Question

我有这两张桌子......

日程安排（gameid，homeid，awayid，weekno，seasonno）

团队（队友，位置，昵称）

下面的这个mysql查询为我提供了所有32支队伍的日程信息......

        $sql = "SELECT
        h.nickname AS home,
        a.nickname AS away,
        h.teamid AS homeid,
        a.teamid AS awayid,
        s.weekno
    FROM schedule s
        INNER JOIN teams h ON s.homeid = h.teamid
        LEFT JOIN teams a ON s.awayid = a.teamid
    WHERE s.seasonno =2014";
$schedule= mysqli_query($connection, $sql);
if (!$schedule) {
die("Database query failed: " . mysqli_error($connection));
} else { 
// Placeholder for data
    $data = array();
    while($row = mysqli_fetch_assoc($schedule)) {
        if ($row['away'] == "") {$row['away']="BYE";}
        $data[$row['homeid']][$row['weekno']] = $row['away'];
        $data[$row['awayid']][$row['weekno']] = '@ '.$row['home'];
    }
}

但是，我只想获取一个特定团队的信息，该团队存储在$ teamid变量中。这应该很容易，对吧？我尝试了很多东西，包括下面的这个（我添加了一个AND语句），但这个仍然输出太多......

$sql = "SELECT
    h.nickname AS home,
    a.nickname AS away,
    h.teamid AS homeid,
    a.teamid AS awayid,
    s.weekno
FROM schedule s
    INNER JOIN teams h ON s.homeid = h.teamid 
    LEFT JOIN teams a ON s.awayid = a.teamid 
WHERE s.seasonno =2014
AND h.teamid=$teamid OR a.teamid=$teamid";
$schedule= mysqli_query($connection, $sql);
if (!$schedule) {
die("Database query failed: " . mysqli_error($connection));
} else { 
// Placeholder for data
    $data = array();
    while($row = mysqli_fetch_assoc($schedule)) {
        if ($row['away'] == "") {$row['away']="BYE";}
        $data[$row['homeid']][$row['weekno']] = $row['away'];
        $data[$row['awayid']][$row['weekno']] = '@ '.$row['home'];
    }
}

以下是上述输出的数组。它更好，但我想要的只是第二个数组（[1]），在这个例子中，它具有Eagles完整的时间表。它没有给我太多其他东西，我想我可以忍受它，只是忽略其他东西，但我宁愿尽可能高效，只得到我需要的东西......

更新：我为Abhik添加了括号（谢谢！）并且它有所帮助，但仍然给予了太多，即...

Array
(
    [1] => Array
        (
            [1] => Jaguars
            [2] => @ Colts
            [3] => Redskins
            [4] => @ 49ers
            [5] => Rams
            [6] => Giants
            [7] => BYE
            [8] => @ Cardinals
            [9] => @ Texans
            [10] => Panthers
            [11] => @ Packers
            [12] => Titans
            [13] => @ Cowboys
            [14] => Seahawks
            [15] => Cowboys
            [16] => @ Redskins
            [17] => @ Giants
        )

    [27] => Array
        (
            [1] => @ Eagles
        )

    [28] => Array
        (
            [2] => Eagles
        )

    [4] => Array
        (
            [3] => @ Eagles
            [16] => Eagles
        )

    [14] => Array
        (
            [4] => Eagles
        )

    [15] => Array
        (
            [5] => @ Eagles
        )

    [3] => Array
        (
            [6] => @ Eagles
            [17] => Eagles
        )

    [] => Array
        (
            [7] => @ Eagles
        )

    [16] => Array
        (
            [8] => Eagles
        )

    [25] => Array
        (
            [9] => Eagles
        )

    [11] => Array
        (
            [10] => @ Eagles
        )

    [7] => Array
        (
            [11] => Eagles
        )

    [26] => Array
        (
            [12] => @ Eagles
        )

    [2] => Array
        (
            [13] => Eagles
            [15] => @ Eagles
        )

    [13] => Array
        (
            [14] => @ Eagles
        )

)

Answer 1

您需要使用括号来分隔条件

WHERE s.seasonno =2014
AND h.teamid=$teamid OR a.teamid=$teamid

到

WHERE 
s.seasonno =2014
AND 
(h.teamid=$teamid OR a.teamid=$teamid)

Answer 2

OR正在减慢查询速度，您可以使用：

 ... WHERE s.seasonno =2014 AND $teamid IN (h.teamid, a.teamid)

或

 ... WHERE s.seasonno=2014 AND $teamid IN (s.homeid, s.awayid)

试图限制mysql查询/数组中返回的数据

2 个答案: