PHP - Problems With Session Variables On .php5 Pages

Hi guys,

I've been going crazy over this and I'm hoping someone can help me. I'm making a website for a friend and I'm stuck on sending the model over to the contact page (from one form to another on a different page using $_POST).

Here is the website : www.ipgetter.com/test

I have successfully posted the Make variable to the contact us page, but I can't get the model sent over (the select boxes have different names).

Example:

Code: [Select]
<select class="form-dropdown" style="width:150px" id="input_2" name="q2_make">
            <option selected="selected" value="Alfa Romeo"> Alfa Romeo </option>
            <option value="Aston Martin"> Aston Martin </option>
            <option value="Audi"> Audi </option>
            <option value="Bently"> Bently </option>
            <option value="BMW"> BMW </option>
            <option value="Chevrolet"> Chevrolet </option>
            <option value="Chrysler"> Chrysler </option>
            <option value="Citroen"> Citroen </option>
            <option value="Ferrari"> Ferrari </option>
            <option value="Fiat"> Fiat </option>
            <option value="Ford"> Ford </option>
            <option value="Honda"> Honda </option>
            <option value="Hyundai"> Hyundai </option>
            <option value="Infiniti"> Infiniti </option>
            <option value="Isuzi"> Isuzi </option>
            <option value="Jaguar"> Jaguar </option>
            <option value="Jeep"> Jeep </option>
            <option value="Kia"> Kia </option>
            <option value="Lamborghini"> Lamborghini </option>
            <option value="Land Rover"> Land Rover </option>
            <option value="Lexus"> Lexus </option>
            <option value="Lotus"> Lotus </option>
            <option value="Maserati"> Maserati </option>
            <option value="Mazda"> Mazda </option>
            <option value="Mclaren"> Mclaren </option>
            <option value="Mercedes-Benz"> Mercedes-Benz </option>
            <option value="MG"> MG </option>
            <option value="Mini"> Mini </option>
            <option value="Mitsubishi"> Mitsubishi </option>
            <option value="Nissan"> Nissan </option>
            <option value="Peugeot"> Peugeot </option>
            <option value="Porsche"> Porsche </option>
            <option value="Renault"> Renault </option>
            <option value="Rolls-Royce"> Rolls-Royce </option>
            <option value="Seat"> Seat </option>
            <option value="Skoda"> Skoda </option>
            <option value="Smart"> Smart </option>
            <option value="Ssangyong"> Ssangyong </option>
            <option value="Subaru"> Subaru </option>
            <option value="Suzuki"> Suzuki </option>
            <option value="Toyota"> Toyota </option>
            <option value="Vauxhall"> Vauxhall </option>
            <option value="Volkswagen"> Volkswagen </option>
            <option value="Volvo"> Volvo </option>
            <option value="Other"> Other </option>
          </select>
That's the make, as for the model, the select boxes have different names so how am I supposed to capture the users selection and POST it?

Code: [Select]
<li class="form-line" id="id_3">
        <label id="label_3" for="input_3"></label>
        <div id="cid_3" class="form-input">
          <select class="form-dropdown" style="width:150px" id="input_3" name="q3_model">
            <option selected="selected" value="147"> 147 </option>
            <option value="159"> 159 </option>
            <option value="8C"> 8C </option>
            <option value="Breva"> Breva </option>
            <option value="Giulietta"> Giulietta </option>
            <option value="Mito"> Mito </option>
            <option value="Other"> Other </option>
          </select>
        </div>
      </li>
      <li class="form-line" id="id_4">
        <label id="label_4" for="input_4"></label>
        <div id="cid_4" class="form-input">
          <select class="form-dropdown" style="width:150px" id="input_4" name="q4_model4">
            <option selected="selected" value="Cygnet"> Cygnet </option>
            <option value="DB9"> DB9 </option>
            <option value="DBS"> DBS </option>
            <option value="Rapide"> Rapide </option>
            <option value="Vantage"> Vantage </option>
            <option value="Virage"> Virage </option>
            <option value="Other"> Other </option>
          </select>
        </div>
      </li>
I want to post only the selected make/model depending on the user selection. So my question is, how can I send the users chosen model selection over to the contact page form where there is more than one different select box? These select boxes are conditional so when you select "other" as make or model make/model, you may type one in. I'd like to send that also.

