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BUGFIX: Allow set objects as properties if the property is not a database field

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git-svn-id: svn://svn.silverstripe.com/silverstripe/open/modules/sapphire/branches/2.4@96663 467b73ca-7a2a-4603-9d3b-597d59a354a9
This commit is contained in:
Andrew O'Neil 2010-01-12 01:40:36 +00:00 committed by Sam Minnee
parent d5a04edf02
commit c65695b807
1 changed files with 9 additions and 8 deletions
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17
core/model/DataObject.php
View File

@ -2041,16 +2041,17 @@ class DataObject extends ViewableData implements DataObjectInterface, i18nEntity
* @param mixed $val New field value
*/
function setField($fieldName, $val) {
// Situation 1: Passing an object
if(is_object($val)) {
if($val instanceof DBField) {
$val->Name = $fieldName;
$this->record[$fieldName] = $val;
} else {
// Situation 1: Passing an DBField
if($val instanceof DBField) {
$val->Name = $fieldName;
$this->record[$fieldName] = $val;
// Situation 2: Passing a literal or non-DBField object
} else {
// If this is a proper database field, we shouldn't be getting non-DBField objects
if(is_object($val) && $this->db($fieldName)) {
user_error('DataObject::setField: passed an object that is not a DBField', E_USER_WARNING);
}
// Situation 2: Passing a literal
} else {
$defaults = $this->stat('defaults');
// if a field is not existing or has strictly changed
if(!isset($this->record[$fieldName]) || $this->record[$fieldName] !== $val) {