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PHP - Passing Parameter Within Redirect
Hello all,
I'm an absolute Newbie to PHP. I'm struggling with a problem that I think is extremely basic to you experts but I just can't get it to work. In short, I have an application that has a link that goes to: Site1/dir/admin?id=(text)&page=(number) I want to redirect to a link that's identical except it's Site2 at the beginning instead of Site 1. What I've attempted so far is this: <?php $GLOBALS[$passid]=$_GET['id']; $GLOBALS[$passpage]=$_GET['page']; header( 'Location: http://site2.com/dir/admin/?page=$passpage&id=$passid'); ?> But it just redirects to http://site2.com/dir/admin/?page=$passpage&id=$passid (with all of that text being literally what is in the URL, not the variable values) I have a feeling I'm doing something small but very basic wrong. The $GLOBALS thing was an afterthought. Outcome is the same with or without that. Any help much appreciated!!! Similar TutorialsThis topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=358276.0 hello this is my first time posting so i'm sorry if i word this wrong or it is a stupid question. i have a simple database made up with a product table with five fields; id, url, name, description, and price. also a categories table with five fields; id, parent_id, siteurl, name, and description. i have everything up and running,but i can't figure out how to pass the product id as a parameter. right now i have "example.php" but i need "example.php?id=1" here is what my code looks like indextest.php Code: [Select] <?php error_reporting(0); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>Car Dealer</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link href="stylesheet.css" rel="stylesheet" type="text/css" /> </head> <body> <?php $usr = "root"; $pwd = ""; $db = "mymarket"; $host = "localhost"; //connect to database $cid = mysql_connect($host,$usr,$pwd); if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); } $category = ""; // setup SQL statement $SQL = " SELECT * FROM categories "; $SQL = $SQL . " WHERE parent_id = '0' "; // execute SQL statement $retid = mysql_db_query($db, $SQL, $cid); // check for errors if (!$retid) { echo( mysql_error()); } else { ?> <div id="layout"> <div id="titlebg"> <div class="title">CAR</div> <div class="title1">DEALER</div> <div class="title2">WE HAVE</div> <div class="title3">THE BEST CARS</div> <div class="title4" align="center"> Links: <a target="_blank" href="http://jigsaw.w3.org/css-validator/" class="title4">CSS VALIDATOR</a></div> </div> <hr id="hrline" /> <div id="gradientbg"> <div id="links"> <h2> <?php // display results echo ("<p><dt><b>$category</b><br>\n"); while ($row = mysql_fetch_array($retid)) { $siteurl = $row["siteurl"]; $name = $row["name"]; echo ("<dd><a href='$siteurl'>$name</a></dd>\n<br>"); } echo ("</dt></p>"); } ?> <h2/> </div> <div id="header"></div> </div> <div id="bodypart"> <div id="catalogue"> <div class="cat">CATALOGUE</div> <div id="cat1"></div> <div id="car1"></div> <div id="cat2"></div> <div id="car2"></div> </div> <div id="line"></div> <div id="ourservices"> <div class="our">OUR SERVICES</div> <div id="pic1"></div> <div class="lorem">Lorem ipsum</div> <div class="txt">At vero eos et accusam sed diam nonumy eirmod Erat, sed diam voluptua. At vero eos et accusam At vero eos et accusam sed diam nonumy eirmod Erat, sed diam voluptua. At vero eos et accusam Consetetur sadi pscing </div> </div> <div id="line1"></div> <div id="aboutus"> <div class="abt">ABOUT US</div> <div id="pic2"></div> <div class="lorem1">Lorem ipsum</div> <div class="txt1">At vero eos et accusam sed diam nonumy eirmod Erat, sed diam voluptua. At vero eos et accusam At vero eos et accusam sed diam nonumy eirmod Erat, sed diam voluptua. At vero eos et accusam Consetetur sadi pscing </div> </div> </div> <hr id="hrline1" /> <div class="foottxt">Copyright © Cars. Design by <a href="http://www.dotcominfoway.com/" class="foottxt">DOT COM INFOWAY</a>.</div> </div> </body> </html> and i would like to link that to this next file att.php Code: [Select] <?php error_reporting(0); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>Smartphone Dealer</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link href="stylesheet.css" rel="stylesheet" type="text/css" /> </head> <body> <?php $usr = "root"; $pwd = ""; $db = "mymarket"; $host = "localhost"; //connect to database $cid = mysql_connect($host,$usr,$pwd); if (!