how to link a database item to a link - php

Question

I am designing a system for a recruitment agency and have a search page where the "applicants" can search for a job and apply, the search results is structured:
-Job_Title
-Job_description
-Apply link
This is the same for every result. Code for the above mentioned:
$num_rows = mysql_num_rows($query) or die(mysql_error());
if ($num_rows > 0) {
while ($row = mysql_fetch_assoc($query)) {
$title = $row['Adv_title'];
$description = $row['Adv_desc'];
echo "<h2><a href='#'> $title</a> </h2>";
echo "$description <br /> <br />";
echo"<a href='../View/applicationForm.php' class='link-button right'><span>Apply</span></a>";
Now my problem is that I don't know how to make the Apply link go to each Jobs respective Application form. How can I make the link different for each advert or link the application form to Adv_title in the database?

Answer 1

In order to identify each job in applicationForm.php, you need to pass a parameter that identifies each job via $_GET. This parameter should be unique for each job, so it could be the id field in database that you mention, or it could be the title, assuming that you are checking title uniqueness whenever you create a new job.
So, if you want to use id, your code would look like this (assuming that your field is named id):
while ($row = mysql_fetch_assoc($query)) {
$title = $row['Adv_title'];
$description = $row['Adv_desc'];
$id = $row['id'];
echo "<h2><a href='#'> $title</a> </h2>";
echo "$description <br /> <br />";
echo"<a href='../View/applicationForm.php?id=$id' class='link-button right'><span>Apply</span></a>";
If, for any reason, you don't want end users to view your ids, you can always encrypt them before showing, and decrypt them in applicationForm.php.

Answer 2

In hyperlink tag give the value of the link to redirect to instead of #.
Like you have Adv_Title and Adv_desc field you will need to keep the link too and then you can fetch it and use it.
while ($row = mysql_fetch_assoc($query)) {
$title = $row['Adv_title'];
$description = $row['Adv_desc'];
$link=$row['Adv_link'];
echo "<h2> $title </h2>";

Answer 3

$num_rows = mysql_num_rows($query) or die(mysql_error());
if ($num_rows > 0) {
while ($row = mysql_fetch_assoc($query)) {
$title = $row['Adv_title'];
$description = $row['Adv_desc'];
$id = $row['Adw_id'];
$link = "/$id/View/applicationForm.php";
echo "<h2> $title </h2>";
echo "$description <br /> <br />";
echo"<a href='../View/applicationForm.php' class='link-button right'><span>Apply</span></a>";

Answer 4

if application from is different according to the job adv, you better should have a field in the database table to consider about different application form according to each Adv_title.
while ($row = mysql_fetch_assoc($query)) {
$title = $row['Adv_title'];
$description = $row['Adv_desc'];
$id = $row['Adw_id'];
//if $row['url_link'] is full url
$link = $row['url_link']; // full link
//if $row['url_link'] is just and identifier
// some conditions goes here
......
echo "<h2> $title </h2>";
echo "$description <br /> <br />";
echo "link";

Answer 5

If you really want to hide the id, you can set a server side session variable holding the current job id. And the pick it up on the page that generates the application form. ALternatively, just include it as part of the URL as manuelpedrera shows.