Help: generate dynamic pages for each row

Hi,

I've tried to do it myself testing a variety of codes but the success doesn't come :(
I want to generate dynamic pages for each row that i have in the db. That's it: one page <=> one row.

--> Add-on to my ask: i want to generate pages like index.php?id=200, without having to make every page myself. That's what i was trying to say: each row in the db has its own page.<--

If someone can tell me how to do it or where to find it i'd be very very happy! :p

Hello,

I'm sorry, but I didn't exactly understand what you are trying to do.

Carl

You will need to have an id for each row. You will then be able to select your data from the table where the id is the id you call the script with:

script.php?id=12

then SELECT * FROM table where id = '$id'

Sean :)

I modify my question above 4 better understanding.

Quote:

---

Originally posted by seanf
SELECT * FROM table where id = '$id'

---

have you tried that?

OK, that's the function that has been nearest of what i'm trying to do:

PHP Code:

---

function club() { 
$result = mysql_query("SELECT * FROM es WHERE zona ='and'"); 
while ($row = mysql_fetch_array($result)) { //gets error here
$output = "<a href=\"$php_self?content=$row[idn]\">".stripslashes($row[nom])."</a><br>"; 
}
echo $output;
} 
function es($id) { 
$result = mysql_query("SELECT * FROM `es`
WHERE ((`zona` ='and') AND (`idn` = '$id'))"); 
while ($row = mysql_fetch_array($result)) { //error here too
$output = "<h3>$row[nom]</h3>"; 
$output .="<br>".stripslashes($row[poblacio])."<br>"; 
} 
return $output; 
} 


---

And the output result errors are:
"Warning: Supplied argument is not a valid MySQL result resource in ..."

Then I call them:

PHP Code:

---

echo club();
echo es($content); 


---

Hmm...

Did you connect to the database???

I alway spit it up into smaller pieces so it is very clear what is happening.. lets see

Lets try this:

Code:

---

function club() {
$sql = "SELECT * FROM es WHERE zona ='and'";
$result = mysql_query($sql)
        or die("No result.");

$row = mysql_fetch_array($result);

while ($row){

$output = "<a href=\"$php_self?content=$row[idn]\">".stripslashes($row[nom])."</a><br>";

echo $output;
}

---

And ofcourse something like that for the other function aswell..

If you still get an error try mysql_error() in the or die part so you get the exact error..

Good luck,

Peanuts

Doing like you've said Peanuts I get this:
"Parse error: parse error in C:\apache\htdocs\edj\es\clubs\es\and15.php on line 86"

and line 86 is after the </html>

:(

Did you close everything off all the echo's all the if - elses ??

You have to make sure all the stuff is closed.. So all { have }
and all the echo " have ";

otherwise dump the complete code here or in a pm ...

Greets

Peanuts

OK, now it does run.

But i don't understand why i get no results, although having the db connections well.
Can the problem be "echo es($content);"???

I've answered your pm!!!