$cid) { echo("ERROR: " . mysql_error() . "\n"); } $category = ""; // setup SQL statement $SQL = " SELECT * FROM categories "; $SQL = $SQL . " WHERE id ='4' OR id = '5' OR id = '6' "; ?> <div id="layout"> <div id="titlebg"> <div class="title">SMARTPHONE</div><br> <div class="title1">DEALER</div> <div class="title2">WE HAVE</div> <div class="title3">NO CONTRACT SMARTPHONES</div> <div class="title4" align="center"> Links: <a target="_blank" href="http://jigsaw.w3.org/css-validator/" class="title4">CSS VALIDATOR</a></div> </div> <hr id="hrline" /> <div id="gradientbg"> <h3> <?php // execute SQL statement $retid = mysql_db_query($db, $SQL, $cid); // check for errors if (!$retid) { echo( mysql_error()); } else { // display results echo ("<p><dt><b>$category</b><br>\n"); while ($row = mysql_fetch_array($retid)) { $siteurl = $row["siteurl"]; $description = $row["description"]; echo ("<dd><a href='$siteurl'>$description</a></dd>\n<br>"); } echo ("</dt></p>"); } ?> </div> <h3> </div> <hr id="hrline1" /> <div class="foottxt">Copyright © Nathan Hein. Template by <a href="http://www.dotcominfoway.com/" class="foottxt">DOT COM INFOWAY</a>. <br> DISCLAIMER: This website was designed for a school project and is not intended for commercial use of any kind.</div> </div> </body> </html> i know i probably have a lot of problems with this code, but the one i can't figure out is passing the parameter and i have been looking at forums, websites, w3schools, everywhere. i;m just not grasping the concept of how to make it work with this. thank you so much for any input Hello, I have below javascript, which I need to put in a new php page. However I need the vistiors to pass in the username to the script either by clicking the link for that user or by entering a value on the website. I am not sure how can I do this in php script and modify below java script to make it work. Any help will be greatly appreciated. <pre class="example"> $(function(){ $(".tweet").tweet({ join_text: "auto", username: "NEED_TO_BE_PARAMETER", avatar_size: 68, count: 25, auto_join_text_default: "said,", auto_join_text_ed: "I", auto_join_text_ing: "I wase", auto_join_text_reply: "I replied", auto_join_text_url: "I was checking out", loading_text: "loading..." }); }); </pre> <div class='query'></div> <script type="text/javascript"> $(function(){ $(".example").each(function(i, e){ eval($(e).text()); }); }); </script> I am experimenting around with PDO, I have a DB handle class and instancing it in a function. The problem is when I want to bind elements in a query, for example I bind a parameter but I can only do it using a varible outside of the object. Code: [Select] <?php $database = new Database("localhost", "fry", "root", ""); $database->set_table("usertest"); $database->set_query("SELECT * FROM usertest WHERE user_name = :user_name"); $value = "matt"; $database->prepare_query(); $database->bind("parameter", ":user_name", $value ,PDO::PARAM_STR, 5); $database->execute(); while($result = $database->fetch()) { echo $result['user_name'].'<br />'; } ?> Is this bad OOP practice? I wanted to put $value = "matt"; into the Database class somehow here is the Database class Code: [Select] <?php class Database{ public $hostname; public $database; public $username; public $password; public $connection; public $prepare; public $query; public $table; public $fetch; public $bind_var; public $bind_val; function __construct($hostname, $database, $username, $password) { $this->hostname = $hostname; $this->database = $database; $this->username = $username; $this->password = $password; $this->Database_connection(); } public function Database_connection() { try { $this->connection = new PDO('mysql:host='.$this->hostname.';dbname='.$this->database, $this->username, $this->password); } catch (PDOException $e) { echo 'Connection failed: ' . $e->getMessage(); } } public function set_query($query) { $this->query = $query; } public function get_query() { return $this->query; } public function set_table($table) { $this->table = $table; } public function get_table() { return $this->table; } public function set_bind_var($var) { $bind_var = $var; } public function set_bind_val($val) { $bind_val = $val; } public function get_bind_var() { return $this->bind_var; } public function get_bind_val() { return $this->bind_val; } public function bind($bindtype, $val, $var, $pdo, $num) { if ($bindtype == "parameter") { $result = $this->prepare->bindParam($val,$var,$pdo); } return $result; } public function prepare_query() { $this->prepare = $this->connection->prepare($this->get_query()); } public function execute() { $this->prepare->execute(); } public function fetch() { return $this->prepare->fetch(); } } ?> Yet agian sorry for the sloppy code, this is mainly an experiment, would be very appriciated if someone could help me out with this problem. Hello All Can someone help me with this: I am trying to create some PHP code to pass 2 variables to a stored procedure. At the moment I have: <?php // Connect to MSSQL and select the database $serverName = "(local)"; $connectionInfo = array( "Database"=>"ashley"); $conn = sqlsrv_connect( $serverName, $connectionInfo); if( $conn === false ) { echo "Could not connect.\n"; die( print_r( sqlsrv_errors(), true)); } /*Call the stored procedure with a named parameter*/ $tsql = 'EXEC CPS_ADD_NEW_USER @agentF = php,@agentS=test'; $stmt = sqlsrv_query($conn, $tsql); /* Free statement and connection resources. */ sqlsrv_free_stmt( $stmt); sqlsrv_close( $conn); ?> Could anyone advise me what I need to be doing and where I am going wrong Thanks in advance Bicky Hi guys I have 2 question. 1. Is it possible to use the a variable in a header redirect function i.e. Code: [Select] $test = 'index.php'; header("location: $test"); 2 assuming $test = 'index.php'; how do I pass the variable from one page to another? Thanks I want to display some specific data from my database (just data from a given date). I first created a page where the user picks the date. After the date is defined it is passed to the next page where the data from that date is displayed. That page includes a form where users can edit the data. The form data is then passed to a third page for processing. All of this works fine. Now I want the user to be taken back to the previous page with the assigned date. On the processing page I am using this statement: Code: [Select] header("Location: /teetimes/adminteetimes.php?teetimedate=$dateID");and on the display page I'm using Code: [Select] $dateID = $_REQUEST[teetimedate];The display page is using a SELECT...WHERE Date='$dateID' statement and $dateID is defined in the processing page. Problem is: I'm redirected from the processing page back to the display page but the date isn't passed, so the correct data isn't being displayed. Can anyone help me pass a variable using a header?? Or show me where my logic is lacking?? Thanks! Hello,
I recently posted here about an issue I was having with my database orientated products page.
I have now run into another problem where say if, /db.php was typed or /db.php?p=IDoNotExist was typed, it returns blank.
I have in my code the desired content to be displayed, but it just doesn't seem to want to make a show.
I was also wondering if it is possible to show different content for whatever the URL is, so for no parameter, the content about the products, and a non existent one, maybe "Product not found"?
Here is my code:
<?php
$db=mysql_connect ("localhost", "webwibco_charlie", "Hello123") or die ('I cannot connect to the database because: ' . mysql_error());
$mydb=mysql_select_db("webwibco_products");
include("header.php");
$status = htmlspecialchars( @$_GET ['p'] );
if ($status == "floorpuzzles") {
echo "<h1>Our Floor Puzzles</h1>";
$sql="SELECT ID, Name, Tags, Description, Category FROM products WHERE Category LIKE '%" . FloorPuzzles . "%'";
$result=mysql_query($sql);
while($row=mysql_fetch_array($result)){
$Name =$row['Name'];
$ID =$row['ID'];
$Description =$row['Description'];
echo "<div class=\"box\">";
echo "<h1>$Name</h1>";
echo "<div class=\"floorbox\"><a href=\"?p=$ID\"><img src=\"images/products/catalogue/big/floorpuzzles/$ID.jpg\" class=\"small\"></a></div>";
echo "<h2>$Description</h2>";
echo "</div>";
}
?>
<?
}else{
if ($status == $_GET["p"]) {
$sql="SELECT ID, Name, Tags, Description, Pieces, Size, Barcode, Category FROM products WHERE ID = '" . $_GET['p'] . "'";
$result=mysql_query($sql);
while($row=mysql_fetch_array($result)){
$Name =$row['Name'];
$ID =$row['ID'];
$Description =$row['Description'];
$Pieces =$row['Pieces'];
$Size =$row['Size'];
$Barcode =$row['Barcode'];
echo "<div class=\"1\">";
echo "<h1>$Name</h1>";
echo "<div class=\"bigbox\">";
echo "<div class=\"floorbox\"><img src=\"images/products/catalogue/big/floorpuzzles/$ID.jpg\" class=\"big\"></div>";
echo "</div>";
echo "</div>";
echo "<div class=\"2\">";
echo "<p>Puzzle Pieces: $Pieces</p>
<p>Puzzle Size: $Size</p>
<p>Barcode: $Barcode</p>";
echo "</div>";
}
}else{
?>
<?
echo"<h1>Our Products</h1>
<p>Our jigsaw puzzles are hand cut by skilled craftsmen and therefore each one is unique with self-correcting pieces. There is a strict quality control process at all stages by our highly experienced staff. The puzzles are durable and provide fun and excitement, enhancing learning and a child’s development.<p>
<p>All of our jigsaws are made using materials from sustainable resources grown in managed forests. Where possible we support companies in the UK and source our components locally, most of our suppliers are in the East Midlands, many in Derbyshire and Nottinghamshire. We keep packaging to a minimum and take our environmental and ethical responsibilities very seriously.</p>
<p>Reducing waste and recycling was a way of life for us before it became fashionable. We are constantly searching for new ideas and consult teachers when developing our jigsaws, which are often used within the national curriculum.</p>
<p>As well as making our own range, we manufacture for leading suppliers to the education market. Check for \"Made in Britain\" and it is probably made by us.</p>
<p>We have a wide variety of products available for viewing, from classic floor puzzles to innovative inset trays. You can take a look at all our products on this page, simply use the navigation buttons to your left.</p>";
}}
include("footer.php");
?>
The final echo is what I wish to be displayed on the URL without or with an invalid parameter.
Here is my site URL: http://www.webwib.co...saws/search.php (note that only the "Floor Puzzles" category has content within it).
Thank you in advance for assistance.
I have the following code just to insert a username and hashed password into the database but somehow I am getting this error and I couldn't find out where I am doing it wrong...can someone please give me a hand?
I tried it in two ways and both errors...
the first few lines are just connecting database which worked fine and a password.php so I can use password_hash() with my php version
$hash = password_hash('xx', PASSWORD_BCRYPT, array('cost' => 10));
$username = 'xx';
$insertQuery = $db->prepare("
INSERT INTO et_todo (username, password)
VALUES (:username, :hash)
");
$insertQuery->execute(array(
'username' => $username,
'password' => $hash
));
also tried
$hash = password_hash('xx', PASSWORD_BCRYPT, array('cost' => 10));
$insertQuery = $db->prepare("
INSERT INTO et_todo (username, password)
VALUES ('xx', :hash)
");
$insertQuery->execute(array(
'username' => 'xx',
'password' => $hash
));
I'm trying to put together a script that redirects visitors based on their IP, user agent and/or referral url. Basically I want the script to scan these three factors from the visitor, if any of them turn out to match my redirect-requirement it redirects the user. I know the code is horribly coded, I'm incredibly new to the php-scene and consider myself a complete noob. As you can see I want redirected visitors to go to google.com and un-redirected to msn.com(examples). Really thankful for all the help I can get! Right now nothing works, any suggestions? <?php function redirect($page) { Header( "HTTP/1.1 301 Moved Permanently" ); header('Location: ' . $page); exit; } $referrals=array('pitchingit.org','referral2'); $badAgents = array("useragent1", "useragent2"); $deny = array("78.105.191..*","100.101.103..*"); if (in_array($_SERVER['HTTP_REFERER'], $referrals, FALSE)) { header("Location: http://www.google.com"); } else { header("Location: http://www.msn.com"); } if(in_array($_SERVER['HTTP_USER_AGENT'],$badAgents)) { redirect("http://www.google.com/"); exit(); } $add=$_SERVER['REMOTE_ADDR']; foreach ($deny as $ip) { if (preg_match("^.$add.*^",$ip)) { redirect("http://www.google.com"); } } redirect("http://www.msn.com"); ?> Code: [Select] <?php $parentesco .=' <ul><li><a href="children.php?category_id='.$idc.'&name= '. $name2 . '">'. $name. ' </a></li></ul>'; ?> in the url string $idc is passing but $name2 is not. $name2 is pickup from a query, I echo that value everywhere in the code and it echo its value but is not passing so how. Help How can one re-direct a visitor, without using a header re-direct? I'd like a page to show up, then after about 5 seconds I need the visitor sent to another page. How can I do this? I have some function or method. Is there a better design patter to implement this?
function myFunction($a=null,$b=null,$c=null,$d=null,$e=null,$f=null,$g=null,$h=null)
{
//Do a bunch of stuff
}
Maybe the second function?
function myNewFunction($data=array())
{
$data=array_merge(array('a'=>null,'b'=>null,'c'=>null,'d'=>null,'e'=>null,'f'=>null,'g'=>null,'h'=>null),$array);
//Do a bunch of stuff
}
Please provide decision making factors why you would use one approach over the other.
