Passing an id variable in a hyperlink....

kmax

I am passing the and id variable in a link like this...
www.domain.com/details.php?id=1

and I need to get the details page to display the results of a certain database column for the id # that is selected...

here is my details page

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

$query = mysql_query("SELECT * FROM products WHERE id='$id'");
print $query;

anyone got any tips?

hermzz

i thought that would work, but if it doesn't try using

list($name, $val) = each($HTTP_GET_VARS);

so that $name would be equall to id and $val would be equal to one in your case..
if you arte using just one variable you can directly use $val..

$query = mysql_query("SELECT * FROM products WHERE id='$val'"); 
print $query

hermzz

actually i made a mistake
the last part should be

$query = mysql_query("SELECT * FROM products WHERE id='$val'"); 
print $query

$val should NOT go in between single quotation marks in any case productos or id might but never a variable like $val or $id

kmax

I am pass a variable for the id in a url...
www.domain.com/list.php?id=1 or 2 or whatever...

Here is what I tried....

<?php

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

list($name, $val) = each($HTTP_GET_VARS);

$query = mysql_query("SELECT * FROM products WHERE id=$val");
print $query;

and I am getting.... Resource id #2

anyone got any suggestions?

ahundiak

Just use $val = $_GET['id'];

kmax

I dont know... maybe it has something to do with the connection to the db... I have been reading about the message Resource id=#2......

For anyone who can help......

Here is the code I am using now...

<?php

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

$val = $_GET['id'];

$query = mysql_query("SELECT * FROM products WHERE id=$val");
print $query;

Thanks

ahundiak

Hmmm.

Your database is called 'products'.

And you have a table called 'products' in it?

Nothing wrong with that but I'll bet you don't have the table created or the user does not have access to it.

Try to echo out mysql_error() after the query.

kmax

Hello, Thanks.
Yes, the db is named products and the table is named products to.

There is only 1 user set up in mysql...... me.

I did this and now do not get an error message...

<?php

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

$val = $_GET['id'];

$query = mysql_query("SELECT * FROM products WHERE id=$val");

while ($a_row=mysql_fetch_array($query))
echo mysql_error();

Any clues Anywone?

ahundiak

Ok. Everyone focused on the $id because of the subject line in the post.

The $query is a result set. It makes no sense to try and print it. Instead, you use mysql_fetch_array to cycle through it and retrieve one row at a time.

You can use print_r($a_row) to see what is in a row.

So is it working now or is there still a problem?

kmax

i think you are right about the fetch_row thing...
I just found something about that when looking up "Resource id=#2"
It is still now working... I have just been trying diffrent stuff.....

This is the latest thing I tried that did not work...

<?php

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

$val = $_GET['id'];

$query = mysql_query("SELECT * FROM products WHERE id=$val");

while ($a_row=mysql_fetch_array($query))
print $query;

Just trying to display a certain filed (details) to corespond to the id number passed in the url....

Thanks

suit4

try this:
while ($a_row=mysql_fetch_array($query)) {
print $a_row["details"];
}

wher details is the name of one of the table columns u trying to get .....

kmax

That did it! thank you very much!
If I knew you I would buy you a pizza!!!

For anyone who ever finds this post and needs an answer...
Here is the solution...

<?php

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

$val = $_GET['id'];

$query = mysql_query("SELECT * FROM products WHERE id=$val");

while ($a_row=mysql_fetch_array($query))
{
print $a_row["details"];
}

Thanks All.

suit4

now u never forget ,that Ressource ID#3 is not an error but a result set, a bit like if '$var' is an array, and u type 'print $var' u get 'Array' as the answer.

Everything is ok then, but u need to 'extract' the result.

by the way, if u use mysql_fetch_object() instead of mysql_fetch_array(), u have to use $row->details;