Last Record in Database

sully

I am new to PHP . I am creating an intranet which shows active projects and allows people to add new projects.

If users want to add a new project, they fill in a form but what I want the form to display in the ProjectNumber field (which is a var char field & not the primarykey) is the last Job Number incremented by 1.

For example if the last ProjectNumber was TS0014, then the default value in the ProjectNumber field should be TS0015 - this way users know what the next job number is. Hope this makes sense.

Sully

tekky

you would do a SELECT for the last jobnumber then you can do

$job = str_replace("TS",'',$jobnumber); //strip chars from job number
$job++; //increment remaining number portion of job number
$jobnumber = "TS$job"; //combine chars and numbers again for complete job number

ahundiak

Just be sure you create an unique index on project number. Otherwise, you run the risk of getting duplicate numbers when two or more people are using the site to add projects.

sully

Sorry but a bit of dumb question coming up

How would you do a select statement to find the last record?

Sully

jayant

are you storing time somewhere when the project was created? if yes, use that in descending order.
if not
and you have some auto_increment field, sort that in descending order
if still not,
then as per current scenario, sort descending the job number field, should suffice

SELECT jobnumber FROM tablename ORDER BY sortfieldname DESC LIMIT 1

use the obtained result from the query to get last job number

sully

Jayant

thanks for that but i'm still not getting the desired result.

The last job number in my table is TS0014. Below is the code i'm using but for some reason in the Job Number field in the form Im getting TS001? WHY?

<?
$sql1 = "SELECT JobNo FROM Projects ORDER by ProjectID DESC LIMIT 1 ";
$result = mysql_fetch_array($results);
$result2[$i]=$result;
echo "$result2";
$job = str_replace("TS",'',$JobNo); //strip chars from job number
$job++; //increment remaining number portion of job number
$JobNo = "TS00$job"; //combine chars and numbers again for complete job number
?>
<input type="text" name="jobno" value="<? echo "$JobNo" ?>">

Please help.