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 🙂
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:
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.