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PHP - Use Same Variable In Two Or More Different Closed Php Tags On Same Page
Ok this is some noob question here
I have a page that has few separate pieces of php code embeded within html code. Each piece is separated with Code: [Select] <?php ... ?> tags, everything works fine. I need to assign a variable in first (from the top) piece of php code. Then I want to be able to use that variable in all others php pieces across this same page.... I searched and searched and first tried to do it like : First piece: Code: [Select] <?php function id() { $id="blah"; } ?> Second piece : Code: [Select] <?php id(); ... echo $id; ?> didn't work, then I tried First place: Code: [Select] <?php $id="blah"; function id() { global $id; } ?> Second piece : Code: [Select] <?php echo $id; ?> After struggling and reading for few days all about global, static, and regular variables i finally decided to ask professionals to help.... please let me know if you need me to make the question description more clear or smn Similar TutorialsHi Everyone, I wonder if somebody could help me. I have been at this for hours and am really struggling. I have a timer which I have set up for a quiz. The quiz opens in a javascript shadow box. When the quiz is opened a session variable for the timer is set up to keep track of the time. What I need is for this session variable to end when the shadow box is closed. I have a onClose hook to peform actions when the shadow box is closed which is in javascript. Obviously I cannot end the session variable in the javascript function due to js being client side and php being server side. Has anyone any suggestions how I can resolve this issue. I need to end this session variable as if the user exits the quiz and then selects another quiz to do the timer uses the current time session rather than assigning the new time value for that particular quiz. Thanks in advance. Edd How can I add html tags in php as a variable? For example, for some reason my code gives me a header error if I use html before the php tags. I just need to add the style in there. I was wondering how I could add a style of any html tag within php as a variable?? Like I was thinking like <?php $style = "<style> div.error { border: 1px dashed #660000; background: #fee; color: #660000; } </style>"; echo $style; ?> But that would probably just print out the code... how can I make it so it wont print out on screen and I can actually use the style? Sorry for the noob question... im kind of an intermeddiate php programmer but the noobs questions still seem to never go away! Some code from my pages ,
Page1 ( Redirecting page )
<html> <title>login_redirect.</title> body> <form name="redirect" action="http://mysite/page2.php" method="post"> <input type="hidden" name="mac" value="$(mac)"> </form> <script language="JavaScript"> <!-- document.redirect.submit(); //--> </script> </body> </html>Page 2 ( select product )
<?php
session_start();
ini_set('display_errors',1);
error_reporting(E_ALL);
include '../lib/config.php';
include '../lib/opendb.php';
// get user mac adres from redirect post page1
$_SESSION['macid'] = $_POST['mac'];
// set $macid for other use ( maybe not needed, am learning )
$macid = $_SESSION['macid'];
// echo $macid does show mac adress, so variable is not empty here
if (!empty($_POST["submit"]))
{
$product_choice = $_POST['accounttype'];
$query= "SELECT AccountIndex, AccountCost, AccountName FROM AccountTypes WHERE AccountIndex='$product_choice'";
$result = mysql_query($query) or die('Query failed. ' . mysql_error());
while($row = mysql_fetch_array($result))
{
$_SESSION['AccountIndex'] = $row['AccountIndex'];
$_SESSION['AccountCost'] = $row['AccountCost'];
$_SESSION['AccountName'] = $row['AccountName'];
}
header('Location: page3.php');
}
// did leave out the other/html/form stuff here
Page 3 ( show Session variables )
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
session_start();
print_r($_SESSION);
?>
Now, on page 3 i do see the right session varables, only the "macid" is empty.
why ?
OK, I'm nearing the end of a very long chat project and I have one last hurdle to jump, I've saved this task to the near end because I thought it would be a problem as I have no idea what to do or where to start. The problem is exactly what is said in the subject title. I have looked into JavaScript options such as "onbeforeunload" but that is just BAD. The login is controlled by sessions, so when the user closes the browser, the session is ended, but data in the database says different, cookies are not an option I'm afraid. I've had an idea that if I create a special account (or a few) and always leave these accounts logged in, they can check if a user has been inactive for a certain period of time, and then run queries to change things, I know this sound odd but maybe I will not have to create these accounts but general accounts that people make will be able to do this for me, without them even knowing. Please help, any ideas, suggestions, logic or methods that you think will help will be great. Thanks. Hello dear friends, let say we have (function) and that function has one variable which is (id) the (id) could be number any number 1 or 4 or even 49999 whatever. how to make closed loop which i mean run this function in unlimited way each time with new id <--- i can do it by set random() but the problem i've is how to make it repeat itself excuted excuted excuted | | | ect let say for example , i will insert random number in at database table Code: [Select] $id = rand(); mysql_query("INSERT INTO comments(id) VALUES ('$id')"); that is it, how then i can repeat it many many times without 1) refresh 2) rewrite it many times thanks in advance I have a class I use to handle my various DB objects. I want to close any connections, if needed, in my destructor. I want to do this even though I already know the connection has been closed previously. What I am looking for is a way to test the $connection variable I have to see if it's in need of closing. I can't figure out how to do this. if (isset($this->connection)) { echo "is set"; } if ($this->connection) { echo "here"; } The above would output "is set" and "here", because in both tests say the already closed connection is still there. Can anyone offer any help? hi everyone, did not know what to make the subject, but here is what I want to do: I have a string, which gets returned to me from a linux app on my server, it looks something like this: Code: [Select] <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD> <META HTTP-EQUIV="CONTENT-TYPE" CONTENT="text/html; charset=utf-8"> <TITLE> </TITLE> <META NAME="GENERATOR" CONTENT="OpenOffice.org 3.2 (Linux)"> <META NAME="AUTHOR" CONTENT="Administrator"> <META NAME="CREATED" CONTENT="20110106;14170000"> <META NAME="CHANGEDBY" CONTENT="HOD"> <META NAME="CHANGED" CONTENT="20110522;16540000"> <STYLE TYPE="text/css"> <!-- @page { margin: 0.26in } P { margin-bottom: 0.15in; direction: ltr; color: #000000; line-height: 0.15in; text-align: justify; widows: 2; orphans: 2 } P.western { font-family: "Arial", sans-serif; font-size: 10pt; so-language: en-US } P.cjk { font-family: "Batang", "바탕", serif; font-size: 10pt } P.ctl { font-family: "Times New Roman", serif; font-size: 10pt; so-language: ar-SA } A:link { color: #0000ff } --> </STYLE> </HEAD> <BODY LANG="en-US" TEXT="#000000" LINK="#0000ff" DIR="LTR" STYLE="border: 5.05pt double #000000; padding: 0.67in 0.92in"> <P>I want this and the tags around it, just not the html, head, body and their closing tags.</P> </BODY> </HTML> within the body, the tags are each styled for example: <p style="color: red"></p> so I cannot just get rid of all html, I want to get only all the content within the body tags, but without the body tags obviously strip_tags does not work as i need, I only want to strip certain tags. If someone can help me with this i will much appreciate it. Hello. I want to display all photos from the database. The problem is that it displays nothing. I added a foreach loop in html but further it not display. I'am beginner, I hope someone help me where I make mistake. I care about separation logic from view.
