在PostgreSQL中組合兩個查詢

Question

我需要在兩個查詢的單個數組中幫助聯合：

在一個查詢相關的缺席成績的學生在一個季度：

$sSqlAsistencia =
         " SELECT
                ca.idcadete,                    
                coalesce(sum(i.cantidad),0) as cantidad

            FROM 
                cadetes ca,
                cursos c,
                cursos_cadetes cc
                left join inasistencias i on i.fk_idcurso = cc.fk_idcurso and i.fk_idcadete = cc.fk_idcadete
            WHERE 
                c.habilitado = true
                and ca.habilitado = true
                and c.fk_idanolectivo = ".$aAnoLectivoBuscar."
                and c.fk_idano = ".$aAnoBuscar."
                and c.fk_iddivision = ".$aDivisionBuscar."
                and cc.fk_idcurso = c.idcurso
                and cc.fk_idcadete = ca.idcadete
                and (EXTRACT(MONTH FROM i.fecha)  in ".$trimestre ." or i.cantidad is null)

            GROUP BY
                ca.idcadete

            ";
    $sSqlInasistencia = $oDB->Query($sSqlAsistencia);

idcadete | cantidad

203      |    4
305      |    0
120      |    10 

然后我得到的一段代碼是學生/學員獎，用於滿足其他問題：

$sSql = " SELECT idcadete, nombre, apellido, matricula
         FROM cadetes
        WHERE idcadete in
        (" . $sSqlPromedioEnCadaMateria . " INTERSECT " . $sSqlPromedioConducta;
        if (strlen($aPromedioGeneralEdFisicaMayorABuscar)) {
          $sSql .= " INTERSECT " . $sSqlPromedioEnEdFisica;
        }


$sSql .=    ")";
    $rsCadetesConPremio=$oDB->query($sSql);

idcadete | nombre | apellido | matricula
203      | adrian | perez    | 212121

嘗試將這兩次咨詢與INNER JOIN聯系起來但是我錯了，因為它應該設法將這兩個查詢組合在一起：

$premio = " SELECT a.idcadete, a.nombre, a.apellido, a.matricula, b.cantidad
        FROM ".$rsCadetesConPremio."a inner join ".$sSqlInasistencia."b on a.idcadete = b.idcadete
        ORDER BY a.idcadete";

- - 錯誤 - - - -

我需要以下結果：$ premio

idcadete | nombre | apellido | matricula| cantidad
203      | adrian | perez    | 212121   |     4

Answer 1

是的，您應該能夠通過將兩者的結果包裝為派生表來加入任意兩個任意查詢。 但是，您需要將兩個派生表括在括號中： FROM ( ...) a INNER JOIN ( ...) b

即：

"SELECT a.idcadete, a.nombre, a.apellido, a.matricula, b.cantidad
        FROM (".$rsCadetesConPremio.") a inner join (".$sSqlInasistencia.") b on a.idcadete = b.idcadete
        ORDER BY a.idcadete"

這是一個簡化的小提琴 。

Answer 2

 $sSqlAsistencia = " SELECT ca.idcadete, ca.nombre, ca.apellido, ca.matricula coalesce(sum(i.cantidad),0) as cantidad FROM cadetes ca, cursos c, cursos_cadetes cc left join inasistencias i on i.fk_idcurso = cc.fk_idcurso and i.fk_idcadete = cc.fk_idcadete WHERE c.habilitado = true and ca.habilitado = true and c.fk_idanolectivo = ".$aAnoLectivoBuscar." and c.fk_idano = ".$aAnoBuscar." and c.fk_iddivision = ".$aDivisionBuscar." and cc.fk_idcurso = c.idcurso and cc.fk_idcadete = ca.idcadete and (EXTRACT(MONTH FROM i.fecha) in ".$trimestre ." or i.cantidad is null) and ca.idcadete in (" . $sSqlPromedioEnCadaMateria . " INTERSECT " . $sSqlPromedioConducta; if (strlen($aPromedioGeneralEdFisicaMayorABuscar)) { $sSqlAsistencia .= " INTERSECT " . $sSqlPromedioEnEdFisica; } $sSqlAsistencia .= ") GROUP BY ca.idcadete ";