我有这个功能:
function view_user_anunt($user) {
$query="SELECT * FROM `anunturi`
FULL OUTER JOIN tranzactie
ON anunturi.tranzactie = tranzactie.id_tranzactie WHERE `anunturi.user`=:code";
$stmt = $this->dbh->prepare($query);
$stmt->bindParam(':code', $user, PDO::PARAM_INT);
$stmt->execute();
foreach ($stmt->fetchAll(PDO::FETCH_ASSOC) as $result)
{
$view[]="
<tr>
<td>".$result['id_anunt']."</td>
<td>".$result['den_tranzactie']."</td>
<td>".$result['den_proprietate']."</td>
<td><a href='#' id='vizualizare'>Select</a></td>
<td><a href='#' id='modificare'>Select</a></td>
</tr>";
}
return $view;
}
和3个表:
anunturi
tranzactie
提示
我需要一个好的查询或ideea来获取anunturi的每一行的事务名称(den_tranzactie)和所有者名称den_proprietate,其中anunturi.user = $ user。
提前致谢...
答案 0 :(得分:0)
这样的东西?
'
SELECT t.den_tranzactie, tip.den_proprietate
FROM
anunturi a
JOIN tranzactie t ON a.tranzactie = t.id_tranzactie
JOIN tip ON a.tip = tip.id_proprietate
WHERE a.user =:code
答案 1 :(得分:0)
试试这个SQL查询
$query="SELECT t1.*, t2.*, t3.*
FROM table1 t1
LEFT JOIN table2 t2 ON t1.id = t2.foreightkey_id
LEFT JOIN table3 t3 ON t1.id = t3.foreightkey_id
WHERE t1.user=:code
";
您可以使用INNER JOIN仅接收所有表中包含数据的行。例如
$query="SELECT t1.*, t2.*, t3.*
FROM table1 t1
INNER JOIN table2 t2 ON t1.id = t2.foreightkey_id
INNER JOIN table3 t3 ON t1.id = t3.foreightkey_id
WHERE t1.user=:code
";
或者如果你想从table1和table2获取连接的所有数据,只从表3中获取连接的数据
$query="SELECT t1.*, t2.*, t3.*
FROM table1 t1
INNER JOIN table2 t2 ON t1.id = t2.foreightkey_id
LEFT JOIN table3 t3 ON t1.id = t3.foreightkey_id
WHERE t1.user=:code
";