Question

我有一个舞蹈比赛网站，每个用户都可以登录并添加舞蹈时刻，在我的html表中，所有用户的所有时刻我都拥有所有数据，但我希望在html列中添加＆＃34;由记录的用户ID＆＃34;添加的每个时刻的舞者数量。

我有这个：

$array_container= array();

while($fila=mysqli_fetch_assoc($sql)){


                    $format_org=str_replace(" ","_",$fila["organization"]);
                    $format_eval=str_replace(" ","_",$fila["EvaluationType"]);
                    $format_test=str_replace(" ","_",$fila["test_name"]);
                                        $CapaEnviar=$format_org.$format_eval;

                              $array_container[] = array($CapaEnviar => $fila['test_name']);

}

echo json_encode($array_container,true);

phpMyAdmin时刻表：有id，俱乐部名称，类别，纪律，部分，以及：

但是这个过程会计算所有用户的所有舞者名字。这个过程的例子：你总共有200个舞者！

我希望这个过程为我计算所有舞者的名字，每个时刻都添加在表单中，而不是所有整个用户的时刻，例如：如果用户john有两个时刻添加：时刻1：5舞者 - 时刻2 ：每个用户都有10位舞者，等等。

Answer 1

让我试着让你以正确的方式（这似乎是一个很长的帖子，但我认为值得初学者阅读它！）。

评论中告诉您规范化您的数据库，如果我是您，如果您希望您的项目长时间正常运行......我会这样做。< / p>

有许多MySQL规范化教程，如果你感兴趣的话，你可以将它自己谷歌...我只是想用你的特定例子来帮助你，我相信你会理解它。

基本上，您必须创建不同的表来存储“不同的概念”，然后在查询数据库时加入它。

在这种情况下，我会创建这些表：

categories，dance_clubs，users和dancers存储“基本”数据。

moments和moment_dancers存储外键以创建数据之间的关系。

让我们看一下内容，以便更好地理解它。

mysql> select * from categories;
+----+---------------+
| id | name          |
+----+---------------+
|  1 | Hip-hop/dance |
+----+---------------+

mysql> select * from dance_clubs;
+----+---------------+
| id | name          |
+----+---------------+
|  1 | dance academy |
+----+---------------+

mysql> select * from users;
+----+-------+
| id | name  |
+----+-------+
|  1 | alex  |
+----+-------+

mysql> select * from dancers;
+----+-------+
| id | name  |
+----+-------+
|  1 | alex  |
|  2 | dan   |
|  3 | mihai |
+----+-------+

mysql> select * from moments;
+----+--------------+---------------+-------------------+
| id | main_user_id | dance_club_id | dance_category_id |
+----+--------------+---------------+-------------------+
|  1 |            1 |             1 |                 1 |
+----+--------------+---------------+-------------------+
          (user alex)  (dance acad..)     (Hip-hop/dance)

mysql> select * from moment_dancers;
+----+-----------+-----------+
| id | moment_id | dancer_id |
+----+-----------+-----------+
|  1 |         1 |         1 | (moment 1, dancer alex)
|  2 |         1 |         2 | (moment 1, dancer dan)
|  3 |         1 |         3 | (moment 1, dancer mihai)
+----+-----------+-----------+

确定！现在我们想从PHP做一些查询。

我们将使用预备语句而不是mysql_ *查询，正如他们在评论中所说的那样。

准备好的陈述的概念起初可能有点难以理解。只需仔细阅读代码并再次查找一些教程;）

列出舞者的简单例子（只是为了理解它）：

// Your connection settings
$connData = ["localhost", "user", "pass", "dancers"];

$conn = new mysqli($connData[0], $connData[1], $connData[2], $connData[3]);
$conn->set_charset("utf8");

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// Here we explain MySQL which will be the query
$stmt = $conn->prepare("select * from dancers");
// Here we explain PHP which variables will store the values of the two columns (row by row)
$stmt->bind_result($dancerId, $dancerName);

// Here we execute the query and store the result
$stmt->execute();
$stmt->store_result();

// Here we store the results of each row in our two PHP variables 
while($stmt->fetch()){
    // Now we can do whatever we want (store in array, echo, etc)
    echo "<p>$dancerId - $dancerName</p>";
}

$stmt->close();
$conn->close();

浏览器中的结果：

好！现在有点困难了！ 列出时刻：

// Your connection settings
$connData = ["localhost", "user", "pass", "dancers"];
$conn = new mysqli($connData[0], $connData[1], $connData[2], $connData[3]);
$conn->set_charset("utf8");

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// Query to read the "moments", but we have their main user and dancers in other tables
$stmtMoments = $conn->prepare("
    select
        moments.id,
        (select name from users where users.id = moments.main_user_id) as main_user,
        (select name from dance_clubs where dance_clubs.id = moments.dance_club_id) as dance_club,
        (select name from categories where categories.id = moments.dance_category_id) as dance_category,
        (select count(*) from moment_dancers where moment_dancers.moment_id = moments.id) as number_of_dancers
    from moments
    ");
// Five columns, five variables... you know ;)
$stmtMoments->bind_result($momentId, $momentMainUser, $momentDanceClub, $momentDanceCategory, $momentNumberOfDancers);

// Query to read the dancers of the "moment" with id $momentId
$stmtDancers = $conn->prepare("
    select
        dancers.name as dancer_name
    from
        dancers join moment_dancers on dancers.id = moment_dancers.dancer_id
    where
        moment_dancers.moment_id = ?
    ");

$stmtDancers->bind_param("i", $momentId);
$stmtDancers->bind_result($momentDancerName);

// Executing the "moments" query
$stmtMoments->execute();
$stmtMoments->store_result();

// We will enter once to the while because we have only one "moment" right now
while($stmtMoments->fetch()){

    // Do whatever you want with $momentId, $momentMainUser, $momentDanceClub, $momentDanceCategory, $momentNumberOfDancers
    // For example:

    echo "<h3>Moment $momentId</h3>";
    echo "<p>Main user: $momentMainUser</p>";
    echo "<p>Dance club: $momentDanceClub</p>";
    echo "<p>Category: $momentDanceCategory</p>";
    echo "<p>Number of dancers: $momentNumberOfDancers</p>";
    echo "<p><strong>Dancers</strong>: ";

    // Now, for this moment, we look for its dancers
    $stmtDancers->execute();
    $stmtDancers->store_result();
    while($stmtDancers->fetch()){

        // Do whatever you want with each $momentDancerName
        // For example, echo it:

        echo $momentDancerName . " ";
    }

    echo "</p>";
    echo "<hr>";
}

$stmtUsers->close();
$stmtMoments->close();

$conn->close();

浏览器中的结果：

就是这样！如果您有任何疑问，请问我！

（如果需要，我可以发布DDL代码以创建包含内容数据的示例数据库）

已修改：已添加dancers表格。已将moment_users重命名为moment_dancers。更改了功能以使脚本适应新的表和名称。

Php / Mysql计数舞者从每个时刻添加问题

1 个答案: