PHP - Stupid Errors

I'm sure this is just a stupid mistake, but can someone please help me figure it out? 

I have one page with a dropdown that is populated from our database, I have submit to pull through to the next page but nothing shows up on the next page (app/approval).  I've tried adding some echo statments, but they all show up blank. 

Code: [Select]
<?
require ('includes/getconfig.inc');
require ('includes/auth.inc');
require ('includes/dbfunctions.inc');
include (dirname(__FILE__).'/../includes/menuhead.inc');
include ('apptable_functions.inc');

$sql="SELECT pol_id, pol_policy_number FROM policies";
$result=myload($sql);
$items = count($result);
$options="";

for($x=0; $x<=$items; $x++)
{
$id=$result[$x]['pol_id'];
$thing=$result[$x]['pol_policy_number'];
    $options.="<OPTION VALUE=\"$id\">".$thing;



}
?>

Select a case number to see the payout approval.
<br>
<form action="app/approval.php" method='post'>
<SELECT NAME=pol_policy_number>
<?=$options?>
</SELECT>
<input name="Submit" type="submit" Value="Submit Policy Number"/>
</form>
<br>
Code: [Select]
<?
require ('includes/getconfig.inc');
require ('includes/auth.inc');
require ('includes/dbfunctions.inc');

$pol_number = $_POST['pol_policy_number'];

echo "$pol_number";
?>

Stupid question probably pretty obvious. but..
I am trying to add a href to an image echo And I can't seem to code this right.

Here is a text link echo call:
Code: [Select]
echo '<div id="nav_image_title" align="center" valign="middle" ><a href="'. $sess->url( URL .'index.php?page=shop.browse&amp;category_id=' . $categories[$kp]['category_child_id']). '">'.$categories[$kp]['category_name'].'</a></div>';
And I want to add that "a href" to this image:
Code: [Select]
echo '<div>'. ps_product::image_tag( $categories[$kp]["category_thumb_image"], "alt=\"".$categories[$kp]["category_name"]."\"", 0, "category") . '</div>';
I not sure if it's just too many hours working today..
But how do I add that link to that image.

Thanks for any help anyone can give..
!!!

I'm not sure what I'm doing wrong here.

I'm trying to get this to show the schedule for each person in the database. It does show the schedule, but it only shows one day in each box, and it's the last day.

So, if the last day is October 28th, it populates every box with the schedule for the 28th.

Code: [Select]
<?php

$query  = "SELECT * FROM employees WHERE day > '$current_date' AND day < '$get_date'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result))
{

$id = $row['id'];
$position = $row['position'];
$pay_rate = $row['pay_rate']; 
$total_hours = $row['total_hours'];
$total_pay = $row['total_pay'];
$start_time = date("g:i a", strtotime($row['start_time']));
$end_time = date("g:i a", strtotime($row['end_time']));
$work_day = date("l", strtotime($row['day']));
$day = date("F j", strtotime($row['day']));
if($end_time == "24:00:00"){
$end_time ="Close";};
$schedule = $start_time ."<br />". $end_time;
$first = $row['first_n'];
$last = $row['last_n'];

if ($work_day == "Monday"){
$name[$last.", ".$first] = $schedule;
}
if ($work_day == "Tuesday"){
$name[$last.", ".$first] = $schedule;
}
if ($work_day == "Wednesday"){
$name[$last.", ".$first] = $schedule;
}
if ($work_day == "Thursday"){
$name[$last.", ".$first] = $schedule;
}
if ($work_day == "Friday"){
$name[$last.", ".$first] = $schedule;
}
if ($work_day == "Saturday"){
$name[$last.", ".$first] = $schedule;
}
if ($work_day == "Sunday"){
$name[$last.", ".$first] = $schedule;
}
}
foreach ($name as $n => $s) {
?>     
      <tr>
        <td align="left" valign="top"><?php echo $n; ?></td>
        <td align="left" valign="top"><?php echo $position; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="center" valign="top"><?php echo $s; ?></td>
        <td align="right" valign="top">&nbsp;</td>
      </tr>
<?
}
?>

Can somebody see what I'm doing wrong?

Thanks in advance

Code: [Select]
$code = array("03","09");
if (!in_array($_POST['CODE'], $code));
message("Sorry, not today");
my $_POST['CODE'] is

<input type=hidden name=CODE value='03'>

still not being validated.. hmm

So like When somones enter my site it shows them as a guest, but then if they havn't clicked anywere for about 5minutes and then refresh or if somone else refresh it will delete from the active users list because it's been over 5 minutes, but then if that same person refreshes again my script wont catch them as a GUEST AGAIN!! It just shows no one online, none of my if's catch him and put him in the ibf_session table!


