[RESOLVED] Passing info in url

[dassyka]

Hi everybody,

I would like to pass a small info in the url for my menu but my script is wrong and i don't know how to solve it! Can somebody see what i'm doing wrong?

Here's my code:

<?php

$db = new mysqli('localhost', 'root', '', 'eurodib');

$query = "SELECT categorie.cat_name, sous_categorie.*
FROM sous_categorie, categorie
WHERE sous_categorie.id_cat=categorie.id_cat
ORDER BY categorie.cat_name";
$result = $db->query($query) or die($db->error);

echo "<ul>\n";
$cat_name = '';
while ($row = $result->fetch_array())
{
if ($row['cat_name'] != $cat_name)
{
// End previous category, if any.
if ($cat_name != '')
{
echo "</ul>\n</li>\n";
}

    // Begin the new category.
    $cat_name = $row['cat_name'];
    echo '<li>' . htmlspecialchars($cat_name) . '<ul>';
}

echo '<li><a href="product.php?$row['sous_cat_name']>' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';

}
// End category, if any.
if ($cat_name != '')
{
echo "</ul>\n</li>\n";
}
echo "</ul>";

?>

And as a result, on my page product.php, when i want to echo the sous_cat_name, here's the result i have:

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\Program Files\wamp\www\index.php on line 44

the line 44 is:

echo '<li><a href="product.php?$row['sous_cat_name']>' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';

Thank you to you if you can help me on that!!

[djjjozsi]

lets see other method, how about this. And use objects after you know more about them.

I'v learned like this...

And on the line: echo '<li><a href="product.php?'.$row['sous_cat_name'].'>' . use " at the $rows!

And, use mysql_fetch_assoc fo fetch the results...

Becouse that writes evrything you've take in...
echo '<li><a href="product.php?$row['sous_cat_name']>' ; echo ' $variable '; writes $variable into ....

<?php

$username = ""; 
$pass = ""; 
$database = ""; 
$host=""; 
$connection = mysql_connect( $host,$username, $pass ); 
if ( ! $connection ) print( "Could not connect" ); 
mysql_select_db( $database ) or print ( "could not open database: ".mysql_error() ); 

$query = "SELECT categorie.cat_name, sous_categorie.*
FROM sous_categorie, categorie
WHERE sous_categorie.id_cat=categorie.id_cat
ORDER BY categorie.cat_name";

$result = mysql_query($query) or die(mysql_error());

echo "<ul>\n";
$cat_name = '';
while ($row = mysql_fetch_assoc($result))
{
	if ($row['cat_name'] != $cat_name)
	{
		// End previous category, if any.
		if ($cat_name != '')
		{
			echo "</ul>\n</li>\n";
		}

	// Begin the new category.
	$cat_name = $row['cat_name'];
	echo '<li>' . htmlspecialchars($cat_name) . '<ul>';
}

echo '<li><a href="product.php?'.$row['sous_cat_name'].'>' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';
}
// End category, if any.
if ($cat_name != '')
{
	echo "</ul>\n</li>\n";
}
echo "</ul>";

?>

www.phpcode.hu

[dassyka]

It's not working! 😕

Maybe i wasn't clear enought in my explanations. I have a menu with categories and sub-categories. When people click on the sub-categories, i want my page product.php to open with the name of my sub-category written on the top.

My menu is displaying very well with my ul li list and here's the code of my working menu:

<?php

$db = new mysqli('localhost', 'root', '', 'eurodib');

$query = "SELECT categorie.cat_name, sous_categorie.*
FROM sous_categorie, categorie
WHERE sous_categorie.id_cat=categorie.id_cat
ORDER BY categorie.cat_name";
$result = $db->query($query) or die($db->error);

echo "<ul>\n";
$cat_name = '';
while ($row = $result->fetch_array())
{
if ($row['cat_name'] != $cat_name)
{
// End previous category, if any.
if ($cat_name != '')
{
echo "</ul>\n</li>\n";
}

    // Begin the new category.
    $cat_name = $row['cat_name'];
    echo "<li>" . htmlspecialchars($cat_name) . "<ul>";
}

echo '<li><a href="#">' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';

}
// End category, if any.
if ($cat_name != '')
{
echo "</ul>\n</li>\n";
}
echo "</ul>";

?>

Now, i want to replace the "#" by the link for my page product.php in adding the sub-category name that was click in the url to retrieve it in the page product.php.

