我在这里有一些 SQL 查询代码,如果我将它放入正确数据库中的 PHPMyAdmin 中,它会显示我想看到的内容,但我希望它显示在 HTML 表中,知道吗?
SELECT
tname AS Team, Sum(P) AS P,Sum(W) AS W,Sum(D) AS D,Sum(L) AS L,
SUM(F) as F,SUM(A) AS A,SUM(GD) AS GD,SUM(Pts) AS Pts
FROM(
SELECT
hteam Team,
1 P,
IF(hscore > ascore,1,0) W,
IF(hscore = ascore,1,0) D,
IF(hscore < ascore,1,0) L,
hscore F,
ascore A,
hscore-ascore GD,
CASE WHEN hscore > ascore THEN 3 WHEN hscore = ascore THEN 1 ELSE 0 END PTS
FROM games
UNION ALL
SELECT
ateam,
1,
IF(hscore < ascore,1,0),
IF(hscore = ascore,1,0),
IF(hscore > ascore,1,0),
ascore,
hscore,
ascore-hscore GD,
CASE WHEN hscore < ascore THEN 3 WHEN hscore = ascore THEN 1 ELSE 0 END
FROM games
) as tot
JOIN teams t ON tot.Team=t.id
GROUP BY Team
ORDER BY SUM(Pts) DESC ;
目前,当我在 PHPMyAdmin 中运行此代码时,这是我得到的结果,这是我想在 html 表中输出的内容:
我正在尝试在此网站上创建此内容:
http://www.artfulsoftware.com/infotree/qrytip.php?id=804
到目前为止,我已尝试运行查询,但没有任何结果;
<?php
$servername = "-";
$username = "-";
$password = "-";
$dbname = "-";
// Create connection
$mysqli = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($mysqli->connect_error) {
die("Connection failed: " . $mysqli->connect_error);
}
$query = "SELECT
tname AS Team, Sum(P) AS P,Sum(W) AS W,Sum(D) AS D,Sum(L) AS L,
SUM(F) as F,SUM(A) AS A,SUM(GD) AS GD,SUM(Pts) AS Pts
FROM(
SELECT
hteam Team,
1 P,
IF(hscore > ascore,1,0) W,
IF(hscore = ascore,1,0) D,
IF(hscore < ascore,1,0) L,
hscore F,
ascore A,
hscore-ascore GD,
CASE WHEN hscore > ascore THEN 3 WHEN hscore = ascore THEN 1 ELSE 0 END PTS
FROM games
UNION ALL
SELECT
ateam,
1,
IF(hscore < ascore,1,0),
IF(hscore = ascore,1,0),
IF(hscore > ascore,1,0),
ascore,
hscore,
ascore-hscore GD,
CASE WHEN hscore < ascore THEN 3 WHEN hscore = ascore THEN 1 ELSE 0 END
FROM games
) as tot
JOIN teams t ON tot.Team=t.id
GROUP BY Team
ORDER BY SUM(Pts) DESC ; ";
if ($stmt = $mysqli->prepare($query)) {
/* execute statement */
$stmt->execute();
/* close statement */
$stmt->close();
}
/* close connection */
$mysqli->close();
?>