Code: [Select]
//check for cookies
//If no username, Is a Guest, Has COOKIE_ID and cookie USER_NAME and COOKIe PASSWORD
if(!isset($_SESSION['user_name']) && !isset($_SESSION['is_guest']) &&
 isset($_COOKIE['user_id']) && isset($_COOKIE['user_name']) && isset($_COOKIE['password'])){
$user_cond = "user_name='{$_COOKIE['user_name']}'";

$sql = "SELECT `id`,`user_name`,`approved` FROM users WHERE
           $user_cond
AND `pwd` = '{$_COOKIE['password']}' AND `banned` = '0'
";
$result = mysql_query($sql) or die (mysql_error());
$num = mysql_num_rows($result);
  // Match row found with more than 1 results  - the user is authenticated.
    if ( $num > 0 ) {
list($id,$full_name,$approved) = mysql_fetch_row($result);
if(!$approved) {
echo "YOU FREAKING HACKER";
exit();
}
$_SESSION['user_id']= $id; 
$_SESSION['user_name'] = $full_name;
$lol = session_id();
//$minute5 = 30 ? (time() - 30) : (time() - 3600);
           // mysql_query("DELETE FROM ibf_sessions WHERE running_time < {$minute5}");
mysql_query("INSERT INTO ibf_sessions (id, member_name, member_id, ip_address, running_time, location, member_group) ".
                       "VALUES ('".$lol."', '{$_SESSION['user_name']}', '{$_SESSION['user_id']}', '{$_SERVER['REMOTE_ADDR']}', '".time()."', ".
                       "'Index', '1')") or die(mysql_error());
unset($_SESSION['is_guest']);
   }
}
$lol = session_id();
//Thanks
if (!$id) { //Create Guest Session ~
if (!isset($_SESSION['user_id']) && @(!$_SESSION['is_guest']) && !isset($_COOKIE['password'])) {
//$minute5 = 39 ? (time() - 30) : (time() - 3600);
//echo $minute5;
           // mysql_query("DELETE FROM ibf_sessions WHERE running_time < {$minute5}");

mysql_query("INSERT INTO ibf_sessions (id, member_name, member_id, ip_address, running_time, location, member_group) ".
                       "VALUES ('".$lol."', 'Guest', '0', '{$_SERVER['REMOTE_ADDR']}', '".time()."', ".
                       "'Index', '0')") or die(mysql_error());
    $_SESSION['is_guest'] = 'yes';
}
}
if (isset($_SESSION['user_id']) && !isset($_SESSION['is_guest'])) {
 mysql_query("UPDATE ibf_sessions SET member_name='{$_SESSION['user_name']}',member_id='{$_SESSION['user_id']}',member_group='1', running_time='".time()."', in_forum='".$f."', in_topic='".$topicid."', location='".$act."' WHERE id='".$lol."'") or die(mysql_error());
  }
else {
mysql_query("UPDATE ibf_sessions SET member_name='Guest',member_id='0',member_group='0', running_time='".time()."', in_forum='".$f."', in_topic='".$topicid."', location='".$act."' WHERE id='".$lol."'") or die(mysql_error());

 
 }

Thanks, it might be confusing but re read

Hi guys,

im needing to grab information out of a mysql database field called "clientname" thats inside a table called "clients"

how do i retrieve the clients name in html and display it?

heres what i have so far, am just needing the code to retrieve...



$host = localhost;
$dbuser = user;
$dbpass = pass;
$dbname = mydb;

$connection = mysql_connect($host, $dbuser, $dbpass);
$db = mysql_select_db($dbname, $connection);

// code to retrieve a particular clients details

 




Cheers,

I have a bunch of quasi-static values that must be available to the application.  Most of them are set based on settings in a configuration file.  Others are based on GET, COOKIE, or SESSION values, and utilize the database to get the actual values.  The values will never be written to or modified by the application, only read.   It seems to me that I could create some sort of superclass which includes static methods and properties, and I could access any value by something like superclass::get('some.value');  I could design the class so that the values are queried from the DB or obtained from a parsed file only the first time they are requested, and for future requests, retrieved from a static property.   That being said, I have been told from more than one person that I am just doing it "wrong".   Please let me know what is wrong about it.   Thank you

all i am doing is trying to left join my column row  "to_id" and "from_id"

i can successfully left join to_id but once i try to add "from_id" to the "ON" clause it makes all my data blank

Code: [Select]
SELECT gold_logs.*,name,star
FROM gold_logs LEFT JOIN ibf_members ON to_id = ibf_members.id
WHERE to_id = 1


That query works fine, but when i try to add my "from_id" column the query works with no error's but no results are shown

Code: [Select]
SELECT gold_logs.*,name,star
FROM gold_logs LEFT JOIN ibf_members ON (to_id = ibf_members.id AND from_id = ibf_members.id)
WHERE to_id = 1

Hi There

I have a simple cms system that I have built but for some reason the login script wont let me login to the cms. Strange thing is the script seems to works on the original database but when I changed the details to a new db it doesnt work.