I know that it's suppose to be something like: product.php?............ and that's where i'm block. Nothing of what i tried worked.

Does someone can help me?

[djjjozsi]

Lets echo the subcategory's ID, and then select in product.php

To solve this thing, give us the sql dump of the database's tables you use,

and we might help to write with you .

Why is bad the code I wrote?

Its running here:

http://phptool.extra.hu/test5/index.php

[dassyka]

My menu look lke this:

# Products

* Crepes Machine
      o Double Crepes Machines
      o Single Crepes Machines
* Dishwasher
      o undercounter dishwasher
* Molds
      o Warner Brother Molds
      o Relief Mats
      o Cake Mold
      o Jelly Flex Mold
* Paninis
      o Sirman Paninis
      o Eurodib Paninis

So my sub-categories names are:
-double crepes machines
-single crepes machines
-undercounter dishwasher
...

and i wrote what you said to me but in my code, it's not working!

What is the sql dump? You have my request in the code i gave at the begining...

What you need else? You have all my code with all the field i need in my request including my sub-categorie name (sous_cat_name)....

Does someone has a solution?

[djjjozsi]

Sql dump
go to http://www.youtube.com/watch?v=BDGjDdRMzTM ,

and see the video, called: How to make an SQL dump/Export
And i tell it what it is ...

Before using my codes, you set the database's config into the first 4 lines?

Becouse i tested, and it is running. But i need the correct structure of your tables, to visible the categories you wanted..

bye!

jjjozsi

dassyka;10889649 wrote:

My menu look lke this:

So my sub-categories names are:
-double crepes machines
-single crepes machines
-undercounter dishwasher
...

and i wrote what you said to me but in my code, it's not working!

What is the sql dump? You have my request in the code i gave at the begining...

What you need else? You have all my code with all the field i need in my request including my sub-categorie name (sous_cat_name)....

Does someone has a solution?

[dassyka]

Here's what i obtained by doing the same that the video...

I hope it can help you to help me!

-- phpMyAdmin SQL Dump
-- version 2.11.6

-- http://www.phpmyadmin.net

-- Serveur: localhost
-- Généré le : Mer 15 Octobre 2008 à 09:13
-- Version du serveur: 5.0.51
-- Version de PHP: 5.2.6

SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";

--

-- Base de données: eurodib

---

--

-- Structure de la table categorie

CREATE TABLE categorie (
id_cat int(10) unsigned NOT NULL auto_increment,
cat_name varchar(50) NOT NULL,
cat_desc varchar(255) NOT NULL,
PRIMARY KEY (id_cat)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;

--

-- Contenu de la table categorie

INSERT INTO categorie (id_cat, cat_name, cat_desc) VALUES
(1, 'Dishwasher', 'Dishwasher are provided by Lamber and very good.'),
(2, 'Crepes Machine', 'Made in France by Krampouz, that is the original one'),
(3, 'Paninis', 'For good and original sandwichs, choose paninis'),
(4, 'Molds', 'For candy, cake in plenty of differents forms');

---

--

-- Structure de la table sous_categorie

CREATE TABLE sous_categorie (
id_sous_cat int(10) unsigned NOT NULL auto_increment,
sous_cat_name varchar(50) NOT NULL,
sous_cat_desc varchar(255) NOT NULL,
id_cat int(10) unsigned NOT NULL,
PRIMARY KEY (id_sous_cat),
KEY id_cat (id_cat)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=10 ;

--

-- Contenu de la table sous_categorie

INSERT INTO sous_categorie (id_sous_cat, sous_cat_name, sous_cat_desc, id_cat) VALUES
(1, 'Sirman Paninis', 'Original from Italy', 3),
(2, 'Eurodib Paninis', 'Paninis made for Eurodib only', 3),
(3, 'Warner Brother Molds', 'funny and original from WB', 4),
(4, 'Relief Mats', 'For decoration original and funny', 4),
(5, 'Cake Mold', 'For cake of all shape', 4),
(6, 'Jelly Flex Mold', 'For jelly candy', 4),
(7, 'Double Crepes Machines', 'to make it double', 2),
(8, 'Single Crepes Machines', 'one at the time', 2),
(9, 'undercounter dishwasher', 'for undercounter', 1);

--

-- Contraintes pour les tables exportées

--
-- Contraintes pour la table sous_categorie

ALTER TABLE sous_categorie
ADD CONSTRAINT sous_categorie_ibfk_1 FOREIGN KEY (id_cat) REFERENCES categorie (id_cat) ON DELETE CASCADE ON UPDATE CASCADE;

[djjjozsi]

I did it with this Dump:
If you click on a product, it writes the details in the same program, you can seee at the last part of the code:

We had a mistake, after the categories link,
<a href="product.php?VARIABLE=SOMETHING"> Something</a>
See into the sourcode!

Hi,
jjozsi

---

http://www.phpcode.hu

<?php

$username = ""; 
$password = ""; 
$database = ""; 
$hoszt=""; 
$connect = mysql_connect( "$hoszt", $username, $password ); 
if ( ! $connect ) print( "Could not connect to the database!" ); 
mysql_select_db( $database) or print ( "Could not open $database database: ".mysql_error() ); 

$query = "SELECT categorie.cat_name, sous_categorie.*
FROM sous_categorie, categorie
WHERE sous_categorie.id_cat=categorie.id_cat
ORDER BY categorie.cat_name";

$result = mysql_query($query) or die(mysql_error());

echo "<ul>\n";
$cat_name = '';
while ($row = mysql_fetch_assoc($result))
{
	if ($row['cat_name'] != $cat_name)
	{
		// End previous category, if any.
		if ($cat_name != '')
		{
			echo "</ul>\n</li>\n";
		}

	// Begin the new category.
	$cat_name = $row['cat_name'];
	echo '<li>' . htmlspecialchars($cat_name) . '<ul>';
}
//	echo '<li><a href="product.php?id_sous_cat='.$row['id_sous_cat'].'">' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';

echo '<li><a href="?id_sous_cat='.$row['id_sous_cat'].'">' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';
}
// End category, if any.
if ($cat_name != '')
{
	echo "</ul>\n</li>\n";
}
echo "</ul>";
if($_GET["id_sous_cat"])
{
$query = "SELECT *
FROM sous_categorie
WHERE id_sous_cat='".mysql_real_escape_string((int)$_GET["id_sous_cat"])." LIMIT 1'";

$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_assoc($result);
print "sous_cat_name:".$row["sous_cat_name"]."<br>";
print "sous_cat_desc:".$row["sous_cat_desc"]."<br>";
}
?>

[dassyka]

Thank you for everything but just forget it, that's not what i want at all.

I just wanted people to click on the menu and open an other page and i wanted to know (on the second page) what they have click on to show the sub-cat details and the products in this sub-category.

I just don't know how to write the url in my link to transport this sub-category that was click because my menu is created with my query... but it just seem that nobody understand what i want to do so i'll find another way somewhere else!

Thank you again for your trying!!🙂

[djjjozsi]

I'll do a video howto do that...

you must send the subcategory's ID number to the next file. And on the second page you select from the table with the details.
so it do what you want, but in the same file you print the menu.
The if( last $_GET .... you find the details of the category.

If you modify this line:
from:

echo '<li><a href="?id_sous_cat='.$row['id_sous_cat'].'">' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';

into

this:
echo '<li><a href="product.php?id_sous_cat='.$row['id_sous_cat'].'">' . htmlspecialchars($row['sous_cat_name']) . '</a></li>';

it will send that category's ID to the product.php.

Hello,
jjozsi

[dassyka]

Thank you, thank you, thousand of thank you!!! It is finaly working like i wanted to!!!!

God bless you!!!!!!
😃

[djjjozsi]

Nice!

Good luck 🙂
jjozsi

dassyka;10889719 wrote:

Thank you, thank you, thousand of thank you!!! It is finaly working like i wanted to!!!!

God bless you!!!!!!
😃