PHP - Linking Menu's That Use Mysql

im finding it hard to link PHP, MySQL and dreamweaver together.

when php document is alaunched within the localhost all the data in the database is present how ever

when the php coding is entered into the html file via dreamweaver no data is present which i think connection to the mysql database cannot be establish

why is this can someone help me please, the php document is attached with all the coding


[attachment deleted by admin]

I have a few tables in a project and i'm linking pages together via PK/FK's.

i have a main table with client names,
ID, CLIENT NAME, ETC

i have a table with communications
COMMID, ID, CONTACT NAMES, ETC

and another of communication history.
COMMHISTORYID, COMMID, DATE, TIME, ETC

I'm not sure how to write the query so that i can view/update information from all 3 tables on the same webpage page using php/mysql. I'd appreciate any assistance you may provide. thanks.

I am tying to make my category menus mysql based but all my sub categories end up under the last category. Here is my code.
Code: [Select]
      <ul class="sf-menu">
                            <li><a href="http://www.mysite.com/index.php">Home</a></li>
                            <?PHP do { ?>
                            <li><a href="news.php?c=<?PHP echo $row_bodynav['id']; ?>"><?PHP echo $row_bodynav['catname']; ?></a>
<?PHP if ($row_bodynav['slug'] == $row_bodynav_sub['parent']) {echo '<ul>'; do { ?><li><a href="news.php?c=<?PHP echo $row_bodynav['id'];?>&sc=<?PHP echo $row_bodynav_sub['id']; ?>"><?PHP echo $row_bodynav_sub['subcatname']; ?></a></li> <?PHP } while ($row_bodynav_sub = mysql_fetch_assoc($bodynav_sub)); echo '</ul>';} ?>
                            </li>
                             
<?PHP } while ($row_bodynav = mysql_fetch_assoc($bodynav)); ?>
                             </ul>

i am storing my menu in the database, i want to be able to output it by priority, heres so far what i have. I have no idea were to start.
Database dump:
--
-- Table structure for table `menu`
--
CREATE TABLE IF NOT EXISTS `menu` (
  `menu_access_lvl` int(2) NOT NULL,
  `priority` int(11) NOT NULL,
  `name` varchar(200) NOT NULL,
  `comment` text NOT NULL,
  `location` text NOT NULL,
  `creator_id` varchar(255) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
--
-- Dumping data for table `menu`
--
INSERT INTO `menu` (`menu_access_lvl`, `priority`, `name`, `comment`, `location`, `creator_id`) VALUES
(0, 1, 'Home Page', 'Home page', 'index.php', 'admin'),
(0, 3, 'Contact', 'Contact', 'index.php?PG=contact', 'admin'),
(0, 2, 'Events & Meetings', 'Events & Meetings', 'index.php?PG=events', 'admin'),
(0, 4, 'About', 'About', 'index.php?PG=about', 'admin'),
(2, 5, 'Admin', 'Admin', 'index.php?PG=admin', 'admin');

And here is the php code displaying it

       //gets the role of the user if set, otherwise role = 0
   if(isset($_SESSION['SESS_MEMBER_ID']))$lvl = $_SESSION['SESS_ROLE']; else $lvl = 0;

   // this loads the menu buttons that correspond to the users role
   $menuqry="SELECT * FROM menu WHERE menu_access_lvl<='$lvl'";
   $menuresult=mysql_query($menuqry);
   
   while($row = mysql_fetch_array($menuresult)){
      echo "<li class=\"menuitem\"><a href=\"".$row['location']."\">".$row['name']."</a></li>";
   }


What this currently displays:
Home
Contact
Events & meetings
About

I want it to be according to priority in the menu like:
Home
Events & meetings
Contact
About

I have a table in my db with all my menus and submenues of my site, I'd like to show them in some kind of menu containing all of them
each entry in db has the columns ID,NAME,TEXT,LEVEL,DEPENDENT
level has a value of 1 or 2 depending if it's an item or a subitem
dependent has a reference to the id of the section it belongs to

Using dreamweaver I created two recordsets based on the level, they are called row_sections (only where level = 1) and row_subsections (only where level = 2)

here is the code I tried to use:

    <?php do { ?>
      <p><a href="test.php?sec=<?php echo $row_sections['id'];?>"><?php echo $row_sections['name']; ?></a></p>  //prints the section text
                    <?php do { ?>
                    <?php
                    if($row_subsections['dependent'] == $row_sections['id']) {
                    echo $row_subsections['name']."<br>"; //tries to print all subitems corresponding to the actual item
                    }
                    ?>
                    <?php } while ($row_subsections = mysql_fetch_assoc($subsections)); ?>
            <?php } while ($row_sections = mysql_fetch_assoc($sections)); ?>

Here's the output:

Quote

section 1 title

subsection 1
subsection 2

section 2 title


both section 1 and 2 should have two different subitems.. where is the problem?

if I add the line <?php echo $row_sections['id']; ?> right before the second <?php do{ ?> it shows the corresponding ID for BOTH sections so I can't understand the problem...

Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike.

In each table their are 2 fields, one INDEX_ID and MODELS.

Under that for rows I have every model bike named.

A quick question before I get onto what I want to do:

Can data be added underneath the rows?

For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit.

To make it easier to understand I'll post the SQL code for the brand Honda:

CREATE TABLE `Honda` (
  `INDEX_ID` int(3) NOT NULL auto_increment,
  `MODELS` varchar(20) collate latin1_general_ci NOT NULL,
  PRIMARY KEY  (`INDEX_ID`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ;

--
-- Dumping data for table `Honda`
--

INSERT INTO `Honda` VALUES(1, 'CR85');
INSERT INTO `Honda` VALUES(2, 'CR125');
INSERT INTO `Honda` VALUES(3, 'CR250');
INSERT INTO `Honda` VALUES(4, 'CRF100');
INSERT INTO `Honda` VALUES(5, 'CRF150');
INSERT INTO `Honda` VALUES(6, 'CRF230');
INSERT INTO `Honda` VALUES(7, 'CRF250X');
INSERT INTO `Honda` VALUES(8, 'CRF250R');
INSERT INTO `Honda` VALUES(9, 'CRF450X');
INSERT INTO `Honda` VALUES(10, 'CRF450R');
INSERT INTO `Honda` VALUES(11, 'CRF50');
INSERT INTO `Honda` VALUES(12, 'CRF70');
INSERT INTO `Honda` VALUES(13, 'CRF80');
INSERT INTO `Honda` VALUES(14, 'XR650');
INSERT INTO `Honda` VALUES(15, 'CR500');
INSERT INTO `Honda` VALUES(16, 'XR100');
INSERT INTO `Honda` VALUES(17, 'XR200');
INSERT INTO `Honda` VALUES(18, 'XR250');
INSERT INTO `Honda` VALUES(19, 'XR400');
INSERT INTO `Honda` VALUES(20, 'XR50');
INSERT INTO `Honda` VALUES(21, 'XR70');
INSERT INTO `Honda` VALUES(22, 'XR80');

Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?:


<?
$connection = mysql_connect("localhost","user","pass");
$fields = mysql_list_fields("dbname", "table", $connection);
$columns = mysql_num_fields($fields);
echo "<form action=page_to_post_to.php method=POST><select name=Field>";
for ($i = 0; $i < $columns; $i++) {
echo "<option value=$i>";
echo mysql_field_name($fields, $i);
}
echo "</select></form>";
?>


Thanks.

Hello everyone,

So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help.

A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right.

Some of this probably doesnt even make sense but I'll show you the kind of things I've tried...


            <select name="gender" size="1" id="gender">
              <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option>
              <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option>
            </select>



OR


     <select name="gender" id="gender">
              <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option>
              <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option>
            </select>


OR



            <select name="gender" size="1" id="gender">
              <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option>
              <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option>
            </select>


OR

            <select name="gender" id="gender">
              <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option>
              <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option>
            </select>


Honestly man, I've got no idea.


The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen.  And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol.

Any help would be tremendous and greatly appreciated.

Thanks very much, Learner

P.S  Man!

Hi all,

I am currently learning PHP and have the homework to produce a function that can delete a row in a MySQL database table by clicking on an item in a drop-down menu in a web page.

The code I have produced up until now is this:

<!DOCTYPE HTML>
<html lang="de">
<head>
	<meta charset="utf-8" />
	<title>E3_Artikel_Löschen</title>

</head>
<body>
<form method = "GET">

<?php
	$anr="";
	
	try {
	$pdo = new PDO ('mysql:dbname=bestelldatenbank;host=localhost;charset=utf8', 'root', '');
	} catch (PDOException $error){
		die ($error->getMessage());
		}		
?>

	<div>
		<p>
			<label for="artikel">Artikel: </label>
				<select id="artikel" name="artikel">
					<?php
						$sqlSelect = "SELECT anr, name FROM artikel ORDER BY anr ASC";
						foreach ($pdo->query($sqlSelect) as $row) {
							echo "<option value=$row[0]>$row[0] | $row[1]</option>\n";
							$anr = $row[0];

						}
					?>
				</select>
			
			<input type = "submit" value = "Delete row" />
			
		</p>
	</div>

<?php
	function artLoeschen($anr) {
		echo "Function called $anr";
		if(isset($_GET[$anr])) {
			$anr = $_GET[$anr];
			$sqlDelete = $pdo->query("DELETE FROM artikel WHERE anr = :anr");
			if ($stmt = $pdo->prepare($sqlDelete)) {
				$stmt->bindParam(':anr', $anr);
				$stmt->execute();
			}
			echo "<h2><b>Artikel gelöscht!</b></h2>";
		}
	}
?>

</form>
</body>
</html>

So, I have observed the following when I run the script in a browser:

1.  The HTML works as expected and I get a drop-down list with the article number and description of each item in the affected table.

