From: Michael Reddick <michael.reddick@gmail.com>
Date: Sat, 13 Jun 2009 22:23:15 +0000 (+0000)
Subject: Added documentation for from => $rs->as_query
X-Git-Url: http://git.shadowcat.co.uk/gitweb/gitweb.cgi?a=commitdiff_plain;h=a0ce462cf118f0da7c960a9e69ea7c6fb50f5f57;p=dbsrgits%2FDBIx-Class-Historic.git

Added documentation for from => $rs->as_query
---

diff --git a/lib/DBIx/Class/ResultSet.pm b/lib/DBIx/Class/ResultSet.pm
index 13d81b4..fad673f 100644
--- a/lib/DBIx/Class/ResultSet.pm
+++ b/lib/DBIx/Class/ResultSet.pm
@@ -3287,9 +3287,21 @@ with a father in the person table, we could explicitly use C<INNER JOIN>:
     # SELECT child.* FROM person child
     # INNER JOIN person father ON child.father_id = father.id
 
-If you need to express really complex joins or you need a subselect, you
+You can select from a subquery by passing a resultset to from as follows.
+
+    $schema->resultset('Artist')->search( 
+        undef, 
+        {   alias => 'artist2',
+            from  => [ { artist2 => $artist_rs->as_query } ],
+        } );
+
+    # and you'll get sql like this..
+    # SELECT artist2.artistid, artist2.name, artist2.rank, artist2.charfield FROM 
+    #   ( SELECT me.artistid, me.name, me.rank, me.charfield FROM artists me ) artist2
+
+If you need to express really complex joins, you
 can supply literal SQL to C<from> via a scalar reference. In this case
-the contents of the scalar will replace the table name asscoiated with the
+the contents of the scalar will replace the table name associated with the
 resultsource.
 
 WARNING: This technique might very well not work as expected on chained