Hi! Code: [Select] $url = "http://www.youtube.com/watch?feature=player_embedded&v=IwPHy17Iu6E"; I have code Code: [Select] if (preg_match("/http:\/\/www.youtube.com\/watch\?v=([0-9a-zA-Z-_]*)(.*)/i", $url, $matches)) { return '<object height="'.$embed_height.'" width="'.$embed_width.'">'. '<param name="movie" value="http://www.youtube.com/v/'.$matches[1].'" />'. '<param name="wmode" value="transparent" />'. '<embed src="http://www.youtube.com/v/'.$matches[1].'&autoplay=1" type="application/x-shockwave-flash" wmode="transparent" height="'.$embed_height.'" width="'.$embed_width.'" />'. '</object>'; }How to remove feature=player_embedded& from $url ? Hi guys, please help me to solve my problem. i want to get the url parameter using /
for now i use index.php?category_id=14
how can i change the ?category_id=14 to index.php/14
thank you so much.
Hello I am experiencing some issues with a function and I cannot identify the problem. The function is not receiving any parameter. The URL values are passed correctly. I used echo to display the parameters outside the function and that worked. I also used echo to display the parameters inside the function and no parameter were displayed. So the function is not receiving parameters. Code: [Select] <?php $find = $_REQUEST['u_find']; $field = $_REQUEST['u_field']; $searching = $_REQUEST['u_search']; echo"(1a)$find, (1b)$field, (1c)$searching";//outside function function test_display($searching, $field, $find) { echo"(2a)$find, (2b)$field, (2c)$searching";//inside function } ?> Hey guys, I'm dealing a url that is similar to hxxp://domain/directory/?#{$id}. What I'm trying to do is extract $id from that. Its not available as a $_GET variable and $_SERVER['REQUEST_URI'] comes up as "/directory/?". Does anyone know how I could retrieve $id ? Thanks Hi all I have an SQL database with products in and I need to write the PHP script to select from the SQl based on values in the URL. The URL looks like this: product.php?category=32&type=90&colour=10 So I can do the script: if (isset($_GET['category']) && isset($_GET['type']) && isset($_GET['colour'])){ $sqlCommand = "SELECT * FROM products WHERE category = '$category' AND type = '$type' AND colour = '$colour'"; } Which works great, I have also got the other versions if the type and colour are empty. What I can't do is get it to select all from the database where the category = 32 and the type = 90 if the URL looks like this: product.php?category=32&type=90&colour= In other words where the colour is set but is empty. I want the script to show all products that have the category of 32 and the type of 90 if this is the case Please help! Many thanks Pete I am trying to create a simple voting form. Everything goes well until I submit and then I get a Warning: mysqli_error() expects exactly 1 parameter, 0 given on line 79 error. I am assuming it is not pulling the ID correctly but as I am new to php and mysqli I cannot exactly say if it the way the code is written or if I am calling the parameter incorrectly in the query. Again I am new to to this so please be gentle. Below is my code. It pulls the drop down list correctly and echo's correctly but I believe my post query to be a little out of wack. Could someone point me in the correct direction? It would be very appreciated.
<form action="businesstype_update.php" method="post">
<?php
if(isset($_POST['voteall'])){
$vote_lg = "update membertest where id={$row_lg['id']} set vote=vote+1";
$run_lg = mysqli_query($con, $vote_lg) or die(mysqli_error());
}
$result_lg = mysqli_query($con, "SELECT id, business FROM membertest WHERE businesstype='large'");
echo "Vote for large business of the year! <SELECT name='business'>\n";
echo "<option>Select a large business</option>";
while($row_lg = $result_lg->fetch_assoc()) {
echo "<option value='{$row_lg['id']}'>{$row_lg['business']}</option>\n";
}
echo "</select></br></br>\n";
echo "<input type='submit' name='voteall' value='voteall'>";
$result_lg->close();
$con->close();
Code: [Select] function fetch_feedback_best() { $sql = "SELECT `like` FROM `feedback` GROUP BY `page` asc LIMIT 10"; while($row = mysql_fetch_assoc($sql)) { $results[] = $row; } return $results; } This doesn't work. My error is "Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in /blah.inc.php on line 105 ". Code: [Select] <?php $results = fetch_feedback_best(); foreach ($results as $result) { ?> echo $result['page']; } ?> |
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