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Fixed Bug lp:1411703 Asign cases even with sample disabled and sample unassigned from questionnaire

This commit is contained in:
Adam Zammit
2015-01-28 14:44:32 +11:00
parent fd89d3917e
commit a90b59bd68
2 changed files with 11 additions and 4 deletions

View File

@@ -96,7 +96,8 @@ function case_status_report($questionnaire_id = false, $sample_id = false, $outc
JOIN questionnaire as q ON (q.questionnaire_id = c.questionnaire_id and q.enabled = 1)
JOIN outcome as o ON (o.outcome_id = c.current_outcome_id AND o.outcome_type_id = 1)
JOIN sample as s ON (s.sample_id = c.sample_id $s)
JOIN sample_import as si ON (s.import_id = si.sample_import_id)
JOIN sample_import as si ON (s.import_id = si.sample_import_id AND si.enabled = 1)
JOIN questionnaire_sample as qs ON (qs.questionnaire_id = q.questionnaire_id AND qs.sample_import_id = s.import_id)
LEFT JOIN `call` as ca ON (ca.call_id = c.last_call_id)
LEFT JOIN outcome as co ON (co.outcome_id = ca.outcome_id)
LEFT JOIN case_queue as cq ON (cq.case_id = c.case_id)
@@ -207,7 +208,7 @@ $outcome_id = false;
print "<label for='questionnaire'>" . T_("Questionnaire") . ":</label>";
display_questionnaire_chooser($questionnaire_id);
print "<label for='sample'>" . T_("Sample") . ":</label>";
display_sample_chooser($questionnaire_id,$sample_import_id);
display_sample_chooser($questionnaire_id,$sample_import_id,false);
if ($questionnaire_id)
case_status_report($questionnaire_id,$sample_import_id,$outcome_id);