PHP - Linking?
I found this code which makes a BMI Calculator (Form) for me however when I click on submit it takes the user back to the index page ie. domain.com/index.php. How do I change it to go to, say, domain.com/calculator.php ? The code is below:
<? /** * @package Module Body Mass Index Calculator for Joomla! 1.5 * @version $Id: mod_bodymassindexcalculator.php 599 2010-03-20 23:26:33Z you $ **/ defined( '_JEXEC' ) or die( 'Restricted access' ); $heightcm=$_POST["heightcm"]; $weightkg=$_POST["weightkg"]; if ($heightcm!="" && $weightkg!="") { $heightm = $heightcm / 100; $bmi=round($weightkg / ($heightm*$heightm),1); echo "Heigth, m: ".$heightm."<br />"; echo "Weigth, kg: ".$weightkg."<br />"; echo "Body Mass Index (BMI): ".$bmi."<br />"; echo "<strong>"; if ($bmi<16.5) {echo "Severely Underweight</strong><br />";} if ($bmi>=16.5 && $bmi<=18.4) {echo "Underweight</strong><br />";} if ($bmi>=18.5 && $bmi<=24.9) {echo "Normal</strong><br />";} if ($bmi>=25 && $bmi<=29.9) {echo "Overweight</strong><br />";} if ($bmi>=30 && $bmi<=34.9) {echo "Obese Class I</strong><br />";} if ($bmi>=35 && $bmi<=39.9) {echo "Obese Class II</strong><br />";} if ($bmi>=40) {echo "Obese Class III</strong><br />";} echo "<br />"; } $domain = $_SERVER['HTTP_HOST']; $path = $_SERVER['SCRIPT_NAME']; $queryString = $_SERVER['QUERY_STRING']; $url = "http://" . $domain . $path; $url3 = "http://" . $domain . $_SERVER['REQUEST_URI']; $mystring1="?"; $s1=strpos($url3,$mystring1); if($s1==0) {$url2=$url3;} if($s1!=0) {$url2=substr($url3,0,$s1);} $path = $url2; //1 foot = 0.3048 meters //1 inch = 2.54 centimeters //1 pound = 0.45359237 kilograms $n1=230; echo "<table style=\"width: 100%\" cellspacing=\"0\" cellpadding=\"0\" align=\"center\"><tr><td valign=\"top\">"; //echo "<h3>BMI Calculator</h3>"; echo "<form action=\"".$path."\" method=\"post\" >"; echo "<strong>Height</strong><br />"; echo "<select name=\"heightcm\" >"; for ($i=30; $i<=$n1; $i++){ echo "<option value=\"$i\">".$i." cm / ".floor($i / 30.48)." ft ".round(($i-(floor($i / 30.48)*30.48)) / 2.54, 1)." in </option>";} echo "</select>"; echo "<br />"; echo "<strong>Weight</strong><br />"; echo "<select name=\"weightkg\" >"; for ($i=30; $i<=$n1; $i++){ echo "<option value=\"$i\">".$i." kg / ".round($i / 0.45359237,2)." pounds </option>";} echo "</select>"; echo "<br />"; //echo "<input name=\"searchterm\" type=text size=\"27\" class=\"ns1\">"; echo "<br />"; echo "<input type=\"submit\" value=\"Calculate\" name=\"B1\">"; echo "</form><br />"; //DON'T REMOVE THIS LINK - DO NOT VIOLATE GNU/GPL LICENSE!!! echo "<a href=\"http://nutritioncaloriecounter.com\">Nutrition Calorie Counter</a>"; //DON'T REMOVE THIS LINK - DO NOT VIOLATE GNU/GPL LICENSE!!! echo "</td></tr></table>"; ?> Similar TutorialsI have been trying to figure out how this is done? One php file and all that changes is the name of the brand and the logo. This has to be clickable from a menu and also if a user changes the brand name it changes to the appropriate name and logo. Any ideas as to how this is done? If you can guide me in the right direction or give me an example of how this is done would be greatly appreciated. Thanks. http://www.drivermanager.com/en/download-confirmation.php?brand=compaq&logo= In a relational database, you "link" tables together. For example, Customer (parent) ---> Product (child) How do you do that with Objects? I have a "User" class and an "AddressBook" class. A User can have zero or more Addresses (in their AddressBook). If I create a User object and an AddressBook object, I need a way to link them up similar to how they are linked up in the database. Make sense?! TomTees Hi: I have a login file where a user goes to a db based on the dbtype selected. Now $dbtype1 links to a db on Server 1 ( that is the server on which this script is running and $dbtype2 links to a db on Server 2 I created a connection-link file connectlink.php as under but while $dbtype1 works without a problem , $dbtype2 gives me an error 'no access to db' and user dbun1@localhost not found on db1.db . What am I missing in the connectlink.php file please ? Thanks login.php =========== <? ...... if($_POST['submit']){ $dbtype = $_POST['dbtype']; if ($dbtype == 'type1') { $section = 'type1'; require("../x/type1/type1.php"); } if ($dbtype == 'type2') { $section = 'type2'; require("../x/type2/connectlink.php"); //the dir 'x' is a common name on both the servers'' } //then it processes $userfile and give this Click <a href="'.$section.'/'.$userFile.'?Userid='.$userid.'"> here ?> connectlink.php ============== <? //this file contains db info, log and checks if user is authorised to access the db - ist check. error_reporting(E_ERROR | E_PARSE | E_CORE_ERROR); $dbusername2='a'; $dbpassword2='b'; $dbname2='c'; $servername = 'ip address of server or localhost ?'; $link2=mysql_connect ("$servername","$dbusername2","$dbpassword2", true); if(!$link2){ die("Could not connect to MySQL");} mysql_select_db("$dbname2",$link2) or die ("could not open db".mysql_error()); $dbusername1='d'; $dbpassword1='e'; $dbname1='f'; $servername = 'localhost'; $link1=mysql_connect ("$servername","$dbusername1","$dbpassword1", true); if(!