URL Parsing/Variable Value Passing

[Anon]

Hello all!

I've never posted here before, but have run into an issue that I am unable to solve on my own. I am fairly new to PHP development, and so am hopeful that someone out there knows what i need to know right now.

Now that the preamble is over, I am passing a variable through the URL, like

www.domain.com/pages/info?id=000001

What I am trying to do is grab that id value and display information from the database based on that value. How can I access that value? Any help you can give me would be greatly appreciated. Thanks!

[Anon]

if the address was www.domain.com/pages/info.php?id=0001

<? echo $id;?>

hope this helps.

Troy Brown wrote:

Hello all!

I\'ve never posted here before, but have run into an issue that I am unable to solve on my own. I am fairly new to PHP development, and so am hopeful that someone out there knows what i need to know right now.

Now that the preamble is over, I am passing a variable through the URL, like

www.domain.com/pages/info?id=000001

What I am trying to do is grab that id value and display information from the database based on that value. How can I access that value? Any help you can give me would be greatly appreciated. Thanks!

[Anon]

www.domain.com/pages/info?id=000001

The id parameter (set here to 000001) gets automagically stored by php in the variable $id -- so you can use $id to access it.

For example...

<?

$sql = "SELECT * FROM dbtable WHERE id = '$id'";

...

?>

Hope this helps =)

[Anon]

Hi! Thanks for the help!

That's pretty much what I figured, except that my query fails for some unknown reason. I echoed the query and do see the variable value, so it is being passed properly. However, I get no results.

What I'm trying to do is pull business listing info from a database and flow it into the page, based on the id value that's being passed. Unfortunately what i get is a blank screen. The code is below:

<?php
include ("template_top.html");

$query="select business_name,firstname,lastname,address1,address2,city,state,zip,email from bp_users where and bp_users.id=$site";
$result=mysql_query($query);

echo($query);
echo($business_name);
echo("<p class=header1>$business_name</p>");
echo("<p class=header1>$address1<br>$address2<br>$city, $state $zip<br><a href='mailto:$email'>$email</a></p>");
$query="select * from bp_user_sites where userid=$id";
$result=mysql_query($query);
echo("<p class=body>$desc1</p>");
echo("<p class=body>$desc2</p>");
echo("<p class=body>$desc3</p>");
echo("<p class=body>Visit our main website at $website</p>");

include ("template_bottom.html");
?>

Anything you can tell me would be phenomenal! Thats in advance for your coding help!

[Anon]

in your query, you will need to use ' ' around the variable:

$query="select business_name,firstname,lastname,address1,address2,city,state,zip,email from bp_users where and bp_users.id='$site'";

you are also going to need to assign the info from the query into variables. for instance:

$result=mysql_query($query);

$i = '0';

while ($i < @mysql_num_rows($result)) {
$business_name[$i] = mysql_results($result, $i, business_name);
$firstname[$i] = mysql_results($result, $i, firstname);
$data_out .= 'Business Name: '.$business_name.'<br>First Name: '.$firstname.'<br>';
$i++;
}

echo $data_out;

Hope this helps!

--
djloc286

[Anon]

I was looking for help on this. I THOUGHT, though, that the 'id' variable would be in the associative arrage, "$HTTP_GET_VARS", and so you would access it via $HTTP_GET_VARS[id].

[Anon]

Conceivably, the variable should also be in that array as well.

(Many times single quotes are also necessary when calling array members by name - IE: $HTTP_GET_VARS['id'] - it may work without it though)

I find it's generally easier to just use the automagically populated $id instead of calling the array member. =)

[Anon]

Hey, cool script to get me here! If only I could program this stuff.

Anyway,
thanks! But what happens if there's a global variable in some included
file that has the same name? I suppose I could experiment, but if you
know, then I don't have to!

[Anon]

There's a problem with your query:

$query="select business_name,firstname,lastname,address1,address2,city,state,zip,email from bp_users where and bp_users.id=$site";

You have 'where and' which will produce a MySQL error. Also, you won't be able to access the data returned just by using $business_name, etc. - that's why you save a reference to the query results in the variable $result. In your case you should probably use mysql_fetch_row to return the row from your result set.

Also, note that when SQL queries don't return any results, you can check to see if the database found an error in your query. The MySQL function for doing so is mysql_error().

-Keegan