Това може да се реши по два начина:
Използване само на SQL:
SQL Fiddle:http://www.sqlfiddle.com/#!2 /19d46/14
SELECT case
when c.text_choice = (select min(text_choice) from choice c2 where c2.id = q.id) then name_question
else ""
end as name_question,
case
when c.text_choice = (select min(text_choice) from choice c2 where c2.id = q.id) then text_question
else ""
end as text_question,
c.text_choice
FROM question as q LEFT JOIN choice as c
ON q.id = c.id
order by q.id, c.text_choice;
Резултат:
Q1 Q1_text C1
C2
C3
Q2 Q2_text C4
C5
C6
Използване и на PHP:
При първоначално гледане това може лесно да се постигне чрез PHP и подреждане на резултатите ви по name_question.
$query = ' SELECT q.name_question, q.text_question, c.text_choice'.
' FROM question as q '.
' LEFT JOIN choice as c ' .
' ON q.id_question = c.id_question ' order by q.name_question;
При показване на резултата:
$current_nameQuestion = "";
foreach ($db->loadObjectList() as $obj){
if($current_nameQuestion==$obj->name_question)
{
echo " ".$obj->text_choice."</br>";
}
else
{
echo $obj->name_question." ".$obj->text_question." ".$obj->text_choice."</br>";
$current_nameQuestion = $obj->name_question;
}
}
Горният PHP код ще отпечата само стойностите на 2 колони, когато name_question
вече не е изведено.