$link1){ die("Could not connect to MySQL");} mysql_select_db("$dbname1",$link1) or die ("could not open db".mysql_error()); $sql = "SELECT * FROM Users WHERE Userid='$userid'",$link2; $result = mysql_query($sql); if ($myrow = mysql_fetch_array($result)){ $login_success = 'Yes'; $status = "On"; .... $sql2= "insert into Log (....) values(.....)",$link2;; $result2 = mysql_query($sql2) or die ('no access to database: ' . mysql_error()); // echo mysql_error(); } } else { $failureMessage = '<p class="data"><center><font face="Verdana" size="2" color="red">Login Failure. You are not authorised to access this database .<br /></font></center></p>'; print $failureMessage; $logoutMessage = 'Click <a href="../NEWDBS/mainlogout.php"> here </a>to logout </p>'; print $logoutMessage; exit; } ?> Hi All, I googled this and there is endless results. I went through a lot of them but couldn't get this working properly. How do I link from within my web site root to files outside the root? It works for me using relative links i.e. ../../phpfiles/includes but that is going to get messy and I can't get a way of doing absolute links to work. If someone could lay that out so a newbie can get it clearly I would really appreciate it! Also - I understand why I should put all of my php files outside the web root but is this a guaranteed way to secure these files other than someone hacking my ftp access? I've looked at a few site hierarchy examples - Am I right that the only pages within the web site root should be template pages with calls to required files (outside the root), session checks, and content includes and all other includes that have php executable code should be outside the root? I really appreciate the advice and insight. Thank you! Hi, I have an issue with a banner I integrated in a page (http://hostelsuites.com) if you go to the page you'll see a banner on the left side , called "Combo Andes". The problem I am encountering is that this banner is linking to inside pages, and it works all fine for me but does not on some other computers, although they are using the same explorer or firefox versions.... I have no idea as to what could trigger this type of error, any help would be muche welcome. Some details that might help : The link is winthin the flash movie The code that brings the banner up depending on the chosen language is <div class="subtitulo"><?= ucfirst($lang["combomendoza"]) ?></div> Thank you very much in advance This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321572.0 I got two lists on a website. One list includes bunch of items and simple info. Another list includes pictures of those items and detailed information. I would like to set it up that way when users clicks on item from list #1 it would automatically jump to an item with the picture and detailed info on a second list. This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=326928.0 is there a way in php to link from the root dir ? like in html you just use the '/' at the start for the link " <a herf="/link.php" ></a> " but i noticed this does not work when using php like include or require. so is there anywey to tell a link to start from root dir? without using the ../../link.php Hi I got three tables (employers , company , and Jobs) Employer table holds info about employer. Company hold info about the company and jobs table holds info about jobs. I was just wandering what would be a good way to link these tables in the database ? Here is the coding for each table. Employer Table Code: [Select] CREATE TABLE IF NOT EXISTS `employers` ( `id` int(11) NOT NULL, `username` varchar(50) NOT NULL, `password` varchar(50) NOT NULL, `email` varchar(50) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; Jobs Table Code: [Select] CREATE TABLE IF NOT EXISTS `Jobs` ( `id` int(11) NOT NULL, `JobTitle` varchar(200) default NULL, `Company` varchar(200) default NULL, `Salary` varchar(30) default NULL, `Description` varchar(2000) default NULL, `CompanyURL` varchar(200) default NULL, `PhoneNumber` varchar(30) default NULL, `Requirements` varchar(2000) default NULL, `JobCategory` varchar(100) default NULL, `JobType` varchar(100) default NULL, `Apply_To` varchar(1000) NOT NULL, `Email` varchar(200) NOT NULL, `modified_at` datetime NOT NULL, `PostedOnDate` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP, `Address` varchar(250) NOT NULL, `State` varchar(200) NOT NULL, `City` varchar(200) NOT NULL, `Country` varchar(100) NOT NULL, `Zipcode` varchar(100) NOT NULL, `JobID` varchar(100) NOT NULL, `WorkExperience` varchar(2000) NOT NULL, `EducationRequirement` varchar(2000) NOT NULL, `WebsitePostedFrom` varchar(200) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; Company Table Code: [Select] CREATE TABLE IF NOT EXISTS `Company` ( `id` int(11) unsigned NOT NULL auto_increment, `CompanyName` varchar(100) NOT NULL, `Address` varchar(100) NOT NULL, `Logo` varchar(100) NOT NULL, `PhoneNumber` varchar(25) NOT NULL, `ContactPerson` varchar(25) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ; <?php $con = database_connect(); $sql = "SELECT * FROM anime1, episode1 WHERE animeid='$animeid'"; $result = mysql_query($sql); while ($row = mysql_fetch_assoc($result)) { $title = $row['title']; $ep = $row['ep']; } ?