New user to PHP. HELP!!!!!!!

[amarafzal]

i have pulled out some data from an MYSQL database fine. Got it to diplay in a table. What i am stuck on is, when a user clicks on a field (which is a hyperlink) from one of many rows retrieved, i want to be able to go to another webpage and display more information from the database for that specific row?

Need some help with what kind of code i need to do this?

PLEASE HELP

[stolzyboy]

post some of your code you have now so we can get an idea of what you are needing

[amarafzal]

HERE IS SOME OF THE PHP CODE I HAVE DONE

<?

$db_name = "web";
$table_name = "jobs";
$connection = @mysql_connect("217.33.218.155", "webadmin", "r2d2c6po") or die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());

$sql = "SELECT job_id, contact_name, contact_phone, date_posted, job_title, job_summary, location, job_status, career_level FROM $table_name";
$result = @($sql,$connection) or die(mysql_error());

while ($row = mysql_fetch_array($result)) {
$job_id = $row['job_id'];
$contact_name = stripslashes($row['contact_name']);
$contact_phone = $row['contact_phone'];
$date_posted = $row['date_posted'];
$job_title = stripslashes($row['job_title']);
$job_summary = stripslashes($row['job_summary']);
$short_summary = substr($job_summary,0,10);
$location = stripslashes($row['location']);
$job_status = stripslashes($row['job_status']);
$salary = $row['salary'];
$career_level = stripslashes($row['career_level']);

if ($date_posted == "0000-00-00") {
$date_posted = "[not specified]";
}

$display_block .= "
<table width=75% height=35 cellpadding=0 cellspacing=1>
<tr bgcolor=#FFEBD7>
<td width=13% height=3 align=left valign=top>$date_posted
<td width=54% align=left valign=top><a href=latest_jobs_detail.php>$job_title</a><br>$short_summary
<td width=15% align=left valign=top>$location
</tr>
</table>";

}

?>

[stolzyboy]

try this or some variant

<a href=latest_jobs_detail.php?job_title=$job_title>$job_title</a>

then on that page do something like

$job_title = $_GET['job_title'];

then a query like

$query = "select * from table where job_title = '$job_title';";
$result = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($result))
{
echo $row['whateverfieldsyouhaveorwantfromdatabase'];
}

[amarafzal]

SHOULD I USE JOB_ID, AS THAT MAKES EACH JOB UNIQUE OR DOES NOT MATTER

[stolzyboy]

yeah i guess i would use job_id, didn't look that close to your fields, but that would be a more narrow result, let us know if you need help further