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Old 03-29-2006, 10:27 PM   · #1
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Show certain tutorials.
Hi,

I have a tutorial site here at www.x6core.com, and people submit tutorials, how would i make it so when the submit the choose a catorogry for the tutorial? I think on say the photoshop page the code could be something like this, but im not sure. Can someone help please? P.S All the records are in a mysql database.
Code:

if $categories == photoshop{
showrecord}


Im not sure how to show all records that meet criteria.


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Old 03-30-2006, 12:12 AM   · #2
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Question
We need to know more ... When the tutorials are posted you you can decide for a category ? If yes then you can select them. If not the single posibility is to look in title after "category" like "photoshop".

Your tutorials are in sql database or are simple text files on harddrive ?

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Old 03-30-2006, 04:51 AM   · #3
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My tutorials are all stored in a mysql database and i want it so when people submit them they can choose the catogoru.
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Old 03-30-2006, 11:28 AM   · #4
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do you have a pre-defined list

if you sign up to http://aysha.hostmatrix.org/ and see the form for submitting a tutorial should show you a good example.
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Old 03-30-2006, 01:14 PM   · #5
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Lightbulb
Your site support adding categorys ? otherwise you need to change your database and all ...

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Old 03-30-2006, 10:11 PM   · #6
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I know i will change the database and everything but i want the predifined list then a page that shows all the ones in which the catogory is photoshop.
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Old 03-31-2006, 12:58 AM   · #7
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OK first what is ur pre-defined list
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Old 03-31-2006, 02:45 AM   · #8
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umm not quite getting wht your saying?

try:

PHP Code:
<a href="/tutorials.php?cat=photoshop">Photoshop</a>


PHP Code:
<?php
$cat = $_GET['cat'];

if ($cat == "photoshop") {
//whtever u can here
}
else
{
echo"Invalid Category";
}
?>
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Old 03-31-2006, 02:58 AM   · #9
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Caution
Is more easy in this way ...

PHP Code:
<?php

function get_articles_cat($cat)
{
//  sql stuff searching for articles from categories $cat
}

$cat = $_GET['cat'];

if ($cat != "") get_articles_cat($cat);

?>


if you trto doit with if's you will need to add a if for each category added

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Old 04-02-2006, 11:43 AM   · #10
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^^ what goes in the orange area?
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Old 04-02-2006, 02:46 PM   · #11
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that is a comment
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Old 04-03-2006, 03:41 AM   · #12
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right, i like the way this one is made and i want to use it:
Code:
<?php 
$cat = $_GET['cat']; 

if ($cat == "photoshop") { 
//whtever u can here 
} 
else 
{ 
echo"Invalid Category"; 
} 
?>


I dont know how to intergrate it with this:
Code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
</head>

<body>
<?php
$connect = mysql_connect("localhost", "dbun", "dbpw");
mysql_select_db("dbname", $connect);

$select = mysql_query("SELECT * FROM `tutorials`", $connect);
while($row = mysql_fetch_array($select, MYSQL_ASSOC)) {
    $id = $row['id'];
    $name = $row['name'];
    $email = $row['email'];
    $url = $row['url'];
	$url2 = $row['url2'];
	$WebsiteName = $row['WebsiteName'];
	$TutorialName = $row['TutorialName'];
    $description = $row['description'];

echo "
    <p>
     <strong>Tutorial Name:</strong> <a href=\"$url2\">$TutorialName</a><br />\n
	<strong>Submitted By:</strong> <a href=\"$url\">$WebsiteName</a><br />\n
    <strong>Description:</strong> $description
    </p>
";
}
?>


I think it should be:
Code:

<?php
$connect = mysql_connect("localhost", "bdun", "dbpw");
mysql_select_db("dbname", $connect);

