how to call the value in mysql then turn into a variable - php

Question

Here is my code
<?php
require("db.php");
$datetoday = date("Y-m-d");
if (isset($_POST['submit']))
{
include 'db.php';
$loginid =$_REQUEST['loginid'];
$result = mysql_query("SELECT * FROM info WHERE id = '$loginid'");
$test = mysql_fetch_array($result);
$testid=$test['id'];
$fnameloginsuccess1=$test['firstname'];
$mnameloginsuccess1=$test['middlename'];
$lnameloginsuccess1=$test['lastname'];
$departmentloginsuccess1=$test['department'];
echo'<input type="text" name="fname" value="<?php echo $fnameloginsuccess1 ?>"/></td>';
if (!$loginid)
{header("location:../index.php"); }
$natureofleave =$_POST['group1'];
$datestart=$_POST['startofleave'];
$dateend=$_POST['endofleave'];
$reason=$_POST['reason'];
$status= 'pending';
mysql_query("INSERT INTO `request`(id,natureofleave,dateofleavestart,dateofleaveend,reasons,datesubmitted,department,status,firstname,middlename,lastname)
VALUES('$log','$natureofleave','$datestart','$dateend','$reason','$datetoday','$departmentloginsuccess1','$status','$fnameloginsuccess1','$mnameloginsuccess1','$$lnameloginsuccess1')");
}
my main problem is i can't put the value of $fnameloginsuccess1, $mnameloginsuccess1','$lnameloginsuccess1',$departmentloginsuccess1 on my database..
but i can "ECHO" them.. some values are working but the 4 values didn't work!!
i already tried fname = $fnameloginsuccess1'; sadly to say it didn't work..
HELP!!

Answer 1

<?php
require("db.php");
$datetoday = date("Y-m-d");
if (isset($_POST['submit']))
{
include 'db.php';
$loginid =$_REQUEST['loginid'];
if (!$loginid) {header("location:../index.php"); }
$result = mysql_query("SELECT * FROM info WHERE id = '$loginid'");
$test = mysql_fetch_array($result);
$testid=$test['id'];
$fnameloginsuccess1=$test['firstname'];
$mnameloginsuccess1=$test['middlename'];
$lnameloginsuccess1=$test['lastname'];
$departmentloginsuccess1=$test['department'];
echo'<input type="text" name="fname" value="'.$fnameloginsuccess1.'"/></td>';
$natureofleave =$_POST['group1'];
$datestart=$_POST['startofleave'];
$dateend=$_POST['endofleave'];
$reason=$_POST['reason'];
$status= 'pending';
mysql_query("INSERT INTO `request` (id, natureofleave, dateofleavestart, dateofleaveend, reasons, datesubmitted,department,status,firstname,middlename,lastname) VALUES('$log','$natureofleave','$datestart','$dateend','$reason','$datetoday','$departmentloginsuccess1','$status','$fnameloginsuccess1','$mnameloginsuccess1','$lnameloginsuccess1')");
}
?>
Consider to use PDO statements as mysql_query is deprecated since PHP 5.5.0 and will be removed in the future.
http://www.php.net/manual/en/function.mysql-query.php
PDO connection examples
http://www.code.rusben.com/php-pdo-connection-with-utf8-compatibility-select-insert-update-delete/

Answer 2

<?php
require_once("db.php");
$datetoday = date("Y-m-d");
if ($_SERVER['REQUEST_METHOD'] == "POST")
{
$loginid = $_REQUEST['loginid'];
if (!$loginid)
{
header("Location: ../index.php");
exit;
}
$result = mysql_query("SELECT * FROM `info` WHERE `id` = '$loginid'");
$user = mysql_fetch_array($result);
$id = $user['id'];
$first = $user['firstname'];
$middle = $user['middlename'];
$last = $user['lastname'];
$dept = $user['department'];
$nature = $_POST['group1'];
$start = $_POST['startofleave'];
$end = $_POST['endofleave'];
$reason = $_POST['reason'];
$status = 'pending';
$sql = <<<SQL
INSERT INTO `request`
(`id`, `natureofleave`, `dateofleavestart`, `dateofleaveend`, `reasons`, `datesubmitted`, `department`, `status`, `firstname`, `middlename`, `lastname`)
VALUES
('$id', '$nature', '$start', '$end', '$reason', '$datetoday', '$department', '$status', '$first', '$middle', '$last');
SQL;
mysql_query($sql) or die ('There was an error processing your data.');
}
?>
A few points I feel the need to point out:
As you "require" the db.php, you should not need to "include" it.
When naming variables, it is best to keep them simple. Easier to debug and track down.
Exit the script after a header redirect. A delay in the header could allow further code to execute.
You can not use PHP tags inside of PHP tags - it just doesn't parse that way
I'd advise to write the SQL outside of the mysql_query() wrapper, since you can then echo out the SQL
Which can't be done if you write direct inside mysql_query()
log isn't defined, so it won't input. I'll assume that should be the users ID and edit to suit.
You had 2 dollar signs in the query (lnameloginsuccess1)
Anyway, if you run the above code and get "There was an error processing your data.", you can debug this pretty easily.
Change
mysql_query($sql) or die ('There was an error processing your data.');
to
mysql_query($sql) or die (mysql_error());
If the error it reports is vague, you tend to get better results running the query direct into the admin panel (PhpMyAdmin and the likes), so do;
On the line before the mysql query, simply add "echo $sql;" and run the page again. Copy the entire output of the query and run in your database admin panel.
If there is no error there, you need to be looking at connection issues - like errors in connection data