PHP - Linking Tables

Hi everyone. I would like to do this:
In page1.php, I need to make a few checkboxes, which when the user clicks "next" button, transfers the data (which checkboxes were clicked) to a database table. Then, when the user clicks a button "next", he is moved to another page, page2.php, from which he is automatically transferred to page3.php. At page3.php those values are gathered from the database, and the user has to click "next" button again, then those values are written into another table, in the same or another database.

How can I do this? I've not much knowledge with neither SQL, neither PHP. I hope you will help me make this work, I'm making a little system for my HTML website. Thanks, if you do!

This portion is kind of stumping me. Basically, I have a two tables in this DB: users and users_access_level (Separated for DB normalization)

users: id / username / password / realname / access_level
users_access_level: access_level / access_name

What I'm trying to do, is echo the data onto an HTML table that displays users.username in one table data and then uses the users.access_level to find users_access_level.access_name and echo into the following table data, I would prefer not to use multiple queries if possible or nested queries.

Example row for users: 1234 / tmac / password / tmac / 99
Example row for users_access_level: 99 / Admin

Using the examples above, I would want the output to appear as such:

Username: Access Name: Tmac Admin
I am not 100% sure where to start with this, but I pick up quickly, I just need a nudge in the right direction.

The code I attempted to create just shows my lack of knowledge of joining tables, but I'll post it if you want to see that I did at least make an effort to code this myself.

Thanks for reading!

Im doing a search system and Im having some problems. I need to search in two tables (news and pages), I already had sucess doing my search system for just one table, but for two tables its not easy to do. I already use a select statment with two tables using UNION because I want to show number of search results, that is number of returned rows of my first sql statment. But now I need to do a select statment that allows me to acess all fields of my news table and all fields of my pages table. I need to acess in my news table this fields: id, title, content, link, date, nViews I need to acess in my pages table this fields: id, title, content, link Im trying to do this also with UNION, but in this case Im not having any row returning. Do you see what I have wrong in my code?   <?php
//first I get my $search keyword
$search = $url[1];

$pdo = connecting();
//then I want to show number of returned rows for keyword searched
$readALL = $pdo->prepare("SELECT title,content FROM news WHERE title LIKE ? OR content LIKE ?
UNION SELECT title,content FROM pages WHERE title LIKE ? OR content like ?");
$readALL->bindValue(1,"%$search%", PDO::PARAM_STR);
$readALL->bindValue(2,"%$search%", PDO::PARAM_STR);
$readALL->bindValue(3,"%$search%", PDO::PARAM_STR);
$readALL->bindValue(4,"%$search%", PDO::PARAM_STR);
$readALL->execute();
//I show number of returned rows
echo '<p>Your search keyword returned <strong>'.$readALL->rowCount().'</strong> results!</p>';
//If dont return any rows I show a error message
if($readALL->rowCount() <=0){
echo 'Sorry but we didnt found any result for your keyword search.';
}
else{
//If return rows I want to show, if it is a page result I want to show title and link that I have in my page table
//if it is a news result I want to show title and link that I have in my news table and also date of news
echo '<ul class="searchlist">';
$readALL2 = $pdo->prepare("SELECT * FROM news WHERE status = ? AND title LIKE ? OR content LIKE ? LIMIT 0,4
UNION SELECT * FROM pages where title LIKE ? OR content LIKE ? LIMIT 0,4");
$readALL2->bindValue(1, '1');
$readALL2->bindValue(2, "%$search%", PDO::PARAM_STR);
$readALL2->bindValue(3, "%$search%", PDO::PARAM_STR);
$readALL2->bindValue(4, "%$search%", PDO::PARAM_STR);
$readALL2->execute();

while ($result = $readALL2->fetch(PDO::FETCH_ASSOC)){
echo '<li>';
echo '<img src="'.BASE.'/uploads/news/'.$result['thumb'].'"/>';
echo '<a href="'.BASE.'/news/'.$result['id_news'].'">'.$result['title'].'</a>';
//if it is a news result I also want to show date on my list
//echo '<span id="date">'.$result['date'].'</span>';
echo '</li>';
}
echo ' </ul>';
//but how can I do my select statement to have access to my news table fields and my page table fields??
}
?>

I am trying to write some data from multiple SQL tables to a page. In the first table is a list of places. I then have more tables that are named after the different places. For example, say my first place in the list is called Place1. I have a table named Place1 with data that corresponds to place1. The data contained in the table named Place1 is a list of things to do in this place. It has 21 columns and each one is something to do in the morning, afternoon, and at night for each day of the week in the place Place1. What I am trying to do is display a sort of weekly calendar as a table on a webpage that lists all of the places in one column and then lists seven days of the week as 7 more columns. Then in each data cell I would want to list the things to do in the morning, afternoon and at night for the certain day of the week and for the place. The problem is that I am creating a CMS to allow other users with no coding knowledge to update events for other places, so I have to display data that could have been altered.

The only solution I know of is to do a while loop that gets all of the place names and while there are still place names, write the place names to the page and set a variable equal to the place name. Inside the while loop I would create another while loop that each time the first while loop is executed uses the variable set in the first while loop to know which table to reference and then make a call to that table pulling out the 21 columns and writing them to the page. Each time the outer while loop executes, it would (hopefully) write the place name, and then set the variable as the current place name so that the inner while loop uses the variable to write the rest of the information about that place.

I don't know if that would even work and if it did, I know it cannot be the best way to do this. I am pretty stuck here and don't really have a good solution so if anyone proposes a solution that is radically different to anything I have done, I am open to everything. Thank you!

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>";




?>

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=326928.0

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

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

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 MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=321572.0

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!

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

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:

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;
  
}


?>

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