Question

我遇到了查询问题。 （我在帖子底部详细介绍）我有一张复选框。我在POST时将复选框发送到PHP文件以更新db。唯一可以判断在checked之前是否存在未选中的POST框，以查询表格中应该检查的内容以及是否未通过{{ 1}}然后它没有被检查。我决定在一个查询中处理所有这些。但是出于某些原因，它并没有处理未经检查的问题。如果我检查一个不优先检查，那么它将保持检查状态（它将在POST中将其标记为已选中）但是，如果我取消选中之前已检查的一个，则会保持检查状态（I希望它取消选中它）。

以下是我创建复选框的代码：

db

以下是我提交的PHP文件的一部分：

echo "<table class='tables' border='1'>";   //start an HTML table

    $fields =array();

    $result = mysqli_query($con, "SELECT u.unit_nickname AS unit, r.id AS reservation,
                                CONCAT(g.fname,' ',g.lname) AS guest, r.arrival, r.departure,
                                r.total_price - COALESCE(p.payment_amt, 0) AS balance,
                                r.checked_in 
                                FROM reservations r
                                JOIN units u ON r.unit = u.id 
                                JOIN guests g ON r.guest = g.id LEFT JOIN
                                (SELECT reservation, SUM(payment_amt) payment_amt
                                FROM payments
                                GROUP BY reservation) AS p
                                ON r.id = p.reservation
                                WHERE r.checked_out = 0");  

    echo '<th></th><th>Unit</th><th>Reservation</th><th>Guest</th><th>Arrival</th>
        <th>Depart</th><th>Balance</th><th>Status</th>';

    while ($row = mysqli_fetch_array($result))
    {       
        $fields[] = $row['unit']; $fields[] = $row['reservation']; $fields[] = $row['guest'];
        $fields[] = $row['arrival']; $fields[] = $row['departure']; $fields[] = $row['balance'];
        $fields[] = $row['checked_in'];

        //rowid == id of table (the row corresponds to the row in the table)
        $rowid = $row['reservation'];

        $count = count($fields);

        echo "<tr class='edit_tr' id='".$rowid."'>";
        echo "<td><input type='checkbox' value='".$fields[1]."' name='checkins[]'"; 
        if ($row['checked_in'] == 1)
            echo "checked='yes'";
        echo "/></td>"; 

        for ($j = 0; $j < $count; $j++)
        {
            echo "<td><span id='".$fields[1]."'>", ($j+1 == $count ? ($fields[$j] == 0 ? "Not Checked In" : "Checked In") : "".$fields[$j]."");
            echo "</span></td>"; 
        }

        echo "</tr>";

        //empty array
        $fields = "";
    }

    echo "</table>";    //close the HTML table

我试图评论得很好，以解释发生了什么。

我唯一看到的问题是我的查询

if(isset($_POST['checkins']))
{
    //open connection
    $con = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

    //set POST to array
    $checkins = array();

    //get all checkins that are not checked out
    $query = "SELECT id FROM reservations WHERE checked_in = 1 AND checked_out = 0";
    $result = mysqli_query($con, $query);
    while ($row = mysqli_fetch_assoc($result)) $checkedins[] = $row;

    //create WHERE clause for new checkins
    $where = " WHERE id =";
     foreach ($_POST['checkins'] as $id) {
        $where .= " $id OR id =";
        array_push($checkins, $id);
    }

    //loop through $checkedins[]
    foreach ($checkedins as $id) {
        //if they were checked in before and are now not checked in, add to list
        if (!in_array($id["id"], $checkins))
            $where .= " ".$id['id']." OR id =";
    }

    //remove last 8 characters from WHERE clause
    $where = substr($where, 0, -8);

    echo $query = "UPDATE reservations
                SET checked_in = CASE 
                WHEN checked_in = 0 THEN 1
                WHEN checked_in = 1 THEN 0 END 
                $where";
    mysqli_query($con, $query);
    mysqli_close($con);
}

我确信这与我的$query = "UPDATE reservations SET checked_in = CASE WHEN checked_in = 0 THEN 1 WHEN checked_in = 1 THEN 0 END $where";声明有关，但无论我多么修补它，我似乎都无法确定它。

应该发生的事情是：我将新CASE的列表放入名为checkins的数组中，并且我还获得了一个check * ed 的列表* ins是在POST之前检查的所有那些名为$checkins的数组。如果它们不在$checkedins列表中，那么我将其添加到$checkins数组中。因此$checkins现在包含要检查和取消选中的所有行。我的查询应该做的是将数组中的所有内容更改为1为0，将所有内容从0更改为1.它正在执行后者。我正在寻求另一半的帮助。

Answer 1

如果checked_in保证为零或一，你可以使用这个小技巧来完全避免`CASE：

SET checked_in = 1 - checked_in
WHERE ...

当1-x为1时，表达式x评估为0，0为x时评估为1。< / p>

Answer 2

我决定做的是拆分查询。我创建了一个$checkin_where字符串和一个$uncheck_where字符串。我在每个中放置了相应的$id s并使用了两个不同的UPDATE查询：

//create WHERE clause for new checkins
    $checkin_where = " WHERE id =";
    $uncheck_where = " WHERE id =";
     foreach ($_POST['checkins'] as $id) {
        $checkin_where .= " $id OR id =";
        array_push($checkins, $id);
    }

    //loop through $checkedins[]
    foreach ($checkedins as $id) {
        //if they were checked in before and are now not checked in, add to list
        if (!in_array($id["id"], $checkins))
            $uncheck_where .= " ".$id['id']." OR id =";
    }

    //remove last 8 characters from WHERE clause
    $uncheck_where = substr($uncheck_where, 0, -8);
    $checkin_where = substr($checkin_where, 0, -8);

    if (strlen($checkin_where) > 5)
    {
        echo $query = "UPDATE reservations
                    SET checked_in = 1
                    $checkin_where";
        mysqli_query($con, $query);
    }

    if (strlen($uncheck_where) > 5)
    {
        echo $query = "UPDATE reservations
                    SET checked_in = 0
                    $uncheck_where";
        mysqli_query($con, $query);
    }

    mysqli_close($con);

CASE语句没有运行两个WHENs

2 个答案: