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PHP - Want To Declare Variable For Homepage
Hi, I have these variables for pages I navigate to because I pass them in the query string. However, I need a variable for the homepage (index.php), before any query string is created. I cant declare a vari
$_GET['intro'] ; $_GET['about'] ; $_GET['contact'] ; However, I need a variable for the homepage (index.php), before any query string is created. I cant declare a variable in index.php because it will be used throughout the entire website. I also cant declare the variable in home.php (the bulk content of the homepage) because I need to use the variable before home.php is included. Is their a trick for this? Thanks! Similar TutorialsCan you set a variable like so: Function something(bob = true) { } Howdy colleagues,
I have noticed some programmers re-declare their variables before using them. Like so:
$name = ''; $name = 'Elizabeth';First of all, why do they do that? Variables are being re-written with each "=" anyways. Secondly, do you believe this is better than just doing: $name = 'Elizabeth';Thank you I'm started learn PHP about 7 days.I want to know>>is this possible to declare a Variable like $password="Phpfreaks"??? I tried these below, Code: [Select] class Site extends Controller { function Site() { public $data = Array(); } } but it fail? Thanks in advanced. hi, cane somebody please explain how to use the declare contruct of PHP some code snippets may help..thanks in advance hi good day, im a little bit confuse. what i want to do is instead of $username = $_POST['username']; is to transform it into OOP? please help me. thanks. According to woorank.com my page shows the following. Language Declared: Missing Detected: en I have used Code: [Select] header('content-type: text/html; charset=utf-8'); but do not know how I declare the language. also how do i specify Language/character encoding how is this done ? I keep running into a bunch of error when I declare the page variable in my model. What syntax do I need to use throughout the class? Code: [Select] <?php if ( ! defined('BASEPATH')) exit('No direct script access allowed'); class Header_Model extends CI_Model{ var $page = substr(end(explode(DIRECTORY_SEPARATOR, $_SERVER['PHP_SELF'])), 0, -4); function get_page_name() { $this->load->library('common'); $page_names = $this->common->page_names(); $title = (array_key_exists($page, $page_names) !== false) ? $page_names[$page]: ''; if (array_key_exists($page, $page_names) !== false) { $title .= " | Jason Biondo"; } return $title; } function get_js_page_file() { if (file_exists("./assets/js/pages/${page}.js")) { $javascript = "<script type=\"text/javascript\" src=\"./js/pages/${page}.js\"></script>"; } return $javascript; } } Hello everyone, i have this page in my website called "index.php": <?php include("include/open_conn.php");?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>name site</title> </head> <body> <div id="container"> <?php include("include/header.php");?> <?php include("include/content.php");?> <?php include("include/footer.php");?> </div> </body> </html> <?php include("include/close_conn.php");?> The content change dinamically with this code "content.php": $pages_dir = 'pages'; if (!empty($_GET['menu'])) { $pages = scandir($pages_dir, 0); unset($pages[0], $pages[1]); $menu = $_GET['menu']; if (in_array($menu.'.php', $pages)) { include($pages_dir.'/'.$menu.'.php'); } So, when i change the page the url will be " index.php?menu=nameofmypage ". But now i want to put an image only in my index.php page, a sort of welcome to the users, but with this codes the image will appear in all the other page because i change only the content. I know that i could create a page named "home.php" and do all i want, but i want to change the first page that the user see when he types the address of my website. thank you! Hello, I never really gave a serious thought to the above question until now that I'm trying to actually launch a live site. Well I'm sure most of you are familiar with this situation, where a user my enter a wrong, non existent url, whose root happens to be your domain name. For example, instead of entering domain.com/greetings, the user may wrongly type domain.com/grewtings . How can I make my site to redirect such erroneous urls to my homepage, instead of to my webhost's homepage with an error message? Appreciate all answers. This topic has been moved to Website Critique. http://www.phpfreaks.com/forums/index.php?topic=346717.0 I Have a protected members section on my site for staff. I would like it so that information specific to that member of staff would appear on their homepage straight after login. Just basic information that is stored on a MySQL database so they can login and check appointments and clients etc for the week etc. is this something that is possible to do in PHP and MySQL? or am I going mad? At the moment I have the entire table on the members homepage and the only way I've figured out to specify a member to only show one users information is to use - (SELECT * FROM Tutors WHERE name = 'XXX'. I don't like this method as it would mean i would have to have a separate files for each member. I was hoping I could do something along the lines of - WHERE name = ' relative to the login name used'. but I am sure I am flogging a dead horse here. So many websites do this kind of thing and I'm surprised after scouring the net for hours and hours that I haven't found any useful information or tutorials on the subject. I am fairly new to programming in PHP and MySQL and am only just getting to grips with the lingo. Maybe I've just been seacrhing the wrong thing, I don't Know. Any help at all would be amazing and much appreciated, a nudge in the right direction would be a start as I'm getting nowhere on my own THANKS This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=357107.0 I am running through a tutorial and I am getting an error based on a the concatication operator on my output. Below is my code. As you can see I am echoing $display_block .= "<p>$title<br>$rec_label<br>.... If I send this as-is I get this error: Notice: Undefined variable: display_block in C:\wamp\www\php\php_mysql\sel_byid.php on line 36 I can resolve the error by placing this before the while loop: $display_block = ""; Is there a better way to output the concatenation.= so I don't need to do this weird fix? while ($row = mysql_fetch_array($result)) { $id = $row['id']; $format = $row['format']; $title = stripslashes($row['title']); $artist_fn = stripslashes($row['artist_fn']); $artist_ln = stripslashes($row['artist_ln']); $rec_label = stripslashes($row['rec_label']); $my_notes = stripslashes($row['my_notes']); $date_acq = $row['date_acq']; if ($artist_fn != "") { $artist_fullname = trim("$artist_fn $artist_ln"); } else { $artist_fullname = trim("$artist_ln"); } if ($date_acq == "0000-00-00") { $date_acq = "[unknown]"; } $display_block .= "<p>$title<br>$rec_label<br>$artist_fullname<br>$my_notes<br>$date_acq<br>$format</p>"; I get this error on my homepage : Warning: implode() [function.implode]: Invalid arguments passed in /home/shqip1/peqini.com/wp-content/themes/sportpress/functions/wpzoom-widgets.php on line 1469 This is the code to modify http://pastebin.com/RiBsyG3c Thanks Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment. My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted." The program should not accept any numbers greater than 100 or any characters. Once I do this, I must take a second number and do a similar thing. Finally, I must have a statement show up at the bottom stating which number is greater. Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters. Any help you can provide will be greatly appreciated! I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> I have just re-installed Xampp and suddenly my sites are now displaying lots of: Notice: Use of undefined constant name - assumed 'name' in ... Notice: Use of undefined constant price - assumed 'price' in ... this is an example of the line its refering too: $defineProducts[1001] = array(name=>'This is a product', price=>123); My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user and $priv are undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working. The code I'm using:
<?php
session_start();
function verify() {
//verify that the user is logged in via the login page. Session_start has already been called.
if (!isset($_SESSION['loggedin'])) {
header('Location: /index.html');
exit;
}
//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users.
$user === $_SESSION['name'];
//Connect to Databse
$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
//SQL Statement to lookup privlege information.
if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
//LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
while ($row = $result->fetch_assoc()) {
$priv === $row["su"];
}
}
// close SQL connection.
mysqli_close($link);
// Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators.
if ($priv !== 1) {
echo $_SESSION['name'];
echo "you have privlege level of $priv";
echo "<br>";
echo 'Your account does not have the privleges necessary to view this page';
exit;
}
}
verify();
?>
I have a form that creates rows of data input textboxes depending on a user input number of things. I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row. All this is working fine. My problem is I need to get the data inserted into this table of textboxes into an array. Here's my code where I attempt to to this (it does not work): Code: [Select] $temp = $_SESSION['Num_Part']; $count = 1; while ($count <= $temp){ $temp2[$count] = "'Participant_P".$count."'"; //echo $temp2[$count]."<br/>"; $temp3[$count]=$_POST[$temp2[$count]]; //here's the problem $temp4[$count] = "'Result_P".$count."'"; $temp5[$count]=$_POST[$temp4[$count]]; //here's the problem //echo $temp4[$count]."<br/>"; $count++; } The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes. Can a variable be used in a POST argument and if so what is the correct syntax? If not, is there some other simple solution to harvest the data into an array. I understand I can harvest by explicitly accessing each key in the post assoc array. But this could be dozens of rows of input fields. Thanks in advance for your help here. I couldn't find anything online re this topic. |
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