was able to work everything. :)