Frustrated!

Anon

Hello all again,
I am trying to write my site in php4 and MySQL now, and I am stuck. What this piece is, is for my "script archive". What I basiclly am trying, is getting links from the MySQL database to show the topics (i.e. Php, C, C++, and so on). I can call the names of the topics fine, but it is the linking part I want to change.

I want the links to be like "http://acidsun.com/ARCHIVE/archive.php?lang=1"

The 1 at the end stands for the id of the topic in the MySQL Database. The code I have is:

/
Connects to the MySQL database.
/
$db = "acidsun";
$link = mysql_connect("localhost");
if(!$link)
die("Couldn't connect to MySQL");
mysql_select_db($db, $link)
or die("Couldn't open the List:" .mysql_error());
$result = mysql_query("SELECT * FROM archive_cats ORDER BY cat_name");

/
End of access to opening MySQL.
Now we want to print out the tables
to show the archive topics.
/

/
This while statement puts the MySQL in a loop so that it will print out all the topics in the database.
/
while ($list = mysql_fetch_array($result))
{
print "<a href=\"http://localhost/ARCHIVE/archive.php?lang=$list[id]\">$list[cat_name]</a>";
}

That is without the html parts only because this is all I have on the page (and the basic html stuff).

Also Note: id = id (auto incremented), cat_name (stands for catagory name, also one of the fields in the MySQL), cat_url (i was trying to do the whole link stuff, also another filed in the database).

Can someone please help me on the right track here?

yojimbo

Ray,
First thing I'd ask is if you tried to use the result data outside of the quotes? meaning:
"blahblah-" . $line[id] . "-blah blah" instead of:
"blahblah-$line[id]-blah blah"

Also..dumb question but is there any data in your table?

Anon

Yes there is data in the database. I put 4 catagories in it. Anyways, I can see the links with the endings on them, I just want to know how I would actually get the links to go to the pages with the content on it.

zplummer

in your archive.php file, just use

SELECT *
FROM archive_cats
WHERE id = $lang

Anon

I can't use that, it keeps conflicting with the while statement I use to show the list of topics. Here is the code I got:

/*Gets the stuff from the archive.
In this case, the php archive.*/

$db = "acidsun";
$link = mysql_connect("localhost");
if(!$link)
die("Couldn't connect to MySQL");
mysql_select_db($db, $link)
or die("Couldn't open the List:" .mysql_error());
$result = mysql_query("SELECT * FROM archive_cats WHERE id=$lang ORDER BY cat_name");

/*End of access to opening MySQL.
Now we want to print out the tables
to show the archive content.*/

while ($list = mysql_fetch_array($result))
{
print "<a href=\"archive.php?$lang\">$list[cat_name]</a><br>";
}

jcornett

Can you please clarify about what problem (exactly) you are currently having? Is it giving you an error, or do you just need help with some coding concepts to make this system work?

I'd be happy to help either way, just want to make sure I'm answering the right question. 😉

John Cornett

Anon

I think I understand what you're trying to do.

If the user goes to:
http://www.domain.com/archive.php
then it'll give them a list of archive topics. These archive topics will have links such as:
http://www.domain.com/archive.php?lang=5
where the 5 would be the article number. When they click on that you want it to display the article. The code you posted would be fine for when they want the list. But now you need code for displaying the actual article. So here's what you do:

<?php

/ first check to see if we need to display the article list or an article. To do this we check to see if $lang is created. /

if(!isset($lang)) { // Begin article list
/ $lang is not set so display list of articles /

/Gets the stuff from the archive.
In this case, the php archive./

$db = "acidsun";
$link = mysql_connect("localhost");
if(!$link)
die("Couldn't connect to MySQL");
mysql_select_db($db, $link)
or die("Couldn't open the List:" .mysql_error());

/ Here we grab all the article names and such to make links to them /
$result = mysql_query("SELECT * FROM archive_cats ORDER BY cat_name");

/End of access to opening MySQL.
Now we want to print out the tables
to show the archive content./

while ($list = mysql_fetch_array($result))
{
print "<a href=\"archive.php?$lang\">$list[cat_name]</a><br>";
}

} // End article list
else { // Begin article
/ or if $lang is set then we want to display the article, so we connect to the database and get the article info and display it /

$db = "acidsun";
$link = mysql_connect("localhost");
if(!$link)
die("Couldn't connect to MySQL");
mysql_select_db($db, $link)
or die("Couldn't open the List:" .mysql_error());

/ Here we grab only the line of data from the selected article /
$result = mysql_query("SELECT * FROM archive_cats WHERE id=$lang");

/End of access to opening MySQL.
Now we want to print out the tables
to show the archive content./

$list = mysql_fetch_array($result) {

/ Now you have all the information for only the selected article in $list. You can grab whatever fields you need and display them here. /

}

} // End article

Anon

Oops copied a mistake you had in what you posted.

You need to change:

print "<a href=\"archive.php?$lang\">$list[cat_name]</a><br>";

to:

print "<a href=\"archive.php?lang=$list[cat_numb]\">$list[cat_name]</a><br>";

assuming $list[cat_numb] is the number of the article.