2.  I can click on an item in the list and it populates the top item in the drop-down list.

3.  When I click delete row, the selected item is not deleted.

4.  There are no error messages returned but the function is not executed (at least not as I would like to expect).

 

I have obviously missed something or made a mistake in my code. 

I would be very grateful for any help...this is driving me mad! :)

Regards,

Kevin

I wonder whether someone can help me please.

I've found http://www.plus2net.com/php_tutorial/ajax-listbox.php tutorial to create a drop down menu using mySQL table data, which, in turn returns a list of results on the page.

Following this tutorial I've put together the tables in my database and the required scripts as shown in the tutorial with the one exception, the "z_db.php" file, which I've assumed to be:

Code: [Select]
<?php mysql_connect("host", "user", "password")or die(mysql_error()); mysql_select_db("database"); ?>

The problem I have, is that when I try and run this, I receive the following error: Quote

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91 which is this line in the search form: echo "</head><body onload="ajaxFunction()";>";.


I must admit I've guessed as to the structure of the 'z_db.php' file should look like because this is not shown so perhaps this is the problem.

I just wondered wether someone could perhaps take a look at this please and let me know where I've gone wrong.


Many thanks and kind regards

I have a mysql table with the structure of
Code: [Select]
ID  Menu_Name   Parent_ID
1   Finance     NULL
2   Business    NULL
3   Investment  1
4   Trading     2
How can I create a html <ul><li> list based on the parent?

your MySQL server version:  5.1.36

Code: [Select]
SELECT * FROM game_weapons

Table: Picture Attached

the EXPLAIN output for your query, if applicable:
I wish to connect a drop down menu's selection of weapons with the correlated table information.

What do I want to happen:
Click on a drop down Menu and have a list of weapons, these weapons are associated with a set number in the database and passed onto the next screen.   (The larger number wins, this part I have figured out).

               <p>
               <select name="weapon2" style="font-size:20px;font-family:Arial;width:275px">
               <option value="power">Power</option>
               <option value="intelligence">Intelligence</option>
               <option value="speed">Speed</option>
               <option value="reserve">Reserve</option>
               </select>
               </p>


I do have the battle code figured out!  (This should be the last step).

Hi all

I need to combine these two scripts:

Firstly, the following decides which out of the following list is selected based on its value in the mySQL table:

<select name="pack_choice">
<option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option>
<option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option>
<option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option>
</select>

Although this works OK, I need it also to show dynamic values like this:

select name="category">
<?php
$listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC");
while($categoryReturned=mysql_fetch_array($listCategories)) {
echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>";
}
?>
</select>

I'm not sure if this is possible?

Many thanks for your help.

Pete

I am working on a project where I want a select form to display information from a MySQL table. The select values will be different sports (basketball,baseball,hockey,football) and the display will be various players from those sports. I have set up so far two tables in MySQL. One is called 'sports' and contains two columns. Once called 'category_id' and that is the primary key and auto increments. The other column is 'sports' and contains the various sports I mentioned. For my select menu I created the following code.


<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","uname","pw")
or die( "You did not successfully connect to the DB!" );

#select the specified database
$rs = @mysql_SELECT_DB ("test", $conn )
or die ( "Error connecting to the database test!");
?>

<html>
<head>Display MySQL</head>
<body>
<form name="form2" id="form2"action="" >
<select name="categoryID">
<?php
$sql = "SELECT category_id, sport FROM sports ".
"ORDER BY sport";

$rs = mysql_query($sql);

while($row = mysql_fetch_array($rs))
{
  echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n  ";
}
?>
</select>

</form>
</body>
</html>




this works great. I also created another table called 'players' which contains the fields 'player_id' which is the primary key and auto increments, category_id' which is the foreign key for the sports table, sport, first_name, last_name.

The code I am using the query and display the desired result is as follows


<html>
<head>
<title>Get MySQL Data</title>
</head>
<body>
<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","uname","pw")
or die( "Err:Db" );

#select the specified database
$rs = @mysql_SELECT_DB ("test", $conn )
or die ( "Err:Db");

#create the query
$sql ="SELECT *
FROM sports
INNER JOIN players ON sports.category_id = players.category_id
WHERE players.sport = 'Basketball'";

#execute the query
$rs = mysql_query($sql,$conn);

#write the data
while( $row = mysql_fetch_array( $rs) )
{
echo ("<table border='1'><tr><td>");
echo ("Caetegory ID:  " . $row["category_id"] );
echo ("</td>");
echo ("<td>");
echo ( "Sport: " .$row["sport"]);
echo ("</td>");
echo ("<td>");
echo ( "first_name: " .$row["first_name"]);
echo ("</td>");
echo ("<td>");
echo ( "last_name: " .$row["last_name"]);
echo ("</td>");
echo ("</tr></table>");
}
?>
</body>
</html>





this also works fine. All I need to do is tie the two together so that when a particular sport is selected, the query will display below in a table. I know I need to change my WHERE clause to a variable. This is what I need help with.

thanks

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>";




?>

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