Any help is appreciated guys, thanks so much.

I'm trying to make a "sort by" script to sort posts. I'm hoping someone can help me figure out a small glitch. For some reason I'm getting multiple variables in the url as if the page doesn't reload. 

The code is

$uri = $_SERVER['REQUEST_URI']; //gets url of current page

      <li><a href="<?php echo $uri; ?>&sort=New">Most Recent</a></li>
        
        <li><a href="<?php echo $uri; ?>&sort=Popular">Popular</a></li>
        
        <li><a href="<?php echo $uri; ?>&sort=Discussed">Discussed</a></li>

if for example I choose to sort by "Popular" the url output will be

Code: [Select]
http://localhost/mysite/single_cat.php?cat=57&sort=Popular
but then when I click let's say "Discussed" a new sort variable is added with the existing one.

Code: [Select]
http://localhost/mysite/single_cat.php?cat=57&sort=Popular&sort=Discussed

and below I'm querying posts based on the sort variable.. for example..

if ($sort == "Popular") {

      //db connection

$sql = "SELECT * FROM posts WHERE cat_id = '$cat' AND views >= 2 ORDER BY views DESC, post_date DESC";

     //while loop, etc

}//end else if


I'm not sure what I did wrong.. How do I get just one "?sort=" url variable?

Hi there,   I want to pass a input variable from login_success.php which will be sent to sqlprocess.php as the variable 'SQLinput';   sqlprocess.php

$link = mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$query = $_REQUEST['SQLinput']; //You don't need a ; like you do in SQL
$result = mysql_query($query);
$numfields = mysql_num_fields($result);

login_sucess.php

<form action="" method="post"
   <label>
         <span>SQL Input :</span>
         <input type ="text" id="message" name="SQLinput" placeholder="Input SQL"></textarea>
     </label> 
	 <label>
         <span>SQL Output :</span>
         <output id="text" id="SQLoutput" ></input>	

<script type="text/javascript" charset="utf-8"> 
			
		// handles the click event for link 1, sends the query
function getOutput() {
  getRequest(
      'sqlprocess.php', // URL for the PHP file
       drawOutput,  // handle successful request
       drawError    // handle error
  );
  return false;
}  
// handles drawing an error message
function drawError () {
    var container = document.getElementById('output');
    container.innerHTML = 'Bummer: there was an error!';
}
// handles the response, adds the html
function drawOutput(responseText) {
    var container = document.getElementById('output');
    container.innerHTML = responseText;
}
// helper function for cross-browser request object
function getRequest(url, success, error) {
    var req = false;
    try{
        // most browsers
        req = new XMLHttpRequest();
    } catch (e){
        // IE
        try{
            req = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            // try an older version
            try{
                req = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                return false;
            }
        }
    }
    if (!req) return false;
    if (typeof success != 'function') success = function () {};
    if (typeof error!= 'function') error = function () {};
    req.onreadystatechange = function(){
        if(req .readyState == 4){
            return req.status === 200 ? 
                success(req.responseText) : error(req.status)
            ;
        }
    }
    req.open("GET", url, true);
    req.send(null);
    return req;
}
</script>

<a href="#" onclick="return getOutput();"><button type="submit" id="search_btn" value="Submit">Submit</button> </a>
<div id="output">waiting for action</div>		

Prior to the JS code - I was able to peform this action - however the JS code enables me to post the query onto the same page. I would like essentially like to query the database through an input box and output the same result on the same page.   Thanks!  

Hi guys,

 

I am having a problem transferring data between pages, using PHP sessions. I have a HTML SELECT drop-down to let user set the number of posts per page to display.

