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PHP - Moved: Linking Back
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http://www.phpfreaks.com/forums/index.php?topic=351259.0 Similar TutorialsThis topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321572.0 This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=326928.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=330834.0 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=354724.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=350924.0 This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=357042.0 This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=349304.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=330185.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=321154.0 Please, take a look to the following code.After clicking Next it goes to overview.php.Why when I click back on my browser to return to this page again, it is not returning back? When I click back I receive "Confirm Form Resubmission" message. After refreshing page it loads page. I guess problem is in "session_start();" part. Something to do with cookies. Please, help me it is very urgent for me. <?php session_start(); echo "<html> <head> <title>Hello World</title> <meta http-equiv='Content-Type' content='text/html; charset=Windows-1252'/> </head>"; require_once ('functions.inc'); if(!isset($_POST['userid'])) { echo "<script type='text/javascript'>"; echo "window.location = 'index.php'"; echo "</script>"; exit; }else{ session_register("userid", "userpassword"); $username = auth_user($_POST['userid'], $_POST['userpassword']); if(!$username) { $PHP_SELF = $_SERVER['PHP_SELF']; session_unregister("userid"); session_unregister("userpassword"); echo "Authentication failed " . "Please, write correct username or password. " . "try again "; echo "<A HREF=\"index.php\">Login</A><BR>"; exit; } } function auth_user($userid, $userpassword){ global $default_dbname, $user_tablename; $user_tablename = 'user'; $link_id = db_connect($default_dbname); mysql_select_db("d12826", $link_id); $query = "SELECT username FROM $user_tablename WHERE username = '$userid' AND password = '$userpassword'"; $result = mysql_query($query) or die(mysql_error()); if(!mysql_num_rows($result)){ return 0; }else{ $query_data = mysql_fetch_row($result); return $query_data[0]; } } echo "hello"; echo "<form method='POST' name='myform' action='overview.php'>"; echo "<input type='submit' value='Next'>"; echo "</form>"; ?> 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>"; ?> 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? I 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= 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! <?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 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 $logoutMessage; exit; } ?> Hi, I have an issue with a banner I integrated in a page (http://hostelsuites.com) if you go to the page you'll see a banner on the left side , called "Combo Andes". The problem I am encountering is that this banner is linking to inside pages, and it works all fine for me but does not on some other computers, although they are using the same explorer or firefox versions.... I have no idea as to what could trigger this type of error, any help would be muche welcome. Some details that might help : The link is winthin the flash movie The code that brings the banner up depending on the chosen language is <div class="subtitulo"><?= ucfirst($lang["combomendoza"]) ?></div> Thank you very much in advance 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 ; 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 |
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