class photo_display
{
private $database;
private $userData;
public function __construct($database)
{
$this->database = $database;
}
public function display()
{
$query = $this->database->connect()->prepare("SELECT photo FROM photo");
$query->execute();
if ($query->rowcount()) {
$row = $query->fetchAll();
}
$this->userData = $row;
}
public function display_Photos(string $colName)
{
return (isset($this->userData[$colName])) ? $this->userData[$colName] : [];
}
}
<div class='container-fluid bg-dark' id='profil'>
<div class='row'>
<div class=" col-12 col-sm-12 col-md-4 col-lg-4 col-xl-4 order-5 order-12 order-6 order-md-4">
<div class="row" id="margin">
<div class="col-xl-12 d-flex justify-content-center" id="test">
<?php foreach ($photo_display->display_Photos('photo') as $value)
{ ?>
<img src ="<?php echo var_dump($value) ?>" />;
<?php }
?>
</div>
</div>
</div>
Hello, How do I get a user's session destroyed when he closes his browser? Thanks in advance i have a script that wrote which updates fields in a mysql database with millions of rows. it just uses a while loop for each row. now the problem is that i forgot to add something to the script, but i already visited the .php file in my browser. i closed the browser and when i refresh the database it still seems to be running even though i closed the browser the script was running on.... is this because the script is still cached in the memory or what? how can i stop it? Hey All, I have a sessions table that is used to see if a user is online. It only gets deleted when they click the Logout button. What I need to do is have it so it gets deleted even if they close the browser. Should I add a column in the sessions table with a time stamp and check on each page for every user to make sure its great than 30 min? Is there a better way to do this, without so many queries? Thanks I made a bug which caused endless loop and endless printing. When I noticed that, I closed the browsers tab, fixed the bug, open a new tab and re-run. I was assuming the server would start from the beginning, but instead I noticed that some variables were active and the program continued from where it stopped. How can I make the program to stop when the client tab is terminated? I GET QuoteFatal error: TPL: [in line 0]: syntax error: file 'm/pg/_categories' does not exist in C:\WebServ\httpd\libs\tpl\class.template.php on line 943 unset($_templatelite_tpl_vars); else: $_templatelite_tpl_vars = $this->_vars; echo $this->_fetch_compile_include($this->_vars['TPLx'].'m/pg/_categories'.$this->_vars['HTML'], array()); $this->_vars = $_templatelite_tpl_vars; WITH FILES http://testynarkotykowe.j13x.pl/index.txt && http://testynarkotykowe.j13x.pl/indexpg.txt HOW TO FIX IT This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=352282.0 I am asking out of curiosity, how does this problem usually get solved, when using a tag system to sort content, and users enter similar tags. For example: flower and flowers... which means the difference between singular and plural. Judging by other sites which are around, this problem is not solved at all, they simply let you add as many tags as you want and then rank the most used tags, it indeed is a solution, but seems more like a walk-around solution to me. Any ideas how somebody could approach to solve this one? EDIT: My suggestion would be an analysis function, which analyzes the entered tag and then suggests the user that similar tags have been already entered and perhaps he wants to choose one of those which have been already entered. This does work though with a high traffic websites what can happen is the following: php and php5 You can end up with two similar tags which are both widely used though have different meanings to the user base. This one could be solved by simply prohibiting similar tags with little additions or changes, and so to speak "forcing" the user to choose something which is already there. May be a way, though it can turn out as a bad solution as well. Do anyone know how i can put the html tags in the same line as the other html tags? Here's an example: Code: [Select] Images | Link | Delete | Enabled On my code, it break the tags to the new line without put on the same line as the other tags. Here's the code: Code: [Select] [code]<?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypass'); define('DB_DATABASE', 'mydbtable'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); if($username == '') { $errmsg_arr[] = 'username ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'PASSWORD ID missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, public FROM members WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo '<p id="images"> <a href="images.php?id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?id='.$row['id'].'">Delete</a> </td> | <p> <p id="test">'.$row['Public'].'</td>''; } } } ?>[/code] Code: [Select] SOLVED I want to use a variable I defined on another page, but I don't know how. Here's the original page with the variable: cart.php Code: [Select] $total += $qty*$row['price']; Here's the new page new page without the variable: checkout.php Code: [Select] //I want the value $total onto this page Thanks Say I have the following text stored in a MySQL database... Code: [Select] <b>Classic Quote from movie</b> and I retrieve it into a variable called $text, how do I properly echo that so that it keeps the bold tags and actually display the text "Classic quote from movie" in BOLD? I'm doing something wrong somewhere along the line (simply doing "echo $text;") because it displays on the page as... Code: [Select] <b>Classic Quote from movie</b> Instead of... Classic Quote from movie Any info on properly storing and echoing back HTML would be very appreciated. I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
I want to include a variable (for example an object $user) in each PHP page. Previously I declared $var in an include file then included that file in every page, however it generates an error in Zend Studio and it's also not very good programming style. I want to know how I can do it any other way. I was thinking of using constants but I don't know if constants should be used for this purpose and also I do not think constants can be defined as objects. Thank you very much for your help. |
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