> keep giving me back error Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\studying\take 2\addin12.php on line 45 I have this code : <img class= "image_frame1" src="images/groupproductimages/<?php echo $offer['cimage'];?>" /> It pulls pics I need the pics to link to urls using this code <a href="index.php?option=com_grouppurchase&view=todaysdeal&id=<?php echo $offer['cid'];?>"> I know I am missing something stupid So I have 3 tables: student, course_student and course. The student table is a student's registration info and course is the different courses the student can register for. The course_student table should contain the id fields of the other 2 which I should then use to link the tables and for example display all student's registered for current courses. Student table has an id field: sno and course table has : cid. Course_student contains both. How do I go about linking the tables so the sno id field of the student table and the cid from course updates with those in the course_table? This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=351259.0 Hi guys, I need your help with my php script. On my script, I'm currently working with image where I can hot-link them from another website while it is on protected. when you click right-mouse button on firefox, you could see something like "view page source" which is disabled. when you click on "save page as", you can save the image as "image.php". when you open them, you would not be able to read due to the image but you can find the real image link in image.php where i want to protect them. here's the currently code: Code: [Select] <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypassword'); define('DB_DATABASE', 'mydbname'); $id = (int)$_GET['id']; $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)); } $qrytable1="SELECT images FROM image_list WHERE id=$id"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { $image = $row['images']; $details = getimagesize($image); header ('Content-Type: ' . image_type_to_mime_type($details[2])); echo readfile($image), "<p id='images'>", $row['images'] . "</p>"; } ?> I guess that there must be a way to protected the url in the php script where i can hide them. It should be easy to modify but I am not sure what line I need to adjust to make it protected. Can you please help me in what line I need to modify in order to protected the hot-linking in my php? I'm working on a site that provides database results. To get to the final report, I'm going through a few pages where the user selects specific options. This builds up arrays that need to be passed from one page to another. As a result the final page has a pretty lengthy variable list. Since I want to provide a link to certain customers where they don't have to go through all the option select stuff, I'm using the GET method on the final page. Due to all the arrays I need to pass, I'm hitting hyperlink address character limitations, and I fear I will eventually hit browser URL length limits. Is there any way to provide links to customers without using the GET method so I could keep the URL length down? i have two tables Loan Application table and Loans table. There is status option and date loan was applied section. After status has been chosen as Pending from the dropdown field and date chosen and loan application saved, it goes under the loans section. Now, admin has to go and approve the loan and pick the date the loan was approved under the loans section but date can be chosen for the approved date but the status cant be changed to approve because it shows pending. Please what can i do so when i select the date for approval , the status should change to approved automatically I have two tables. Let's call the first one items and the second one item categories. Now the items table would look something like this: id(auto_increment id) name description categoryid The item categories table would look something like this: categoryid name description An entry in the items table would look something like this: categoryid = 1,2 name = Thing description = something id = 1 Say I want to retrieve records that contain "1" in the categoryid column. How would I do that? I wonder if anyone can help me, I have a MySQL table of football teams (team_name; link; description; rival_1; rival_2; rival_3;) where each 'team_name' has its own page with the relevant information from the table and is named as the link field from the table. Within each team page I also want to display the rival_1,2 and 3 teams with links to their pages. I know how to just display the information but am struggling with picking up the link from the relevant row in the table. Is this possible or would I be better off just placing a <a href> link around each rival team when I enter it into the MySQL table?? I hope this makes sense, any help would be very much appreciated. Thanks |