<form method="POST" style="margin:auto;position:relative;width:35%;padding:17px;top:47px;">
          			<span>select posts-per-page:</span>
          			<select name="ppp">
          				<option selected disabled>--select ppp--</option>
          				<option value=10>10</option>
          				<option value=15>15</option>
          				<option value=20>20</option>
          				<option value=50>50</option>
          				<option value=100>100</option>
          			</select>
          			<input type="submit" name="submit" value="OK"/>
          		</form>

I want to transfer the value selected by user to a session so that I can load the value in other pages. The value of posts per page is carried by variable $ppp; so, file1 contains the code

$_SESSION['ppp'] = $ppp;

That seems to work great; the system updates correctly the session's value as user submits different numbers of posts-per-page.

 

The problem is in page2.php: it doesn't update to the new user's selection; I already have a session active in that page so I don't need session_start(). To call the session, I use the line

$ppp=$_SESSION['ppp'];

When user changes the selection from 50 to 100 posts per age, for example, the value of page1.php changes as expected, but the value transferred to page2.php doesn't change, appears to remain with the default value the page loads with. I don't know if the problem is that I already have a session going. Can you have more than one session running simultaneously in a page?

Can anybody help? Thanks in advance!

 

Sept

I am creating a webpage for my golf league and some of the information(variables) such as Name, Address and email, must be reused across multiple pages for different outputs.  May question is "Can I, or is there a way to reUse variables across multiple pages.   "Do I need to create a session and make everything session variables"?  "Should I link to a database and call the information when needed"? 

Any explanation would be greatly appreciated. 

Hi

Can some steer me in the right direction.  I am having issues subtotaling 2 session variables:

<td>'.$_SESSION['itemqty'][$key].'*'.$_SESSION['itemprice'][$key].'</td>

Can you advise if I am doing this right?

Cheers

Say I have a dropdown menu consisting of user specific options. I need to be able to set a session variable depending on the option chosen & then re-access it on the following page. Can someone give me a crash course on how to utilize session variables?

they recently reset my website so my php.ini file has been adjusted. One of my sites is no longer passing a session variable to the next page. I am sure its one of the settings being reset. here is the data from my phpinfo file. Anything that looks off to anyone?

session
Session Support    enabled
Registered save handlers    files user sqlite
Registered serializer handlers    php php_binary wddx

Directive   Local Value   Master Value
session.auto_start   Off   Off
session.bug_compat_42   On   On
session.bug_compat_warn   On   On
session.cache_expire   180   180
session.cache_limiter   nocache   nocache
session.cookie_domain   no value   no value
session.cookie_httponly   Off   Off
session.cookie_lifetime   0   0
session.cookie_path   /   /
session.cookie_secure   Off   Off
session.entropy_file   no value   no value
session.entropy_length   0   0
session.gc_divisor   100   100
session.gc_maxlifetime   1440   1440
session.gc_probability   1   1
session.hash_bits_per_character   4   4
session.hash_function   0   0
session.name   PHPSESSID   PHPSESSID
session.referer_check   no value   no value
session.save_handler   files   files
session.save_path   /home/users/web/b535/ipw.plevtech/public_html/cgi-bin/tmp   /home/users/web/b535/ipw.plevtech/public_html/cgi-bin/tmp
session.serialize_handler   php   php
session.use_cookies   On   On
session.use_only_cookies   Off   Off
session.use_trans_sid   0   0

...anyone know anything about Internet Explorer not printing values from session variables.  It works on Firefox but not on Internet Explorer.  The session variable value works find on the web page but click on Print preview and it dissapears and will not print.

Hi.  I'm brand new to this forum, so sorry if I do something wrong...

I'm someone who has thought myself to code in php to add dynamic objects to my basic static websites.  I have had no formal education.  This is not my first project however. 

I'm having problems with session variables working fine up to a point when I call a new file and then all my session variables are lost.  I have tried everything, but cannot keep them alive when I get into this or any other new file.

I display a form, get user data submitted, then display a next form and after the user submits this next form all the session variables are lost.

Anyone have any idea what is killing my session variables?

Thanks in advance.

Hi,

I am facing problems in destroying the session variables in my logout file  .I have used session_id() in my php files for ex $sid = session_id().Now when I click on logout I naviagte to my index file but the session variable $sid still remain.I have chkd it.I have kept session_start() function in a file and I am including that file in every php by require_once function.The file is palced in Includes folder..is it causing the problem?? please help me out

Also can I use session_start() more than once in a single php file?

