Drupal & PHP/MySQL : How to compare fields with current user's field? - php

Question

I've an issue, which is about php syntax/mysql in drupal:
Let's say that userA has created a content type called "test" where he filled the field field_example with value "xxx". Afterwards, another user, userB has created another content and filled the same field field_example with the same value "xxx".
I'd like to know how is it possible to display a view only with the node created where the field field_example is the same for the current user ? I don't have (and i don't want) a user reference in the content type "test" i'm using.
I've looked through View PHP Filter, but i'm wondering how to compare values of field ? Here's my attempt [i'm not an expert in PHP as you'll might notice :) ] :
<?php
$a="SELECT uid FROM users WHERE uid=%d";
/* ??? How to create $b which can get value of field_example from content_type_test from current user which is logged in ? */
$b="";
$c="SELECT field_example FROM content_type_test";
if ($b==$c){
echo "Ok, I've what I want :) ";
}
?>
Any help would be greatly appreciated since it's been a while i'm looking for information about this query...
Thanks all :)

Answer 1

I don't have a Views module solution.
One thought that comes to mind is what if User A submits multiple nodes, which example value would you use then? Also, what version of Drupal are you working with?
Assuming that a user will only ever submit one content of this type and you are running Drupal 6 (my guess from code examples), then it might look something like this:
<?php
// current user
global $user;
// select this user's node
$nid = db_result(db_query("SELECT nid FROM {node} WHERE type = 'my_content_type' AND status = 1 AND uid = %d", $user->uid));
// if this node loads fine, then proceed with the rest
if ($node = node_load($nid)) {
// find nodes with the same example value, which do not belong to the current user
$result = db_query("SELECT nid FROM {node}
INNER JOIN {content_type_test} ctt ON n.vid = ctt.vid
WHERE n.status = 1 AND ctt.field_example_value = '%s' AND uid <> %d
ORDER BY n.created DESC", $node->field_example[0]['value'], $user->uid);
// loop through results
while ($row = db_fetch_object($result)) {
$node = node_load($node->nid);
// append the teaser output (if this is what you want to do)
$output .= node_view($node, TRUE);
}
// print the output
print $output;
}
else {
print t('No other content found.');
}
?>
If users submit more than one of those content types, then that would have to be another answer to avoid from writing a novel here. There's a couple ways to approach that.
Also if this was Drupal 7, I'd be using different functions as well.

Answer 2

I assume you want something like this:
function _get_list() {
global $user; // This is global variable, contains information for current logged in user.
$results = db_query("SELECT ctt.field_example FROM {node} n JOIN {content_type_test} ctt ON n.nid = ctt.nid AND n.vid = ctt.vid WHERE n.uid = %d", $user->uid);
while($row = db_fetch_object($results)){
//$row->nid.. etc..
}
}