I have made the necessary changes for the connecting to the db etc but it just wont let me login.

Can anyone help

The url is drmonlinemarketing.com/cms2/loginadmin.php

try login with username demo
                                      password test


thanks

shalli

Okay, So I have a dropdown box:
Code: [Select]
<select name="imgsize">
 <option value="small">480 x 360</option>
 <option value="medium">720 x 540</option>
 <option value="big">1200 x 900</option>
 <option value="large">1440 x 1080</option>
 </select>
and I have code that gets the size:
Code: [Select]
$resizeimg = $_REQUEST["imgsize"];
But I cannot get it to work with an if statement:
Code: [Select]
if ($resizeimg == "large")
    {
    }

It just comes up with about 50 lines of gibberish. I have used error_reporting(-1); but that is not coming up with any errors.
Could anyone help please. Its probably so simple, but my mind is drawing a blank

Hi there,

Can anyone show me how I can grab the value "c87cc576092fe1a0f5ac3d50091694b4" out of this string, it changes.

Also its coming from a socket a $buffer variable, so How can I do this without grabbing everything else aswell, ie all the data after it too

Quote

$buffer = fgets($fp, 4096);      

   /*** recieve data looks like this - <dd><input type="hidden" name="sid" value="c87cc576092fe1a0f5ac3d50091694b4" />

    fclose($fp);
}

fclose($Handle);



If someone could show me how to do this, I think you can do it with the split function, i just dont know how. if someone could show. i'd be SO greatful

How do I display the actual errors in IIS 7.5?

If I miss a semicolon, I get:

"HTTP Error 500.0 - Internal Server Error"

absolutely useless. In prior versions, I could see the line and get to the PHP error.

How do I display PHP errors?

I've added:

set_ini('display_errors', '1');

but it doesn't help.

i am trying to make a page to upload music to my website and add the information of the songs into a database, but i cannot get php to let me upload a file bigger than 2mb even after i went into the php.ini and changed the max_upload_size and a few others to 99mb.
So i was wondering if i could put my music in a folder on my website and pull the data from the items in the folder and make database entries(probably doubtful) or if i could get some help with getting php to upload bigger files?

thanks,
Adam

<?php
$result = mysql_query ("SELECT genre FROM genres ORDER BY genre");
 

while ($row = mysql_fetch_row($result)) {
    echo "<tr>";
for ($i =0; $i<5; i++) {
foreach ($row as $genre)
        echo "<td><input type='submit' name='listMovies' value='" .$genre ."'></td>"; 

}  

echo "</tr>";
}
?>

ok i'm getting a single line of submit buttons for this output. i want there to be 5 buttons in a row.  it seems like i have done this before with the for loop i have, but something is not working right.  i know it will have to be a for loop, i just can't figure out where to start it.  i know there are 1000 different ways to get the buttons on the screen, but i want to do it with the submit buttons.  the best i can get is 5 columns of the same thing i'm outputting now.

thanks in advance, i know this is a no-brainer.  i just don't have a brain atm

     $sql1 = "INSERT INTO `playerdb` ( `ID` , `User_ID` , `User_PW` , `Rx` , `Ry` , `Rz` , `Tx` , `Ty` , `Level` , `XP_total` , `XP_spent` , `qp` , `class` , `Base_h` , `Base_t` , `Base_ar` , `Base_al` , `Base_l` , `Base_b` , `Bday` , `Creation_time` , `Last_login` , `Admin` , `Items` , `Worn` , `att` , `att_xp` , `agi` , `agi_xp` , `dex` , `dex_xp` , `vit` , `vit_xp` , `mag` , `mag_xp` , `skill_name` , `skill_xp` ) VALUES ('', '$_POST['uid']', '$_POST['upw']', '$Rx', '$Ry', '$Rz', '$Tx', '$Ty', '1', '0', '0', '0', '$_POST['classa']', '$_POST['haira']', '$_POST['bodya']', '$_POST['armsa']', '$_POST['armsa']', '$_POST['legsa']', '$_POST['bootsa']', '$_POST['age_yeara']-$_POST['age_montha']-$_POST['age_daya']', NOW( ) , NOW( ) , '0', '', '', '5', '0', '5', '0', '5', '0', '5', '0', '5', '0', '$skilllvls', '$skillexps')";