"SELECT * FROM tabel WHERE id='$id'

[daybyday]

Hey, have used a long time on this... i hope you guys can help me

- im new in the php / mysql world.

so here it is:

i want to make a list on my first page that post links (list.php)

list.php works fine, i get a list of links from my database, and when i click on it it goes to test.com/klub.php?id= (and the id for the tabel) fx. test.com/klub.php?id=000000018.

so i think the problem is in my klub.php, here i want to select * from siteinfo1 where id = the same id as i get from clicking the link on the other site xD...

and then post the infomation for the 'name' of the link i clicked

well here is my code i think the problem is

$result = mysql_query("SELECT * FROM siteinfo1 WHERE id='$id'");

for if i change the id=$id to id=000000018 it post the info for the table whit that id nr... but then i cant see all the other info i need form 000000019 and so on xD

i hope you understand my stupidity and know how to fix my problem thx...

 

_________________________________________

list.php

_________________________________________

 

$result1 = mysql_query("SELECT * FROM siteinfo1") or die (mysql_error());

 

while($row1 = mysql_fetch_array($result1))

{

echo "<A href=\"klub.php?id={$row1['id']}\">{$row1['navn']}</A>";

echo "</br>";

}

 

_________________________________________

klub.php

_________________________________________

 

$result = mysql_query("SELECT * FROM siteinfo1 WHERE id='$id'");

 

while($row = mysql_fetch_array($result))

{

echo "$row[id]";

echo "</br>";

echo "Navn: ";

echo "$row[navn]";

echo "</br>";

}

[Drummin]

Looks like you're missing grabbing the value of id.  Add this to the top of klub.php.

$id=$_GET['id'];

[daybyday]

Looks like you're missing grabbing the value of id.  Add this to the top of klub.php.

$id=$_GET['id'];

 

Hahahaha ... no way ... i used hours in this

THX it works like a charm

[Maq]

@daybyday, in the future, please place

tags around your code.