My logout file is :

<?php
        session_start();
   session_destroy();
   header("location:index.php");
   exit();
?>

Hi All! I've written up a script for my website. It\ is basically a virtual job quest. My queries are all correct it just isn't registering the variable for the session. It is $-SESSION[theid']. I want to be bale to use it in my table but I get an error. How do I write this in my SQL query for it to work. The page (when no errors), doesn't show my data. Here is my ocde:

Code: [Select]
<?php
session_start();
include("config536.php");
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css" />
</head>

<?php
if(!isset($_SESSION['username'])) {
echo "<ubar><a href=login.php>Login</a> or <a href=register.php>Register</a></ubar><content><center><font size=6>Error!</font><br><br>You are not Logged In! Please <a href=login.php>Login</a> or <a href=register.php>Register</a> to Continue!</center></content><content><center><font size=6>Messages</font><br><br></center></content>";
}

if(isset($_SESSION['username'])) {
echo "<nav>$shownavbar</nav><ubar><img src=/images/layout/player.gif><a href=status.php>$showusername</a>.......................<img src=/images/layout/coin.gif> $scredits</ubar><content><center><font size=6>Basic Quests</font><br><br>";

$startjob = $_POST['submit'];
$jobq = "SELECT * FROM jobs WHERE username='$showusername'";
$job = mysql_query($jobq);
$jobnr = mysql_num_rows($job);

if($jobnr == "0") {
?>
<form action="<?php echo "$PHP_SELF"; ?>" method="POST">
<input type="submit" name="submit" value="Start Job"></form>
<?php
}
if(isset($startjob)) {
$initemidq = "SELECT * FROM items ORDER BY RAND() LIMIT 1";
$initemid = mysql_query($initemidq);
while($ir = mysql_fetch_array($initemid)) {
$ids = $ir['itemid'];
}
mysql_query("INSERT INTO jobs (username, item, time, completed) VALUES ('$showusername', '$ids', 'None', 'No')");
$wegq = "SELECT * FROM items WHERE itemid='$ids'";
$weg = mysql_query($wegq);
while($wg = mysql_fetch_array($weg)) {
$im = $wg['image'];
$nm = $wg['name'];
$id = $wg['itemid'];
}
$_SESSION['theid'] = $id;

echo "<font color=green>Success! You have started this Job!</font><br><br>Please bring me this item: <b>$nm</b><br><br><img src=/images/items/$im><br><br><br>";
echo $_SESSION['theid'];
}

if($jobnr == "1") {
$finish = $_POST['finish'];
$okgq = "SELECT * FROM items WHERE itemid='$yes'";
$ok = mysql_query($okgq);
while($ya = mysql_fetch_array($ok)) {
$okname = $ya['name'];
$okid = $ya['itemid'];
$okimage = $ya['image'];
}
echo "Where is my <b>$okname</b>?<br><br><img src=/images/items/$okimage><br><br><br>";
echo $_SESSION['theid'];
?>
<form action="<?php echo "$PHP_SELF"; ?>" method="POST">
<input type="submit" name="finish" value="I have the Item"></form>
<?php
}
}

if(isset($finish)) {
$cinq = "SELECT * FROM uitems WHERE theitemid='$_SESSION[theid]'";
$cin = mysql_query($cinq);
$connr = mysql_num_rows($cin);
if($connr != "0") {
echo "<font color=green>Success! You have the item.</font>";
} else {
echo "<font color=red>Error! You do not have my item!</font>";
}

}
?>
. I basically just want to know how I can set this session as a variable. Also..I have a user login on every page and I want to be able to destroy JUST THE "theid" session and NOT the username session. How would I do that too? thanks for the help in advance!

Hi,

I'm having a wierd problem related to session variables. It looks like variables would get duplicated inside same session, which of course should not be possible.

Background:
This problem exists on one page script which creating a quote request system where users can select different products and get quote for those. Quote system creates quote ID per customer and all products selected in one session should go under the same quote ID. Problem is that sometimes when user selects products second quote ID is created and after that products are randomly added into both of quotes. While trying to solve this problem I added simple logs to see what's going on, those are added below.