$select = mysql_query("SELECT * FROM `tutorials`", $connect);
while($row = mysql_fetch_array($select, MYSQL_ASSOC)) {
    $id = $row['id'];
    $name = $row['name'];
    $email = $row['email'];
    $url = $row['url'];
	$url2 = $row['url2'];
	$WebsiteName = $row['WebsiteName'];
	$TutorialName = $row['TutorialName'];
    $description = $row['description'];
    $cat = $row['cat'];
}
if ($cat == "photoshop") { 
//echo statement??
echo "
<p>
     <strong>Tutorial Name:</strong> <a href=\"$url2\">$TutorialName</a><br />\n
	<strong>Submitted By:</strong> <a href=\"$url\">$WebsiteName</a><br />\n
    <strong>Description:</strong> $description
    </p>";
}
else {
echo "There are no tutorial in this category";
{
?>


Is this correct??
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Old 04-03-2006, 04:04 AM   · #13
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i think what you want to do is

SELECT * FROM tutorials WHERE cat = 'photoshop'

then loop through the results.
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Old 04-03-2006, 04:39 AM   · #14
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Thanks, ive done a copy with my code and get this error:
Code:
Parse error: syntax error, unexpected $end in /home/gibbons/public_html/photoshop.php on line 37


In
Code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
</head>

<body>
<?php
$connect = mysql_connect("localhost", "dbun", "dbpw");
mysql_select_db("dbname", $connect);

$select = mysql_query("SELECT * FROM `tutorials`", $connect);
while($row = mysql_fetch_array($select, MYSQL_ASSOC)) {
    $id = $row['id'];
    $name = $row['name'];
    $email = $row['email'];
	$category = $row['category'];
    $url = $row['url'];
	$url2 = $row['url2'];
	$WebsiteName = $row['WebsiteName'];
	$TutorialName = $row['TutorialName'];
    $description = $row['description'];
	
if ($category == "Photoshop"){
echo " <p>
     <strong>Tutorial Name:</strong> <a href=\"$url2\">$TutorialName</a><br />\n
	<strong>Submitted By:</strong> <a href=\"$url\">$WebsiteName</a><br />\n
    <strong>Description:</strong> $description
    </p>";
	}
else {
echo "Sorry, There are currently no tutorials in this category.";
}
?>
</body>
//LINE 37 BELOW
</html>
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Old 04-03-2006, 05:46 AM   · #15
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you're missing the closing } for the while($row = mysql_fetch_array... statement.

generally when you get an error like that on the last line that's what it is.
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Old 04-03-2006, 06:12 AM   · #16
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I have done this and its has worked the WHERE cat=photoshop.

Where can i put a statement if there is no tutorials??
Last edited by webmonkey : 04-03-2006 at 06:41 AM.
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Old 04-03-2006, 08:44 AM   · #17
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also now what has happened is that it wont display the site and tutorial name, check here: http://www.x6core.com/?id=photoshoptutorials But all i have changed is the where cat = photoshop, if you want to see my code its here:
Code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
</head>

<body>
<?php
//Connect goes here, but deleted user and pw.

$select = mysql_query("SELECT * FROM `tutorials` WHERE category = 'photoshop'", $connect);
while($row = mysql_fetch_array($select, MYSQL_ASSOC)) {
	$category = $row['category'];
    $id = $row['id'];
    $name = $row['name'];
    $email = $row['email'];
	$url = $row['url'];
	$url2 = $row['url2'];
	$WebsiteName = $row['WebsiteName'];
	$TutorialName = $row['TutorialName'];
    $description = $row['description'];

echo "
    <p>
     <strong>Tutorial Name:</strong> <a href=\"$url2\">$TutorialName</a><br />\n
	<strong>Submitted By:</strong> <a href=\"$url\">$WebsiteName</a><br />\n
    <strong>Description:</strong> $description
    </p>
";
}
?>

</body>
</html>
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Old 04-03-2006, 10:30 PM   · #18
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can anyone help me please?
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Old 04-03-2006, 11:26 PM   · #19
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after you run the mySQL query add this:
PHP Code:
if (mysql_num_rows($select) >0) {//there are tutorials in this cat.

//print out your stuff

} else {//nothing in this cat.

    echo 'Sorry, no tutorials in this category'; //replace with message of your choice.
}


i'm not sure if this is what you're looking for or not. hope it helps.
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Old 04-04-2006, 02:17 AM   · #20
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Fixed thanks everyone.
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