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Fix the foreign key definition for the contacts table.

Mark Crane %!s(int64=12) %!d(string=hai) anos
pai
achega
a29eb419db
Modificáronse 1 ficheiros con 3 adicións e 1 borrados
  1. 3 1
      app_config.php

+ 3 - 1
app_config.php

@@ -93,7 +93,9 @@
 		$apps[$x]['db'][$y]['fields'][$z]['type']['pgsql'] = 'uuid';
 		$apps[$x]['db'][$y]['fields'][$z]['type']['sqlite'] = 'text';
 		$apps[$x]['db'][$y]['fields'][$z]['type']['mysql'] = 'char(36)';
-		$apps[$x]['db'][$y]['fields'][$z]['key']['type'] = 'primary';
+		$apps[$x]['db'][$y]['fields'][$z]['key']['type'] = 'foreign';
+		$apps[$x]['db'][$y]['fields'][$z]['key']['reference']['table'] = 'v_contacts';
+		$apps[$x]['db'][$y]['fields'][$z]['key']['reference']['field'] = 'contact_uuid';
 		$apps[$x]['db'][$y]['fields'][$z]['description']['en-us'] = '';
 		$z++;
 		$apps[$x]['db'][$y]['fields'][$z]['name'] = 'contact_type';