Problem seems to be caused by duplicated session variables. At somepoint it looks like session variables are not defined any more and new quote ID is created. After that the old and new session variables seems to be used randomly.

To demonstrate the problem I added session variable called 'counter' which is defined if it does not exist and incremented always when page is reloaded by submitting the form. As you can see from below logs the 'counter' variable seems to have duplicate values, see the quote value to separate two different session variable "sets".


<?php session_start();
echo "\n Current session id: ".session_id();
echo "\n _SESSION['quote']: ".$_SESSION['quote'];
$_SESSION['counter'] = isset($_SESSION['counter'])? $_SESSION['counter'] +1 : 0;
echo "\n _SESSION['counter']: ".$_SESSION['counter'];


Output when page is reloaded(form submitted):
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376
 $_SESSION['quote']: 
 $_SESSION['counter']: 0
 set  _SESSION['quote']: 984
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376
 $_SESSION['quote']:
 $_SESSION['counter']: 0
 set  _SESSION['quote']: 985
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376
 $_SESSION['quote']: 985
 $_SESSION['counter']: 1
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 985 
 $_SESSION['counter']: 2 
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 984 
 $_SESSION['counter']: 1 
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 985 
 $_SESSION['counter']: 3 
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 984 
 $_SESSION['counter']: 2 


Could anyone explain how the 'counter' can behave such way inside same session (based on session ID)?
This is far beyond my knowledge and I would highly appreciate any advice or tip how to solve this porblem. Thanks.

BestRegards,
Laowai

I am finally making the upgrade from PHP 5.3 to PHP 7.3.  When converting over the code I noticed that session some sessions are not being processed.  My test code for this is below:

 

session_start(); 
include ("../includes/connect.inc");

echo "hiiiiiiii: ".$currentDate;
echo "He  ".$_SESSION['user'];


if (isset($_SESSION['user']))
{
	$user = $_SESSION['user'];
	$currentDate = date("Y/m/d");

}
else
{
	echo "Session not started: ";
}

 

My include is listed below:

$link = mysqli_connect('localhost', 'root', ''); 
if (!$link) { 
    die('Could not connect: ' . mysql_error("Connection error message")); 
} 
mysqli_select_db($link, 'dbname') or die (mysqli_error($link)); 



date_default_timezone_set('America/Chicago');

define ("TZ_CORRECTION", 43200);

$bkcolor = "#FFFFFF";

$currentDate = (Date("Y-m-d"));

I am not getting the message "Session not started or the current date from the include".   

If it matters I am using WAMP as my testing server

Any help is greatly appreciated.

Hello! I'm a bit of a novice on PHP variables still and perhaps I'm trying to run before I can walk.

I have an SQL database that I am able to successfully write to without any problems. I have also created a session variable for the user id number that I am able to retrieve on a later page. My problem is that I have hit a wall when it comes to retrieving the data from the database based upon the session variable user id.  Here is my code and error:

$_SESSION['ID_ASSIGNED']=$_POST[KSU_ID]; //this line takes the form data and successfully assigns it to the session variable on an earlier page

echo $_SESSION['ID_ASSIGNED']; //that line works normally.

Here's my problem code (the line starting with $result ). The line works fine if I input: $result = mysql_query("SELECT * FROM $usertable WHERE KSU_ID='some id number'");:

<?php

$result = mysql_query("SELECT * FROM $usertable WHERE KSU_ID=$_SESSION['ID_ASSIGNED']");

$row = mysql_fetch_array($result);

   echo "<br />Participant Number:";
   echo $row['ID'];

   echo "<br />Name: ";
   echo $row['lname'];
   echo  ",&nbsp;";
   echo $row['fname'];

   echo "<br />Student Number:";
   echo $row['KSU_ID'];

   echo "<br />Age: ";
   echo $row['age'];

   echo "<br />Gender: ";
   echo $row['gender'];

   echo "<br />Grade: ";
   echo $row['grade'];

mysql_close($con);
?>